Have you ever been to a corporate manager's or CEO's office and noticed a Isaac Newton's pendulum (or Newton's cradle) sitting on their desk? When you lift one metal sphere and release it, it collides with the stationary spheres, which transmits a force through the stationary spheres that propels the sphere at the end forwards and upwards. Then that moving sphere reverses direction and collides with the virtually stationary spheres, which subsequently repeats the effect in the opposite direction. How and why does this happen?
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| This image shows a Newton's cradle with two balls of equal weight and perfectly efficient elasticity. After the left ball is pulled backwards and released, it collides with the right ball, which transfers all the kinetic energy to the right ball. Since the balls have identical weight, the right ball would move at the same velocity, which demonstrates all the momentum and energy are transferred. As the kinetic energy is transformed into potential energy, the right ball reaches the same height as the initial ball and the cycle repeats. Any energy lost to vibration, heat and sound are ignored. |
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| Here is an idealised Newton's cradle with 5 balls when there are no energy losses and a tiny separation between the balls, except for when a pair is colliding. |
In 1999, Professor Walter Lewin was "a strong believer in physics" that "he was willing to put his life on the line". In his famous lecture about classical mechanics, he risked his life to demonstrate the conservation of energy in front of a curious and concerned college class at MIT. This sort of teaching is rare in university / college and he is an example of education made interesting and exciting with some anticipation. What he demonstrated is the physics of the pendulum. But how do they work?
What is mechanical energy?
In physical sciences, mechanical energy is defined as the sum of potential energy and kinetic energy, which is the macroscopic energy associated with a system. Potential energy is measured by the position of the system’s components and the kinetic energy is described as the energy of motion.Em = EU + EK
— EU = Potential energy depends on the position of an object subjected to a conservative force. It is defined as the object’s ability to do work, which increases as the object is moved in the opposite direction of the direction of the force.
If F represents the conservative force and x the position, the potential energy of the force between the 2 positions x1 and x2 is defined as the negative integral of F from x1 to x2.
EK = 0.5*m*v2
Jain (2009) expounded the principle of conservation of mechanical energy states the mechanical energy of that body or system remains constant, provided that a body or system is subjected only to conservative forces. When conservative forces moves an object from one point to another, the work done by the conservative force is independent of the path, and vice versa for non-conservative forces.
What are examples of mechanical energy conversion?
— An electric motor converts electrical energy into mechanical energy.
— A generator converts mechanical energy into electrical energy.
— A hydroelectric power plant converts the mechanical energy of water in a storage dam into electrical energy.
— An internal combustion engine is a heat engine that burns fuel, which converts chemical energy into mechanical energy. From this mechanical energy, the internal combustion engine often generates electricity.
— A steam engine converts the heat energy of steam into mechanical energy.
— A turbine converts the kinetic energy of a stream of gas or liquid into mechanical energy.
What is the conservation of mechanical energy?
According to the principle of conservation of mechanical energy, the mechanical energy of an isolated system remains constant in time, as long as the system is free of friction and other non-conservative forces. Although frictional forces and other non-conservative forces are present in the real world, their effects on the system are minuscule. This means the the principle of conservation of mechanical energy can be used for fair approximations. Although energy cannot be created or destroyed in an isolated system, it can be converted from one form to another.
i. Swinging Pendulum
https://en.wikipedia.org/wiki/Pendulum
A pendulum is a weight suspended from a pivot that swings freely from side to side. When it is displaced sideways from its resting, equilibrium position, it is subject to a restoring force due to gravity. This accelerates the pendulum back towards the equilibrium position. When the pendulum is released, the restoring force acting on the pendulum’s mass causes it to oscillate about the equilibrium position, swinging back and forth. |
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This diagram illustrates the “simple gravity pendulum" model, assuming no friction or air resistance. |
i. Simple gravity pendulum
This idealised mathematical model of a pendulum involves a weight (or bob) on the end of a massless cord suspended from a pivot, without friction. After a weight is initially pushed, it swings back and forth at a constant amplitude. Since these pendulums are subject to friction and air drag, so the amplitude of their swings declines. This model assumes that:
— The rod or cord on which the bob swings is massless, inextensible and always remains taut.
— The bob is a point mass
— Motion occurs only in 2 dimensions, i.e. the bob does not trace an ellipse but an arc.
— The motion doesn’t lose energy to friction or air resistance
— The gravitational field is uniform
— The support, which the cord is attached to, doesn’t move.
[1] d2θ/dt2 + (g/l)*sin(θ) = 0
— g = Acceleration due to gravity— l = Length of the pendulum
— θ = Angular displacement
1. How is [1] derived from “force”?
Consider all the forces acting on a simple pendulum.
— The path of the pendulum sweeps out an arc of a circle, with the angle (θ) swept out measured in radians.
— The blue arrow indicates the gravitational force acting on the bob
— The violet arrow indicates the same force resolve into components parallel and perpendicular to the bob's instantaneous motion.
— The direction of the bob’s instantaneous velocity always points along the red axis. This indicates the tangential axis since the bob’s direction is always tangent to the circle.
Recall Newton’s 2nd law:
F = m*a
— F = Sum of forces on the object
— m = mass
— a = acceleration
Since we are focused on changes n speed, the bob is forced to stay in a circular path, meaning Newton's 2nd law can only be applied to the tangential axis only.
The short violet arrow indicates the tangential component of the gravitational axis, and determining its magnitude requires trigonometry. Hence:
F = -m*g*sin(θ) = m*a
a = -g*sin(θ)
— g = Acceleration due to gravity near the Earth’s surface
— The negative sign implies θ and a are always pointing in opposite directions.
For instance, when the pendulum swings further to the right, we would expect it to accelerate back toward the left.
Consider all the forces acting on a simple pendulum.
— The blue arrow indicates the gravitational force acting on the bob
— The violet arrow indicates the same force resolve into components parallel and perpendicular to the bob's instantaneous motion.
— The direction of the bob’s instantaneous velocity always points along the red axis. This indicates the tangential axis since the bob’s direction is always tangent to the circle.
Recall Newton’s 2nd law:
F = m*a
— F = Sum of forces on the object
— m = mass
— a = acceleration
Since we are focused on changes n speed, the bob is forced to stay in a circular path, meaning Newton's 2nd law can only be applied to the tangential axis only.
The short violet arrow indicates the tangential component of the gravitational axis, and determining its magnitude requires trigonometry. Hence:
F = -m*g*sin(θ) = m*a
a = -g*sin(θ)
— g = Acceleration due to gravity near the Earth’s surface
— The negative sign implies θ and a are always pointing in opposite directions.
For instance, when the pendulum swings further to the right, we would expect it to accelerate back toward the left.
s = l*θ,
v = ds/dt = l*(dθ/dt)
a = d2s/dt2 = l*(d2θ/dt2)
Therefore,
In order the compare the approximation to the full solution, let the period of a pendulum of length 1 m on Earth (g = 9.80665 m/s2) at initial angle 10 degrees:
4*(1*m/g)0.5 *K*[sin(10o/2)] ~ 2.0102s
The linear approximation would be evaluated as:
2*π*(1*m/g)0.5 ~ 2.0064s
The difference between these 2 values is much less than that caused by the variation of g with geographical location. This allows us to calculate the elliptic integral further on.
— Legendre polynomial solution for the elliptical integral
Given [3] and the Legendre polynomial solution for the elliptic integral:
— n!! = Double factorial, an exact solution to the period of a pendulum is:
This graph shows the relative errors using the power series for the period. T0 is the linear approximation and T2 to T10 include respectively the terms up to the 2nd to the 10th powers.
— Power series solution for the elliptic integral
Another formulation of the above solution can be evaluated if the following Maclaurin series:
sin (θ0/2) = (θ0/2) - (θ03/48) + (θ05/3840) - (θ07/645120) + …
is used in the Legendre polynomial solution above. The power series becomes:
T = (2*π)*(l/g)0.5 *[1 + (θ02/16) + (11*θ04/3072) + (173*θ06/737280) + (22931*θ08/1321205760) + (1319183*θ010/951268147200) + (233526453*θ012/2009078326886400) + … ]
— Arithmetic-geometric mean solution for elliptic integral
Given [3] and the arithmetic-geometric mean solution of the elliptic integral:
K(k) = [0.5*π]/[M(1 - k, 1 + k)]
— M(x,y) = Arithmetic-geometric mean of x and y.
This leads to an alternative and faster-converging formula for the period:
T = (l/g)0.5 *(2*π)/M(1, cos(θ0/2))
This first iteration of this algorithm gives out:
T1 = (2*T0)/(1 + cos(θ0/2))
This approximation has the relative error of less than 1% for angles up to 96.11 degrees. Since (1+cos(θ0/2))/2 = cos2(θ0/4), the expression can be simplified as:
T1 = T0*sec2(θ0/4)
The 2nd order expansion of sec2(θ0/4) simplifies to T ~ T0*(1 + θ02/16).
A 2nd iteration of this algorithm gives:
Carvalhaes & Suppes stated that this 2nd approximation for θ0 < 163o lead to a relative error of less than 1%.
What is the approximate formulae for the non-linear pendulum period?
The approximate formulae for the increase of the pendulum period with amplitude is categorised as follows:
— q = exp(-π*K’/K) : Elliptic nome
— ω = 2π/T : Angular velocity
If ε is defined above, q can be approximated using the expansion:
q = ε + 2*ε5 + 15*ε9 + 150*ε13 + 1707*ε17 + 20910*ε21 + …
Note that for θ0* < π, we have ε < 0.5, thus the approximation is applicable even for large amplitudes.
Examples
(Left to right, from top to bottom)
a. Initial angle of 0o , a stable equilibrium
b. Initial angle of 45o .
c. Initial angle of 90o .
d. Initial angle of 135o
e. Initial angle of 170o
f. Initial angle of 180o
g. Pendulum with just barely enough energy for a full swing
h. Pendulum with enough energy for a full swing
iii. Compound pendulum
Also known as a physical pendulum, a compound pendulum is a swinging rigid body that is free to rotate about a fixed horizontal axis. This setup implies the rod has mass and can stretch in size, i.e. an arbitrarily shaped rigid body swinging by a pivot. This means the pendulum’s period correlates with its moment of inertia (I) around the pivot point.
The equation of torque is:
τ = I*α
— α = Angular acceleration
— τ = Torque
The torque generated by gravity is:
τ = -m*g*L*sin(θ)
— m = Mass of the body
— L = Distance from the pivot to the object's centre of mass
— θ = Angle from the vertical
The formula for a takes the same form as the conventional simple pendulum. This means the period of oscillation can be evaluated as:
T = 2*π*[1/(m*g*L)]0.5
And frequency can be expressed as:
f = 1/T = (1/2π)*[(m*g*L)/I]0.5
We take the initial angle into consideration (for large amplitudes), then the formula for α becomes:
α = θn = -(m*g*L*sin(θ))/I
And the period becomes:
T = (4*K)*[sin2(θ0/2)]*[1/(m*g*L)]0.5
— θ0 = Maximum angle of oscillation (with respect to the vertical
— K(k) = Complete elliptic integral of the first kind
Huygens (1673) found that the appropriate equivalent length (L) for evaluating the period of any compound pendulum is the distance from the pivot to the centre of oscillation. This point is located beneath the centre of mass at a distance from the pivot labelled as the radius of oscillation, which depends on the mass distribution of the pendulum. Glasgow (1885) worked out the the centre of oscillation to be adjacent to the centre of mass if a large proportion of the pendulum’s mass is concentrated in a relatively small bob compared to the pendulum length.
The radius of oscillation or equivalent length (L) of any physical pendulum can be calculated as:
L = I/(m*R)
— I = Moment of inertia of the pendulum about the pivot point
— m = Mass of the pendulum
— R = Distance between the pivot point and the centre of mass.
If we substitute this equation into the formula (T ~ 2*π*(L/g)0.5, then the period (T) of a compound pendulum for sufficiently small oscillations can be calculated as:
T = 2*π*(1/(m*g*R))0.5
T = 2*π*[(m*L2/3) / (m*g*(L/2))]0.5
T = 2*π*[(2*L)/(3*g)]0.5
This demonstrates a rigid rod pendulum has the same period as a simple pendulum that is 2/3 of its length.
In 1673, Huygens proved that the pivot point and the centre of oscillation are interchangeable. If we flipped the pendulum upside down and swung it from a pivot located at its previous centre of oscillation, it demonstrates the same period as before and the old pivot point becomes the new centre of oscillation. In 1817, Kater used this principle to produce a type of reversible pendulum, namely Kater pendulum, in order to detect accurate measurements of the acceleration due to gravity.
Describe the physical interpretation of the imaginary period
The Jacobian elliptic function expresses the position of a pendulum as a function of time. It is a doubly periodic function with a real period and an imaginary period. The real period is defined as the time it takes the pendulum to go through one full cycle. Appell (1978) highlighted the physical interpretation of the imaginary period: if θ0 is the maximum angle of one pendulum and (180 - θ0) is the maximum angle of another, then the real period of each is the magnitude of the imaginary period of the other.
What are coupled pendulums?
When pendulums are coupled together, they affect each other’s motion, either through a direct connection (e.g. bobs connected by a spring), or through motions in a supporting structure (such as a tabletop). The equations of motion for 2 identical simple pendulums coupled by a spring connecting the bobs are evaluated using Lagrangian Mechanics. The kinetic energy of this system can be worked out using the formula:
EK = 0.5*m*L2 (*θ12 + *θ22)
— m = Mass of the bobs
— L = Length of the strings
— θ1, θ2 = Angular displacements of the 2 bobs from equilibrium.
The potential energy of this system is given by the following formula:
Ep = m*g*L*[2 - cos(θ1) - cos(θ2)] + 0.5*k*L2*(θ2 - θ1)2
— g = Gravitational acceleration
— k = Spring constant
— L(θ2 - θ1) = Displacement of the spring from its equilibrium position, which assumes the small angle approximation.
The Lagrangian can be calculated to be:
L = 0.5*m*L2*(*θ12 + *θ22) - m*g*L*[2 - cos(θ1) - cos(θ2)] + 0.5*k*L2*(θ2 - θ1)2
This leads to the following set of coupled differential equations:
**θ1 + (g/L)*sin(θ1) + (k/m)*(θ1 - θ2) = 0
**θ2 + (g/L)*sin(θ2) - (k/m)*(θ1 - θ2) = 0
When we add and subtract these 2 equations in turn, and then apply the small angle approximation, this yields 2 harmonic oscillator equations in the variables (θ1 + θ2) and (θ1 - θ2):
**θ1 + **θ2 + (g/L)*(θ1 + θ2) = 0
**θ1 - **θ2 + (g/L + 2*(k/m))*(θ1 - θ2) = 0
along with the corresponding solutions:
θ1 + θ2 = A*cos(ω1*t + α)
θ1 - θ2 = Β*cos(ω2*t + β)
— ω1 = (g/L)0.5
— ω2 = (g/L + 2*k/m)0.5
— A, B, α, β = Constants of integration
If we express the solutions in terms of θ1 & θ2 alone, then:
θ1 = 0.5*A*cos(ω1*t + α) + 0.5* Β*cos(ω2*t + β)
θ2 = 0.5*A*cos(ω1*t + α) - 0.5* Β*cos(ω2*t + β)
If the bobs weren’t provided an initial push, then condition *θ1(0) = *θ2(0) = 0 requires α = β = 0, which yields:
A = θ1(0) + θ2(0)
B = θ1(0) - θ2(0)
Describe the history of the pendulum
One of the earliest known uses of a pendulum was a 1st-century seismometer device invented by Han Dynasty Chinese scientists Zhang Heng. Needham described its function as swaying and activating a series of levers after the disturbance by the tremor of an earthquake. Then a small ball would be released by a lever and fall out f the urn-shaped device into 1 of 8 metal toad's mouths below. Each of the 8 points of the compass signifies a direction of the earthquake’s original location.
I. 1602: Galileo’s Research
II. 1656: The Pendulum clock
III. 1673: Huygen’s Horologium Oscillatorium
IV. 1721: Temperature compensated pendulums
V. Foucault Pendulum
In 1851, the Foucault pendulum was the first device to demonstrate the Earth's rotation without the involvement of celestial observations, which generated a "pendulum mania". The rate of precession is exaggerated in this figure.
VI. 1930: Decline in use
How was the pendulum used to measure time?
From its discovery around 1582 until development of the quartz clock in the 1930s, the pendulum was the world's standard for accurate timekeeping for roughly 3 centuries. In the 17th and 18th centuries, free-swinging seconds pendulums were extensively used as precision timers in scientific experiments.
i. Clock pendulums
1. Blackburn pendulum
Known as a harmonograph, it is a mechanical apparatus that uses pendulums to create a geometric image, which illustrates simple harmonic motion. The device is named after Hugh Blackburn, who described it in 1844. However, it was first discussed by James Dean and analysed mathematically by Nathaniel Bowditch in 1815. It involves a bob being suspended from a string that in turn hangs from a V-shaped pair of strings. When the pendulum oscillates simultaneously in 2 perpendicular directions with different periods, the bob consequently follows a path resembling a Lissajous curve.
Video of a harmonograph in action:
https://www.youtube.com/watch?v=Y9W-XsIQFE0
The movements of damped pendulums allows the harmonograph to delineate its figures. Its motion is described by the following equation:
x(t) = A*sin(t*f + p)*e-d*t
— f = Frequency
— p = Phase
— A = Amplitude
— d = damping
— t = time
If a pendulum can move about 2 axes (in a circular or elliptical shape), due to the principle of superposition, the motion of a rod attached to the bottom of the pendulum along 1 axis can be described by the following equation:
x(t) = A1*sin(t*f1 + p1)*e-d1*t + A2*sin(t*f2 + p2)*e-d2*t
A typical harmonograph with 2 pendulums move the pen by 2 perpendicular rods connected to these pendulums. Therefore, the path of the harmonograph figure drawn is described by the following equations:
x(t) = A1*sin(t*f1 + p1)*e-d1*t + A2*sin(t*f2 + p2)*e-d2*t
y(t) = A3*sin(t*f3 + p3)*e-d3*t + A4*sin(t*f4 + p4)*e-d4*t
2. Ballistic pendulum
https://en.wikipedia.org/wiki/Ballistic_pendulum
https://www.youtube.com/watch?v=dBAi74A8Td8
This device measures a bullet’s momentum, from which it can calculate its velocity and kinetic energy. Before the invention of modern chronographs, ballistic pendulums were used to directly measure the projectile velocity.
The first ballistic pendulum was invented in 1742 by English mathematician Benjamin Robins (1707-1751), and its function was published in his book New Principles of Gunnery. This revolutionised the science of ballistics because it was the first method to accurately measure the velocity of a bullet.
Most physics textbooks provide a simplified method of calculation of the bullet's velocity that uses its mass and pendulum and the height of the pendulum’s deflection to calculate the amount of energy and momentum in the pendulum and bullet system. Nonetheless, Robins’ calculations used a measure of the period of oscillation to determine the rotational inertia of the system.
Consider the motion of the bullet-pendulum system from the instant the pendulum is struck by the bullet. We can calculate the initial velocity of the bullet-pendulum system using conservation of mechanical energy (kinetic energy + potential energy), if the acceleration due to gravity (g) and the final height of the pendulum (h) are both known. Let the initial velocity be v1, the masses of the bullet and pendulum be mb & mp respectively.
The initial kinetic energy of the system is Ksystem = 0.5*(mb + mp)*v12
There is only 1 conserved quantity (the energy), and no conserved momenta. Therefore, the 2 generalised momenta can be expressed as:
When these equations are inverted, it leads to:
The remaining equations of motion can be written as:
These last 4 equations are explicit formulas for the time evolution of the system given its current state. Therefore, it isn’t possible to advance and integrate these equations analytically in order to yield formulas for θ1 and θ2 as functions of time. However the Runge Kutta method of similar techniques may make it possible to perform this integration numerically.
What is chaotic motion?
Animations show the double pendulum undergoes chaotic motion, demonstrating its sensitivity to initial conditions.
The boundary of the central white region is defined in part by energy conservation with the following curve:
3*cos(θ1) + cos(θ2) = 2
Within the region defined by this curve, it is defined by:
3*cos(θ1) + cos(θ2) > 2
This means it is energetically impossible for either pendulum to flip, whereas the pendulum can flip outside this region. However, it is difficult to determine when it will flip. Alex Small (2013) observed similar behaviour for a double pendulum composed of 2 point masses.
The lack of a natural excitation frequency meant the double pendulum systems in seismic resistance designs can be applied to buildings. The building itself is the primary inverted pendulum, and a secondary mass is connected to complete the double pendulum.
7. Double inverted pendulum
https://en.wikipedia.org/wiki/Double_inverted_pendulum
This device is a combination of the inverted pendulum and the double pendulum. The double inverted pendulum is unstable, which leads to it falling down unless it is controlled in some way. 2 main methods of controlling a double inverted pendulum include movement of the base, or application of a torque at the pivot point between the 2 pendulums.
8. Doubochinski’s pendulum
https://en.wikipedia.org/wiki/Doubochinski%27s_pendulum
This is a classical oscillator that interacts with a high-frequency field, which allows it to develop a discrete set of stable regimes of oscillation. Each regime oscillates at a frequency near to the proper frequency of the oscillator, but each with a distinct, "quantised" amplitude. he phenomenon of amplitude quantisation in this coupled system was first discovered by the Doubochinski brothers, Danil and Yakov, in 1968–69
This figure is a schematic of the Doubochinski’s pendulum. The system is composed of 2 interacting oscillatory processes:
— A pendulum arm with a natural frequency on the order of 0.5–1 Hz.
— A stationary electromagnet (solenoid) positioned under the equilibrium point of the pendulum's trajectory and supplied with alternating current of fixed frequency, typically in the range of 10–1000 Hz.
The mechanical pendulum arm and solenoid are configured to allow the pendulum arm to interact with the oscillating magnetic field of the solenoid only over a limited portion of its trajectory. This portion is called the “zone of interaction”, which is located outside of which the strength of the magnetic field drops off rapidly to zero.This spatial inhomogeneity of the interaction is key to the quantised behaviour and other unusual properties of the system.
This figure illustrates the first 4 quantised amplitudes of Doubochinski’s pendulum at a magnetic field frequency f=50 Hz. Solenoid is located at the bottom of the pendulum.
When the pendulum is released at any arbitrary starting position, its motion develops into 1 of a discrete set of stable oscillation modes. The figure depicts sharply differing amplitudes of the pendulum with approximately the same period of oscillation, which is close to the pendulum's undisturbed period. Many studies found that in each mode, the energy lost due to friction in the pendulum's motion is compensated by an average net energy transfer from the oscillating magnetic field. The stability of each amplitude-mode is maintained by a constant self-adjustment of the phase relationship between the pendulum and the high-frequency field. The pendulum’s interaction with field allows extraction of the required energy for the compensation of its frictional losses for a given period. Several studies found the pendulums compensates for the changes in the magnetic field strength through slight changes in the phase of its entry into the zone of interaction, whilst maintenance of the amplitude and frequency is occurring.
The table below shows the values of the quantised amplitudes, and the corresponding energies of the quantised modes. They are essentially independent of the strength of the alternating current supplied to the electromagnet, over a long range. This means the applied frequency is directly proportional to the array of stable modes that become accessible to the pendulum.
9. Elastic pendulum
https://en.wikipedia.org/wiki/Elastic_pendulum
Also known as a spring pendulum, this physical system consists of a mass connected to a spring o produce a motion that pertains both a simple pendulum and a one-dimensional spring-mass system.
When the spring compresses, the shorter radius causes the spring to move faster due to the conservation of angular momentum. The figure shows 2 DOF elastic pendulum with polar coordinate plots.
The spring has the rest length (l0), which stretches to length (x). The angle of oscillation of the pendulum is θ.
The Lagrangian (L) is:
L = T - V
— T = Kinetic energy
— V = Potential energy
Hooke’s law is the potential energy of the spring itself:
Vk = 0.5*k*x2
— k = Spring constant
On the other hand, the height of the mass determines the potential energy from gravity. For a given angle and displacement, the potential energy is calculated as:
Vg = -g*m*(l0 + x)*cos(θ)
— g = Gravitational acceleration
The kinetic energy is calculated by:
T = 0.5*m*v2
— v = Velocity of the mass
Relating v to the other variables requires writing v as a combination of a movement along and perpendicular to the spring:
T = 0.5*m*(x*2+ (l0 + x)2 * θ*2)
So the Lagrangian becomes:
L = T - Vk - Vg
L[x,x*,θ,θ*] = 0.5*m*(x*2 + (l0 + x)2 * θ*2) - 0.5*k*x2 + g*m*(l0 + x)*cos(θ)
With 2 degrees of freedom, for x and θ, the equations of motion can be found using 2 Euler-Lagrange equations:
For x:
m*(l0 + x)*θ*2 - k*x + g*m*cos(θ) - m*x** = 0
Isolating x** leads to:
x** = (l0 + x)*θ*2 - k*x/m + g*cos(θ)
And for θ:
-g*m*(l0 + x)*sin(θ) - m*(l0 + x)2 *(θ**) - 2*m*(l0 + x)*(x*)*(θ*) = 0
Isolating θ** gives:
θ** = -[g/(l0 + x)]*sin(θ) - [(2*x*)/(l0 + x)]*(θ*)
The motion of the elastic pendulum is depicted by 2 coupled ordinary different equations, which can be solved numerically. Furthermore, analytical methods can be used to study the intriguing phenomenon of order-chaos-order in this system.
10. Foucault pendulum
https://en.wikipedia.org/wiki/Foucault_pendulum
https://en.wikipedia.org/wiki/List_of_Foucault_pendulums
Invented by French physicist Léon Foucault in 1851, this device was conceived as an experiment to demonstrate the Earth's rotation.
This photo is the Foucault’s pendulum in the Panthéon, Paris.
Foucault Pendulum at the Chicago Museum of Science and Industry:
https://www.youtube.com/watch?v=iqpV1236_Q0
Foucault Pendulum at the Panthéon, Paris:
https://www.youtube.com/watch?v=59phxpjaefA
The first public exhibition of a Foucault pendulum took place in February 1851 in the Meridian of the Paris Observatory. A few weeks later, Foucault suspended a 28-kilogram (62 lb) brass-coated lead bob with a 67-metre long (220 ft) wire from the dome of the Panthéon, Paris. The period of the pendulum was 2*π*(l/g)0.5 = 16.5 seconds. Because the latitude of its location was φ = 48°52’, the plane of the pendulum's swing traverses a full circle in approximately 24h 56’/ sin(φ) = 31.8 hours (31 hours 50 mins), rotating clockwise approximately 11.3° per hour.
The original bob used in 1851 at the Panthéon was moved in 1855 to the Conservatoire des Arts et Métiers in Paris. In 1902, a 2nd temporary bob was installed at the same Paris conservatory for the 50th anniversary.
During the museum reconstruction in the 1990s, the original pendulum was temporarily displayed at the Panthéon (1995), but was later returned to the Musée des Arts et Métiers before its reopening in 2000. On April 6, 2010, the cable suspending the bob in the Musée des Arts et Métiers fractured, causing irreparable damage to the pendulum bob and to the marble flooring of the museum. Since this incident, the original, yet-to-be-repaired pendulum bob is now displayed in a separate case adjacent to the current pendulum display. Since 1995, an exact copy of the original pendulum has been operating under the dome of the Panthéon, Paris since 1995.
Discuss the mechanics of the Foucault Pendulum
At either geographic pole, the plane of oscillation of a pendulum remains fixed relative to the distance masses of the universe while Earth rotates underneath it, which takes 1 sidereal day to complete a full rotation. Therefore, the plane of oscillation of a pendulum at the North Pole relative to earth rotates clockwise during a full day and vice versa at the South Pole.
If we suspend a Foucault pendulum at the equator, the plane of oscillation remains fixed relative to Earth. At other latitudes, the plane of oscillation precesses relative to Earth, but more slowly than at the pole. Therefore the angular speed (ω, clockwise degrees per sidereal day) is proportional to the sine of the latitude (Ψ):
ω = 360° *sin(Ψ) / day
— Latitudes north and south of the equator are defined as positive and negative, respectively.
e.g. A Foucault pendulum at 30° south latitude, viewed from above by an earthbound observer, rotates counterclockwise 360° in 2 days.
This is an animation of a Foucault pendulum on the northern hemisphere, with the Earth's rotation rate greatly exaggerated. The green trace shows the path of the pendulum bob over the ground (a rotating reference frame), while in any vertical plane. The actual plane of swing appears to rotate relative to the Earth. The wire should be as long as possible—lengths of 12–30 m (40–100 ft) are common.
Foucault used a gyroscope in 1852 to demonstrate rotation directly rather than indirectly via the swinging pendulum. The inner gimbal of the Foucault gyroscope was balanced on knife edge bearings on the outer gimbal, which was suspended by a fine, torsion-free thread. This allowed the lower pivot point to carry almost no weight. The gears were organised to allow the gyro to spin 9,000–12,000 revolutions per minute before they were positioned correctly. This provided sufficient time to balance the gyroscope and carry out 10 minutes of experimentation. At least 3 more replicates of the Foucault gyro were constructed in convenient travelling and demonstration boxes, which still survived in the UK, France and the USA.
This diagram shows a Foucault pendulum at the North Pole, where it swings in the same plane as the Earth rotates beneath it.
Nobel laureate Heike Kamerlingh Onnes developed a fuller theory of Foucault pendulum for his doctoral thesis (1879). It concluded geometrical imperfection of the system or elasticity of the support wire would cause interference between two horizontal modes of oscillation. This may explain Onnes' pendulum going over from linear to elliptic oscillation in an hour.
Since air resistance damps the oscillation, some Foucault pendulums in museums incorporate an electromagnetic or other drive to keep the bob swinging. A 2009 report highlighted another main engineering problem is the preferred direction of swing, which can be alleviated with a 1-meter Foucault pendulum.
11. Furuta pendulum
https://en.wikipedia.org/wiki/Furuta_pendulum
https://www.hindawi.com/journals/jcse/2011/528341/
https://www.hindawi.com/journals/mpe/2010/742894/
13. Grotta Giganta horizontal pendulums
https://www.researchgate.net/publication/236655991_The_Grotta_Gigante_horizontal_pendulums_-_Instrumentation_and_observations
https://en.wikipedia.org/wiki/Grotta_Gigante_horizontal_pendulums
This is a schematic drawing of the horizontal pendulum of grotta gigante. The rod with the mass rotates in the horizontal plane about the virtual rotation axis. The angle φ of the virtual rotation axis with the plumb line is essential for the amplification factor of the pendulum.
14. Inertia wheel pendulum
This pendulum consists of an inertia wheel that is used as a pedagogical problem in control theory, which deals with the control of dynamical systems in engineered processes and machines.
Equations of motion
Euler – Lagrange formalism was used to derive the equations of motion of the inertia wheel pendulum.
The kinetic energy of the system is:
The potential energy of the system is:
The Lagrangian is:
A classical model of the torque developed by the DC motor is:
Due to the coupling by the rubber belt, the torque applied to the disk joint is:
The angular velocity of the disk is:
-- N = The ratio between the diameter of the inertia disk and the diameter of the motor pulley, known as the reduction ratio.
Thus, the torque applied to the disk is:
If we combine derivative and the Lagrangian, this yields:
-- τf1 & τf2 = Friction torques at the joints of the pendulum and the disk, respectively.
15. Inverted pendulum
https://en.wikipedia.org/wiki/Inverted_pendulum
Therefore, the inverted pendulum accelerates away from the vertical unstable equilibrium in the direction initially displaced, and the acceleration is inversely proportional to the length.
The pendulum is assumed to consist of a point mass (m) attached to the end of a massless rigid rod of length (l) attached to a pivot point at the end opposite the point mass.
The torque due to gravity providing the net torque is given as:
The moment of inertia for a point mass:
- v1 = Velocity of the cart
- v2 = Velocity of the point mass m.
- v1 & v2 can be expressed in terms of x and θ by writing the velocity as the first derivative of the position.
If we simplify the equation for v2:
The kinetic energy equation would be:
The generalised coordinates of the system are θ and x, each having a generalised force. On the x axis, the generalised force Qx can be calculated through its virtual work:
- Rx & Ry = Reaction forces at the joint
- FN = Normal force applied to the cart.
The top equation solves for the horizontal reaction force, whereas the bottom equation solves only depends on the vertical reaction force thus solves for the normal force.
In order to complete the equations of motion, we need to compute the acceleration of the point mass attached to the pendulum. The position of the point mass can be defined in inertial coordinates:
If we take 2 derivatives, it yields the acceleration vector in the inertial reference frame.
Next, we use Newton's 2nd law to write the 2 equations in the x-direction and the y-direction. Note that the reaction forces are positive as applied to the pendulum and negative when applied to the cart, due to the application of Newton's 3rd law.
The first equation allows us to compute the horizontal reaction force in case the applied force (F) is unknown, whereas the second equation solves for the vertical reaction force.
into:
This yields:
The pendulum equation of motion written in vector form is:
If we dot x^B with the acceleration of the pendulum on the RHS of the equation, then this yields:
If we combine a the LHS with the RHS and divide through by m, this yields:
Example real world applications of the inverted pendulum include:
16. Invariable pendulums
An invariable pendulum measuring gravity in Madras, India, in 1821.
17. Kapitza’s pendulum
https://en.wikipedia.org/wiki/Kapitza%27s_pendulum
This is a drawing of a Kapitza pendulum. It consists of a motor that rotates a crank at a high speed, a crank that vibrates a lever arm up and down, which the pendulum is attached to with a pivot.
The following notation will be used:
Energy equations
This means the effective potential has two minima if (a*v)2 > 2*g*l, or equivalently, (a/l)*(v/ω0) > √2. The first minimum is in the same position (x,y) = (0, -l) as the mathematical pendulum and the other minimum is in the upper vertical position (x,y) = (0, l). Although the upper vertical position is unstable in a mathematical pendulum, it can be stable in Kapitza's pendulum.
Rotating solutions
When the pendulum rotates around the pivot point at the same frequency that the pivot point is driven, we can derive 2 rotating solutions in this pendulum (1 clockwise and 1 counterclockwise). We can shift to the rotating reference frame using φ --> φ' + v*t to yield equation for φ:
If we consider the limit in which v is significantly higher the proper frequency ω0, we can work out the rapid-v slow-φ0' limit leads to the equation:
There is a stable equilibrium at φ0' = 0 and an unstable equilibrium at φ0' = π.
18. Kater’s pendulum
https://en.wikipedia.org/wiki/Kater%27s_pendulum
Kater used this pendulum to accurately determine the period by using the clock pendulum by the method of coincidences, which involves timing the interval between the coincidences when the 2 pendulums were swinging in synchronism. Furthermore, he determined the mean length of the solar seconds pendulum at London, at sea level, at 17 °C (62 °F), swinging at vacuum, was 0.9941 metres (39.1386 inches). This is equivalent to a gravitational acceleration of 9.81158 m/s2. Since the biggest margin of error from the mean was 7.1 μm (0.00028 inches), it translated to a precision of gravity measurement of 0.7×10-5 (7 milligals). Therefore, in 1824, the British Parliament officially made Kater's measurement of the seconds pendulum a backup standard of length for defining the yard if the yard prototype was destroyed.
19. Metronome
https://en.wikipedia.org/wiki/Metronome
What are the different types of metronomes?
20. Paraconical pendulum
https://www.researchgate.net/figure/The-left-panel-is-adapted-from-1-on-paraconical-pendulum-experiment-and-the-right-is_fig2_310899774
21. Persoz pendulum
https://en.wikipedia.org/wiki/Persoz_pendulum
22. Quantum pendulum
https://en.wikipedia.org/wiki/Quantum_pendulum
https://phys.org/news/2016-05-quantum-swinga-pendulum.html
Describe the Schrödinger equation
Using Lagrangian mechanics from classical mechanics, we can develop a Hamiltonian for the system. A simple pendulum has 1 generalised coordinate (the angular displacement, φ) and 2 constraints (the length of the string and the plane of motion). The kinetic and potential energies of the system can be computed as:
This results in the Hamiltonian
The time-dependent Schrödinger equation for the system is:
To solve the time-independent Schrödinger equation to find the energy levels and corresponding eigenstates, we need to change the independent variable as follows:
This is simply Mathieu's differential equation:
The solutions to this equation are Mathieu equations.
What are the solutions?
Provided that q are countably many special values of a, known as characteristic values, the Mathieu equation accepts solutions that are periodic with period 2π. The characteristic values of the Mathieu cosine, sine functions respectively are written an(q), bn(q), where n is a natural number. The periodic special cases of the Mathieu cosine and sine functions are often written CE(n,q,x), SE(n,q,x) respectively:
The energies of the system (E) for even/odd solutions respectively, are quantised based on the characteristic values found by solving the Mathieu equation.
The general solution of the above differential equation for a given value of a and q is a set of linearly independent Mathieu cosines and Mathieu sines, which are even and odd solutions respectively. In general, the Mathieu functions are aperiodic. However, for characteristic values of an(q), bn(q), the Mathieu cosine and sine become periodic with a period of 2π.
For positive values of q, the following is true:
Below are first few periodic Mathieu cosine functions for q = 1:
Note that the green line CE(1,1,x) resembles a cosine function, but with flatter hills and shallower valleys.
23. Rayleigh-Lorentz pendulum
https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Lorentz_pendulum
-- E = Average energy averaged over an oscillation
This leads to:
The last equation illustrates that angular momentum around the vertical axis, |Lz| = l*sin(θ) x m*l*sin(θ)φ* is conserved. The factor m*l2 * sin2(θ) will contribute in the Hamiltonian formulation below.
The azimuth (φ) is omitted from the Lagrangian because it is a cyclic coordinate, which suggests that its conjugate momentum is a constant of motion.
This allows Hamilton's equations to provide time evolution of coordinates and momenta in 4 first-order differential equations.
The trajectory of the mass on the sphere can be computed from the expression for the total energy.
Note that the component of angular momentum Lz = m*l2 * sin2(θ) is a constant of motion, independent of time.
This leads to an elliptic integral of the first kind for θ.
as well as an elliptic integral of the third kind for φ.
https://www.youtube.com/watch?v=0GAdMAm1-3o&ab_channel=AmritaVlab
This is a drawing of Coulomb's torsion balance. From Plate 13 of his 1785 memoir.
-- I = Moment of inertia
Note this is the a simple harmonic oscillator with equation of motion:
θ = θ0 cos(ω*t + φ)
ω = √(κ/I)
28. Wilberforce pendulum
https://en.wikipedia.org/wiki/Wilberforce_pendulum
What is its frequency?
29. Airy’s coal pit experiments
https://www.mezzacotta.net/100proofs/archives/120
30. Von Sterneck and Mendenhall gravimeters
31. Double pendulum gravimeters
ii. Irreversibile processes
https://en.wikipedia.org/wiki/Irreversible_process
A loss of mechanical energy in a system often eventuated in an increase of the system's temperature. Physicist James Prescott Joule was the first to experimentally demonstrate how a certain amount of work done against friction produced a definite quantity of heat, which was conceived as the random motions of the particles that comprise matter. This highlights the equivalence between mechanical energy and heat in collisions. In inelastic collisions, the decrease in mechanical energy may have been transformed into kinetic energy of the constituent particles and temperature increases. I’ll discuss thermodynamics in another post.
iii. Satellites
https://en.wikipedia.org/wiki/Vis-viva_equation
A satellite of mass (m) at a distance (r) from the centre of Earth possesses both kinetic energy (K) and gravitational potential energy (U). Thus, mechanical energy (Emechanical) of the satellite-Earth system can be calculated with:
Emechanical = U + K
Emechanical = -G*(M*m)/r + 0.5*m*v2
If the satellite is in circulate orbit, the energy conservation equation can be further simplified into:
Emechanical = -G*(M*m) / (2*r)
In circular motion, Newton's 2nd Law of motion can be expressed as:
G*(M*m)/r2 = (m*v2)/r
https://en.wikipedia.org/wiki/Projectile_motion
This graph shows the components of the initial velocity of parabolic throwing.
This graph shows the trajectories of a projectile with air drag and varying initial velocities.
This photo shows parabolic water motion trajectory.
Consider a projectile being launched with an initial velocity v(0) = v0, which can be expressed as the sum of horizontal and vertical components as follows.
v0 = v0xi + v0yj
The components v0x & v0y can be evaluated if the initial launch (i.e., elevation) angle (θ) is known:
v0x = v0*cos(θ)
v0y = v0*sin(θ)
Describe the kinematic quantities of projectile motion
In projectile motion, the horizontal motion and the vertical motion occur independently of each other. This principle of compound motion was established by Galileo in 1638, which was used to prove the parabolic form of projectile motion. An example of a ballistic trajectory with constant acceleration is a space ship in the absence of other forces. In contrast, the space ship’s acceleration on Earth changes magnitude with altitude and direction with latitude/longitude, leading to an elliptic trajectory. At higher speeds the trajectory can also be circular, parabolic or hyperbolic (unless it is distorted by other celestial objects).
This figure shows the independence of horizontal and vertical components of a projectile's velocity.
i. Acceleration
In projectile motion, there is only acceleration in the vertical direction (due to free fall), so the velocity in the horizontal direction is constant, being equal to v0*cos(θ). The components of the acceleration are:
— ax = 0
— ay = g
ii. Velocity
The horizontal component of the object’s velocity remains unchanged throughout the motion. Since the acceleration due to gravity is constant, the vertical component of the object’s velocity changes linearly. We can integrate accelerations in the x and y directions to solve for the components of velocity at any time t, as follows:
— vx = v0*cos(θ)
— vy = v0*sin(θ) - g*t
Using Pythagorean theorem would yield the magnitude of the velocity:
v = (vx2 + vy2)0.5
iii. Displacement
At any time t, the projectile's horizontal and vertical displacement are:
— x = v0*t*cos(θ)
— y = v0*t*sin(θ) - 0.5*g*t2
The magnitude of the displacement is:
Δr = (x2 + y2)0.5
If we eliminate t between the equations for x and y, then we can obtain the following equation:
y = tan(θ)*x - x2*g/[2*v02*cos2(θ)]
Since g, θ, v0 are constants, the above equation is of the form
y = a*x + b*x2
— a & b = Constants
The above equation is a parabola, so the path is parabolic, hence the axis of the parabola is vertical.
If the projectile’s position (x,y) and launch angle (θ or α) are known, the initial velocity can be found solving for v0 in the aforementioned parabolic equation:
v0 = [(g*x2) / (x*sin(2*θ) - 2*y*cos2(θ)]0.5
How do we calculate the time of flight or total time of the whole journey?
The total time (t) for which the projectile is airborne is called the time of flight.
y = v0*t*sin(θ) - 0.5*g*t2
After the flight, the projectile returns to the horizontal axis (x-axis), so y = 0.
0 = v0*t*sin(θ) - 0.5*g*t2
v0*t*sin(θ) = 0.5*g*t2
v0*sin(θ) = 0.5*g*t
t = 2*v0*sin(θ) / g
— Note that air resistance is negligible on the projectile.
If the starting point is at height, y0, with respect to the point of impact, the time of flight is:
t = d / (v*cos(θ)) = {v*sin(θ) + [(v*sin(θ))2 + 2*g*y0]0.5} / g
The above equation can be simplified to
t = {v*sin(θ) + [(v*sin(θ))2]0.5}/g
= [v*sin(θ) + v*sin(θ)] / g
= 2*v*sin(θ) / g
= 2*v*sin(45) / g
= [2*v*0.5*20.5] / g
t = (v*20.5)/g
— θ = 450
— y0 = 0
How do we calculate the maximum height of the projectile?
The maximum height an object can reach is known as the peak of the object's motion. The increase in height lasts until vy = 0, which is expressed as:
0 = v0*sin(θ) - g*th
The time it takes to reach the maximum height (h):
th = (v0*sin(θ))/g
The vertical displacement of the maximum height of projectile can be calculated by the following equations:
h = v0*th*sin(θ) - 0.5*g*th2
h = (v02 * sin2(θ)) / 2g
Describe the relationship between horizontal range and maximum height
The relation between the range (R) on the horizontal plane and the maximum height (h) reached at 0.5*td is:
h = (v02 * sin2(θ)) / 2g
R = (v02 * sin(2θ)) / g
h/R = [(v02 * sin2(θ)) / 2g]*[g / (v02 * sin(2θ))] = sin2(θ) / [4*sin(θ)*cos(θ)]
h = R*tan(θ) / 4
How do we calculate the maximum distance of a projectile?
https://en.wikipedia.org/wiki/Range_of_a_projectile
If we assume the Earth’s surface is fairly level, has a uniform gravitational field, then we can predict the range of a launched projectile if the initial conditions are known. If there is no air resistance and no change in gravitational acceleration, then the projectile motion is ideal.
If we launch a projectile at 45o relative to the ground, it will travel the furthest horizontally.
d = (v2 * sin(2θ))/g
The range becomes maximum when sin(2θ) is equal to 1 (i.e. 2θ = 90o , so θ = 45 o ).
i. Flat ground
When y0 equals zero, the horizontal position of the projectile is:
x(t) = v*t*cos(θ)
In the vertical direction:
y(t) = v*t*sin(θ) - 0.5*g*t2
To compute the time it takes for the projectile to return to the same height it originated, we let tg be any time when the height of the projectile is equal to its initial value.
0 = v*t*sin(θ) - 0.5*g*t2
If we factor in: t = 0 or t = (2*v*sin(θ)) / g
Since t = T = time of flight, then T = (2*v*sin(θ)) / g
The first solution corresponds to the moment the projectile first launches, while the second solution can be used to determine the range of the projectile. If we substitute the value for t into the horizontal equation, we get:
x = [2*v2 * cos(θ)*sin(θ)] / g
Next we apply the following trigonometric identity:
sin(x + y) = sin(x)*cos(y) + sin(y)*cos(x)
If x = y, then:
sin(2θ) = 2*sin(θ)*cos(θ)
That means we can simplify the solution to:
d = (v2 * sin(2θ)) / g
If θ = 45o, then
dmax = v2/g
ii. Rough ground
If y0 is not zero, then the equations of projectile motion becomes:
x(t) = v*t*cos(θ)
y(t) = y0 + v*t*sin(θ) - 0.5*g*t2
To solve for t, the position (y) of the projectile needs to be zero.
0 = y0 + v*t*sin(θ) - 0.5*g*t2
Next we apply the quadratic formula to yield 2 solutions. After several steps of algebraic manipulation, we yield:
t = [v*sin(θ) / g] + [(v2 * sin2(θ) + 2*g*y0)0.5 / g]
Since the square root needs to be a positive number, and the velocity and the sine of the launch angle is assumed to be positive, then the solution with the greatest time can be obtained when the positive of the plus or minus sign is used.
t = [v*sin(θ) / g] + [(v2 * sin2(θ) + 2*g*y0)0.5 / g]
The projectile range would be:
d = (v*cos(θ)/g) * [v*sin(θ) + (v2 *sin2(θ) + 2*g*y0)0.5]
or, equivalently
d = (v2/2g)*{1+ [1 + (1 + (2*g*y0)/(v2 * sin2(θ))]0.5}*sin(2θ)
The range can be maximised at any height when the angle:
θ = arccos[(2*g*y0 + v2) / (2*g*y0 + 2*v2)]0.5
If we want to check the limit as y0 approaches 0, it can be written as: How do we calculate the maximum distance of a projectile?
iii. Angle of impact
The angle (ψ) at which the projectile lands on the ground is expressed as:
tan(ψ) = -vy(td) / vx(td) = (v2 * sin2(θ) + 2*g*y0)0.5 / (v*cos(θ))
To achieve maximum range, the above equation becomes:
tan2(ψ) = (2*g*y0 + v2) / v2 = C + 1
We can rewrite the original solution for θ to yield:
tan2(θ) = (1 - cos2θ) / cos2θ = v2 / (2*g*y0 + v2) = 1 / (C + 1)
If we multiply the above equation with tan2(ψ), we get:
tan2(ψ)*tan2(θ) = [(2*g*y0 + v2) / v2] * [v2 / (2*g*y0 + v2)] = 1
Because of the trigonometric identity:
tan(θ + ψ) = [tan(θ) + tan(ψ)] / [1 - tan(θ)*tan(ψ)]
— This means that (θ + ψ) must be equal to 90 degrees.
This graph shows trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s2. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed. t = time from launch, T = time of flight, R = range and H = highest point of trajectory (indicated with arrows).
How is work energy theorem applied to projectile motion?
According to the work-energy theorem, the vertical component of velocity is:
vy2 = (v0*sin(θ))2 - 2*g*y
— Note it ignores aerodynamic drag and assumes that the landing area is at uniform height 0.
How we calculate the angle of reach?
The “angle of reach” is the angle (θ) at which a projectile must be launched in order to go a distance (d) given the initial velocity (v).
sin(2θ) = g*d/v2
There are 2 solutions to this equation:
(1) θ = 0.5*arcsin(g*d/v2)
(2) θ = 90o - 0.5*arcsin(g*d/v2)
How we calculate the angle required to hit a certain coordinate (x,y)?
If we launch a projectile from the origin (0,0) with an initial speed (v), the angle(s) of launch (θ) required to reach a target at range x and altitude y are:
tan (θ) = {[v2 + (v4 - g*(g*x2 + 2*y*v2))0.5] / (g*x)}
The 2 roots of the equation correspond to the 2 possible launch angles, assuming they aren’t imaginary, in which case the initial speed is insufficient in reaching the point (x,y) selected. Nevertheless, we can work out the angle of launch needed without the restriction of y = 0.
We can also evaluate the launch angle required to achieve the lowest possible launch velocity. If the 2 solutions above equal one another, this suggests that the quantity under the square root sign is zero. So, solving the quadratic equation for v2 yields:
v2 /g = y + (y2 + x2)0.5
This gives out the angle:
θ = arctan[y/x + (y2 /x2 + 1)0.5]
If we designate the angle whose tangent is y/x by α, then:
tan(θ) = (sin(α) + 1)/cos(α)
tan(π/2 - θ) = cos(α)/(sin(α) + 1)
cos2(π/2 - θ) = 0.5*(sin(α) + 1)
2*cos2(π/2 - θ) -1 = cos(π/2 - α)
Hence:
θ = 0.5*π - 0.5*(π/2 - α)
This means the launch should be at the angle halfway between the target and Zenith (vector opposite to Gravity).
This graph illustrates vacuum trajectory of a projectile for different launch angles. Launch speed is the same for all angles, 50 m/s if "g" is 10 m/s2.
How do we calculate the total path length of the trajectory?
If we know the height of launch and landing is identical and ignore air resistance, we can compute the length of the parabolic arc (L) traced by a projectile using the following formula:
L = [(v2 * cos2(θ))/2g] * [2*sec(θ)*tan(θ) - ln|(1 - sin(θ))/(1 + sin(θ))|
— v = Initial velocity
— θ = Launch angle
— g = Acceleration due to gravity as a positive value.
The above equation is obtained through evaluation of the arc length integral for height-distance parabola between the bounds initial and final displacements (i.e. between 0 and the horizontal range of the projectile).
Describe the trajectory of a projectile with air resistance
If air resistance is accounted for, it directly affects the velocity of the particle (Fa ∝ v→), which is only valid at Reynolds Number below about 1000. Since air has a kinematic viscosity around 0.15 cm2/s , the product of speed and diameter must be less than about 150 cm2/s.
At higher values of Reynolds number, the force of air resistance is proportional to the square of the particle's velocity.
The free body diagram illustrates a projectile under the effects of air resistance and gravity. It assumes air resistance to be in the direction opposite of the projectile's velocity. It’s more realistic to use the relationship Fair = -k*v2 to achieve an analytic solution. Nevertheless, Fair = -k*v is used because of the initial assumption of direct proportionality, which indicates the difference between the air resistance and the velocity being a constant arbitrary factor with units of N*s/m.
This graph shows the trajectories of a mass thrown at an angle of 70°:
— Black = Without drag
— Blue = With Stokes drag
— Green = With Newton drag
The relationships representing the motion of the particle are derived by Newton’s 2nd law, both in the x and y directions.
— x direction: ΣF = -k*vx = m*ax
— y direction: ΣF = -k*vy - m*g = m*ay
This implies that:
[1] ax = -k*vx /m = d*vx /dt
&
[2] ay = (-k*vy - m*g)/m = -k*vy /m - g = dvy /dt
The solution to equation [1] is an elementary differential equation, therefore the steps leading to a unique solution for vx and, subsequently, x will not be enumerated. Given the initial conditions vx = vxo (vxo is the x component of the initial velocity) and sx = 0 for t = 0:
[1a] vx = vxo*e-k*t/m
[1b] sx = (m*k)*vxo*(1 - e*-k*t/m)
Extensively solving [2] requires the initial conditions vy = vyo & sy = 0 when t = 0:
[2a] - g = dvy/dt + k*vy/m
We can solve the first order, linear, non-homogeneous differential equation via an integrating factor e ∫k/m dt .
[2b] et*k/m *(dvy/dt + vy*k/m) = et*k/m *(-g)
[2c] (vy*et*k/m)’ = et*k/m *(-g)
[2d] ∫(vy*et*k/m)’ dt = vy*et*k/m = ∫et*k/m *(-g) dt
[2e] vy*et*k/m = (m*k)* et*k/m *(-g) + C
[2f] vy = -m*g/k + C*et*k/m
Integration of [2f] yields the following equation:
[3] sy = -m*g/k + (m/k)*(vyo + m*g/k)*et*k/m + C
Solving for our initial conditions gets us the following functions:
[2g] vy(t) = -m*g/k + (vyo + m*g/k)*et*k/m
[3a] sy(t) = -m*g/k - (m/k)*(vyo + m*g/k)*et*k/m- (m/k)*(vyo + m*g/k)
Equation [3a] can be simplified into:
[3b] sy(t) = -(m*g/k)*t + (m/k)*(vyo + m*g/k)*(1 - et*k/m)
Example: Terminal velocity of a baseball
m = 0.145 kg (5.1 oz)
v0 = 44.7 m/s (100 mph)
g = -9.81 m/s2 (-32.2 ft/s2)
vt = -33.0 m/s (-73.8 mph)
k = m*g/vt = 0.145*(-9.81)/(-33) = 0.0431 kg/s
θ = 45o
The more realistic trajectory Fair = -k*|v|*v can not be calculated analytically, but only by numerical simulations.
ax = -k*vx2/m = dvx/dt
vx = dsx/dt
— If vxo is not 0, vx = 1/(1/vxo + k*t/m).
— If vxo is 0, vx = 0
sx = m*ln(1 + vxo*k*t)/k
Since acceleration is negative while the velocity is positive and vice versa, a projectile fired upwards requires the absolute value to be taken of the vertical velocity. This would make an analytical solution for vertical position more complex.
Where g0 is gravitational acceleration set to some constant, such as standard gravity:
ay = (-k*|vy|*vy)/m - g0 = dvy/dt for constant gravity
A more complex solution:
ay = (-k*|vy|*vy)/m - g0*(r /(r+sy))2 = dvy/dt for gravity as a function of height above a planet's surface.
— g0 = Planet’s gravity at the surface
— r = Planet’s radius
How do we calculate the time of flight with air resistance?
The total time of the journey in the presence of air resistance can be determined by solving the equation sy(t) = 0. The Lambert W function is required to solve sy(t) = 0 in the case of zero air resistance. Since the equation [3b] is of the form c1*t + c2 + c3 *ec4*t = 0, it can be transformed into an equation solvable by the W equation. In closed form, the total time of flight can be evaluated algebraically by the following equation:
t = (m/k) + vyo/g + (m/k)*W*{-[1 + k*vyo/(m*g)*e-(1 + k*vyo(m*g))]
How does numerical integration yield the solution?
Solving the problem of motion of a projectile with air resistance modelled as Fair = -k*v2 follows. Assumptions include constant gravitational acceleration and air resistance dictated by the following drag equation: FD = 0.5*c*ρ*A*v2
— FD = Drag force
— c = Drag coefficient
— ρ = Air density
— A = Cross sectional area of the projectile
Consider a projectile of mass (m) launching from a point (x0, y0), with an initial velocity (V0) in an initial direction that makes an angle (ψ0) with the horizontal. The air resistance experienced by the projectile is dictated by the formula Fair = -k*v2, which acts tangentially to the path of travel at any point. If we apply Newton’s 2nd of motion in the x-direction, we get:
[1] fx = -k*V2 *cos(ψ) = m*(d2x/dt2)
— V = (u2 + v2)0.5
— u & v = Horizontal and vertical components of the velocity V respectively.
If μ = k/m, du/dt = d2x/dt2, and cos(ψ) = u/V, then equation [1] becomes:
[1a] du/dt = -μ*u*(u2 + v2)0.5
In the y-direction:
[2] fy = -k*V*sin(ψ) - m*g = m*(d2y/dt2)
If μ = k/m, dv/dt = d2y/dt2, and sin(ψ) = u/V, then equation [2] becomes:
[2a] dv/dt = -μ*u*(u2 + v2)0.5 - g
Since we know that dv/du = (dv/dt) / (du/dt), we can divide equation [2a] by equation [1a] to obtain:
[2b] dv/dt = v/u + g/[μ*u*(u2 + v2)0.5]
If we introduce a quantity q such that v = q*u, then we get:
[2c] dv/du = d(q*u)/du = q + u*(dq/du)
Notice that if equation [2b] = equation [2c], then it can be rewritten as: u*(dq/du) = g/[μ*u*(u2 + v2)0.5]. If we separate the variables and integrate them, we yield:
[2d] (g/μ)* ∫(u-3) du = ∫ (1+q2)0.5 dq
The left hand side of equation [2d] is ∫(u-3) du = -1/(2*u2). Meanwhile, for the right-hand side of equation [2d], let q = sinh(Q), such that 1 + u2 = 1 + sinh2(Q) = cosh2(Q) and, dq = cosh(Q) dQ. Hence, ∫(1+q2)0.5 dq = ∫cosh2(Q) dQ. Also cosh2(Q) = 0.5*(1+cosh(2Q))
This yields ∫cosh2(Q) dQ = 0.5*[Q + 0.5*sinh(2Q)].
Equation [2d] becomes -g/(d*μ*u2) = 0.5*[Q + 0.5*sinh(2Q)] - Ã
For which;
u = (g/μ)0.5 * 1/[A - (Q + 0.5*sinh(2Q))0.5
Since v = u*q = u*sinh(Q)
v = (g/μ)0.5 *sinh(Q) / [A - (Q + 0.5*sinh(2Q))]0.5
If we denote λ = Α - (Q + 0.5*sinh(2Q), this means:
[2e] u = (g/μ)0.5 * (1/λ0.5)
[2f] v = (g/μ)0.5 * [sinh(Q)]/λ0.5
At the start of the motion, t = 0 and V02 = u02 + v02 .
Thus, tan(ψ0) = v0/u0 = q0 = sinh(Q0), such that A = g/(μ*u02) + [Q0 + 0.5*sinh(2*Q0)]
As the motion proceeds, u —> 0 and v —> (a negative constant), i.e. q —> ∞ & Q —> -∞.
This means that, λ —> ∞ & 1/λ0.5 —> 0. Hence:
Note that in equations [2e] & [2f], as u —> 0, v —> (g/μ)0.5 .
When a state of dynamic equilibrium is achieved under vertical free fall, the opposing forces of gravity and drag are equalised, i.e. k*vt2 = m*g.
[2g] vt = (g/μ)0.5
In equation [1a], if we substitute the values for u and v from equations [2e] and [2f], this leads to:
du/dt = (-g/λ)*cosh(Q)
Moreover, du/dQ = [1/(2*λ1.5)]*(g/μ)*(1 + cosh(2Q))
Since dQ/dt = (du/dt)/(du/dQ), this means:
dt/dQ = -cosh(Q) / [(g*μ)0.5 * λ0.5]
[2h]
Since dx/dQ = (dx/dt)/(dQ/dt) = (-1/μ)*(cosh(Q)/λ)
[2i]
And dy/dQ = (dy/dt)/(dQ/dt) = (-1/2μ)*(sinh(2Q)/λ)
[2j]
I. Time of flight
To determine the time of flight (T), we set y to 0 in equation [2j] and then solve for the value of the variable Q.
[2k]
Equation [2h] with QT substituted for Q yields:
[2l]
II. Horizontal range
Equation [2i] gives the horizontal range R as:
[2m]
III. Maximum height
At the highest point of the projectile path ψ = 0, and Q = 0, it gives the maximum height from equation [2j].
[2n]
Describe lofted trajectory
A special case of a ballistic trajectory for a rocket is a lofted trajectory, which has an apogee greater than the minimum-energy trajectory to the same range. Generally speaking, the rocket travels higher by it expends more energy to get to the same landing point. A lofted trajectory is preferred for various reasons such as increasing distance to the horizon to give greater viewing/communication range or for changing the angle with which a missile will impact on landing.
Lofted trajectories of North Korean missiles Hwasong-14 and Hwasong-15.
Eddie Woo’s videos on projectile motion
Therefore,
l*(d2θ/dt2) = -g*sin(θ)
d2θ/dt2 + (g/l)*sin(θ) = 0
τ = r*F = dL/dt
Let’s define the torque on the pendulum bob using the force due to gravity.
τ = l*Fg
— l = Length vector of the pendulum
— Fg = Force due to gravity
Consider the magnitude of the torque on the pendulum:
|τ| = -m*g*l*sin(θ)
— m = Mass of the pendulum
— g = Acceleration due to gravity
— l = Length of the pendulum
— θ = Angle between the length vector and the force due to gravity
We can rewrite the angular momentum formula:
L = r*p = m*r*(ω*r)
Consider the magnitude of the angular momentum:
|L| = ω*m*r2 = (m*l2)*(dθ/dt)
and its time derivative:
(d/dt)*|L| = (m*l2)*(d2θ/dt2)
Since τ = dL/dt, comparing the magnitudes yields us:
-m*g*l*sin(θ) = (m*l2)*(d2θ/dt2)
Hence:
d2θ/dt2 + (g/l)* sin(θ) = 0
3. How is [1] derived from “energy”?
It can be obtained via the principle of conservation of mechanical energy. Remember that any object falling a vertical distance (h) acquires kinetic energy that is equal in magnitude to the potential energy lost during the fall. i.e. Gravitational potential energy is converted into kinetic energy, which is expressed as:
ΔU = m*g*h
Since the change in kinetic energy (body starting from rest) is expressed as:
ΔK = 0.5*m*v2
Since no energy is lost, the gain in one form must be equal to the loss in the other form:
0.5*m*v2 = m*g*h
The change in velocity for a given change in height is given as:
v = (2*g*h)0.5
Using the arc length formula above, this equation can be rewritten in terms of dθ/dt.
v = l*(dθ/dt) = (2*g*h)0.5
dθ/dt = (2*g*h)0.5 / l
— h = Vertical distance the pendulum fell
This diagram illustrates the trigonometry of a simple gravity pendulum.
If a pendulum begins its swing from an initial angle (θ0), then the vertical distance from the screw (y0) can be calculated using:
y0 = l*cos(θ0)
Similarly, for y1, we get:
y1 = l*cos(θ)
Then h is the difference between the 2 above formulae:
y1 - y0 = h = l*[(cos(θ) - cos(θ0)]
In terms of dθ/dt, this leads to:
dθ/dt = [(2*g/l)*(cos(θ) - cos(θ0))0.5 [2]
The above equation is known as the “first integral of motion”, which provides the velocity in terms of the location and an integration constant associated with the initial displacement (θ0). If we apply the chain rule, we can differentiate this equation with respect to time to evaluate the acceleration:
ii. Period of oscillation (Small-angle approximation)
https://en.wikipedia.org/wiki/Pendulum_(mathematics)
Although the above differential equation is difficult to solve, we can’t express a solution that can be written in terms of elementary functions. If we restrict the size of the oscillation's amplitude, we can obtain a form of the solution. Assume that the angle is much less than 1 radian (often cited as less than 0.1 radians, about 6°), or θ << 1, then we substitute for sin(θ) into [1] by applying the small-angle approximation:
sin(θ) ~ θ,
We then get the equation for a harmonic oscillator:
d2θ/dt2 + (g/l)*θ = 0.
The error due to the approximation is of order, θ3 (from Taylor expansion for sin (θ).
Given the initial conditions θ(0) = θ0 and dθ/dt(0) = 0, then we yield the solution:
θ(t) = θ0*cos(t*(g/l)0.5), θ0 << 1
This describes simple harmonic motion.
— θ0 = Amplitude of the oscillation i.e. maximum angle between the rod of the pendulum and the vertical.
The period of the motion, the time for a complete oscillation (outward and return) is expressed as:
T0 = (2*π)*(l/g)0.5, θ0 << 1
The above equation is known as Christiaan Huygens’s law for the period. Under the small-angle approximation, the period is independent of the amplitude (θ0). This is a property of isochronism that Galileo discovered during this time. The rule of thumb for pendulum length is:
T0 = (2*π)*(l/g)0.5—> l = (g/π2)*(T02/4)
If SI units are used (i.e. measured in metres and seconds), and the measurement is assumed to occur on the Earth's surface, then g ~ 9.81 m/s2 and g/π2 ~ 1.
Therefore, a relatively reasonable approximation for the length and period can be:
l ~ (T02/4) ,
T0 ~ 2*l0.5
-- T0 = Number of seconds between 2 beats (1 beat for each side of the swing).
— l = Metres
This graph shows small-angle approximation for the sine function: for θ ~ 0 we find sin(θ) ~ θ.
What is the arbitrary-amplitude period?
For amplitudes beyond the small angle approximation, we can compute the exact period by first inverting the equation for the angular velocity obtained from the energy method [2]:
and then integrating over 1 complete cycle:
T = t*(θ0—> 0 —> -θ0—> 0 —> θ0)
or twice the half-cycle:
T = 2*t*(θ0—> 0 —> -θ0)
or 4 times the quarter-cycle:
T = 4*t*(θ0—> 0)
leading to:
Notice that this integral diverges as θ0 approaches the vertical limit:
Hence a pendulum with sufficient energy to reach a vertical position won’t actually achieve it.
This integral can be rewritten in terms of elliptic integrals:
— F = Incomplete elliptic integral of the first kind defined by:
Substitution can express this more concisely as:
sin(u) = sin(θ/2) / sin(θ0/2)
— θ is expressed in terms of u.
T = (2*T0/π)*K(k), where k = sin(θ0/2). [3]
— K = The complete elliptic integral of the first kind defined by:
τ = r*F = dL/dt
Let’s define the torque on the pendulum bob using the force due to gravity.
τ = l*Fg
— l = Length vector of the pendulum
— Fg = Force due to gravity
Consider the magnitude of the torque on the pendulum:
|τ| = -m*g*l*sin(θ)
— m = Mass of the pendulum
— g = Acceleration due to gravity
— l = Length of the pendulum
— θ = Angle between the length vector and the force due to gravity
We can rewrite the angular momentum formula:
L = r*p = m*r*(ω*r)
Consider the magnitude of the angular momentum:
|L| = ω*m*r2 = (m*l2)*(dθ/dt)
and its time derivative:
(d/dt)*|L| = (m*l2)*(d2θ/dt2)
Since τ = dL/dt, comparing the magnitudes yields us:
-m*g*l*sin(θ) = (m*l2)*(d2θ/dt2)
Hence:
d2θ/dt2 + (g/l)* sin(θ) = 0
3. How is [1] derived from “energy”?
It can be obtained via the principle of conservation of mechanical energy. Remember that any object falling a vertical distance (h) acquires kinetic energy that is equal in magnitude to the potential energy lost during the fall. i.e. Gravitational potential energy is converted into kinetic energy, which is expressed as:
ΔU = m*g*h
Since the change in kinetic energy (body starting from rest) is expressed as:
ΔK = 0.5*m*v2
Since no energy is lost, the gain in one form must be equal to the loss in the other form:
0.5*m*v2 = m*g*h
The change in velocity for a given change in height is given as:
v = (2*g*h)0.5
Using the arc length formula above, this equation can be rewritten in terms of dθ/dt.
v = l*(dθ/dt) = (2*g*h)0.5
dθ/dt = (2*g*h)0.5 / l
— h = Vertical distance the pendulum fell
This diagram illustrates the trigonometry of a simple gravity pendulum.
If a pendulum begins its swing from an initial angle (θ0), then the vertical distance from the screw (y0) can be calculated using:
y0 = l*cos(θ0)
Similarly, for y1, we get:
y1 = l*cos(θ)
Then h is the difference between the 2 above formulae:
y1 - y0 = h = l*[(cos(θ) - cos(θ0)]
In terms of dθ/dt, this leads to:
dθ/dt = [(2*g/l)*(cos(θ) - cos(θ0))0.5 [2]
The above equation is known as the “first integral of motion”, which provides the velocity in terms of the location and an integration constant associated with the initial displacement (θ0). If we apply the chain rule, we can differentiate this equation with respect to time to evaluate the acceleration:
ii. Period of oscillation (Small-angle approximation)
https://en.wikipedia.org/wiki/Pendulum_(mathematics)
Although the above differential equation is difficult to solve, we can’t express a solution that can be written in terms of elementary functions. If we restrict the size of the oscillation's amplitude, we can obtain a form of the solution. Assume that the angle is much less than 1 radian (often cited as less than 0.1 radians, about 6°), or θ << 1, then we substitute for sin(θ) into [1] by applying the small-angle approximation:
sin(θ) ~ θ,
We then get the equation for a harmonic oscillator:
d2θ/dt2 + (g/l)*θ = 0.
The error due to the approximation is of order, θ3 (from Taylor expansion for sin (θ).
Given the initial conditions θ(0) = θ0 and dθ/dt(0) = 0, then we yield the solution:
θ(t) = θ0*cos(t*(g/l)0.5), θ0 << 1
This describes simple harmonic motion.
— θ0 = Amplitude of the oscillation i.e. maximum angle between the rod of the pendulum and the vertical.
The period of the motion, the time for a complete oscillation (outward and return) is expressed as:
T0 = (2*π)*(l/g)0.5, θ0 << 1
The above equation is known as Christiaan Huygens’s law for the period. Under the small-angle approximation, the period is independent of the amplitude (θ0). This is a property of isochronism that Galileo discovered during this time. The rule of thumb for pendulum length is:
T0 = (2*π)*(l/g)0.5—> l = (g/π2)*(T02/4)
If SI units are used (i.e. measured in metres and seconds), and the measurement is assumed to occur on the Earth's surface, then g ~ 9.81 m/s2 and g/π2 ~ 1.
Therefore, a relatively reasonable approximation for the length and period can be:
l ~ (T02/4) ,
T0 ~ 2*l0.5
-- T0 = Number of seconds between 2 beats (1 beat for each side of the swing).
— l = Metres
This graph shows small-angle approximation for the sine function: for θ ~ 0 we find sin(θ) ~ θ.
What is the arbitrary-amplitude period?
For amplitudes beyond the small angle approximation, we can compute the exact period by first inverting the equation for the angular velocity obtained from the energy method [2]:
and then integrating over 1 complete cycle:
T = t*(θ0—> 0 —> -θ0—> 0 —> θ0)
or twice the half-cycle:
T = 2*t*(θ0—> 0 —> -θ0)
or 4 times the quarter-cycle:
T = 4*t*(θ0—> 0)
leading to:
Notice that this integral diverges as θ0 approaches the vertical limit:
Hence a pendulum with sufficient energy to reach a vertical position won’t actually achieve it.
This integral can be rewritten in terms of elliptic integrals:
— F = Incomplete elliptic integral of the first kind defined by:
Substitution can express this more concisely as:
sin(u) = sin(θ/2) / sin(θ0/2)
— θ is expressed in terms of u.
T = (2*T0/π)*K(k), where k = sin(θ0/2). [3]
— K = The complete elliptic integral of the first kind defined by:
In order the compare the approximation to the full solution, let the period of a pendulum of length 1 m on Earth (g = 9.80665 m/s2) at initial angle 10 degrees:
4*(1*m/g)0.5 *K*[sin(10o/2)] ~ 2.0102s
The linear approximation would be evaluated as:
2*π*(1*m/g)0.5 ~ 2.0064s
The difference between these 2 values is much less than that caused by the variation of g with geographical location. This allows us to calculate the elliptic integral further on.
 |
| This graph illustrates the deviation of the "true" period of a pendulum from the small-angle approximation of the period. The “true” value can be obtained numerically by evaluating the elliptic integral. |
— Legendre polynomial solution for the elliptical integral
Given [3] and the Legendre polynomial solution for the elliptic integral:
This graph shows the relative errors using the power series for the period. T0 is the linear approximation and T2 to T10 include respectively the terms up to the 2nd to the 10th powers.
— Power series solution for the elliptic integral
Another formulation of the above solution can be evaluated if the following Maclaurin series:
sin (θ0/2) = (θ0/2) - (θ03/48) + (θ05/3840) - (θ07/645120) + …
is used in the Legendre polynomial solution above. The power series becomes:
T = (2*π)*(l/g)0.5 *[1 + (θ02/16) + (11*θ04/3072) + (173*θ06/737280) + (22931*θ08/1321205760) + (1319183*θ010/951268147200) + (233526453*θ012/2009078326886400) + … ]
— Arithmetic-geometric mean solution for elliptic integral
Given [3] and the arithmetic-geometric mean solution of the elliptic integral:
K(k) = [0.5*π]/[M(1 - k, 1 + k)]
— M(x,y) = Arithmetic-geometric mean of x and y.
This leads to an alternative and faster-converging formula for the period:
T = (l/g)0.5 *(2*π)/M(1, cos(θ0/2))
This first iteration of this algorithm gives out:
T1 = (2*T0)/(1 + cos(θ0/2))
This approximation has the relative error of less than 1% for angles up to 96.11 degrees. Since (1+cos(θ0/2))/2 = cos2(θ0/4), the expression can be simplified as:
T1 = T0*sec2(θ0/4)
The 2nd order expansion of sec2(θ0/4) simplifies to T ~ T0*(1 + θ02/16).
A 2nd iteration of this algorithm gives:
 |
| The top graph shows the potential energy and phase portrait of a simple pendulum. Note that the x-axis, being angle, wraps onto itself every 2π radians. |
What is the approximate formulae for the non-linear pendulum period?
The approximate formulae for the increase of the pendulum period with amplitude is categorised as follows:
— ‘Not so large-angle’ formulae, i.e. those yielding good estimates for amplitudes below π/2 radians, which is a natural limit for a bob on the end of a flexible string. With respect to the exact period, the deviation increases monotonically with amplitude, making it unsuitable for amplitudes closest to π radians. Lima (2006) proposed the following formula:
T ~ -T0*[ln(a)/(1 - a)], where a = cos(θ0/2).
— ‘Very large-angle formulae, i.e. those which approximate the exact period asymptotically for amplitudes near to π radians, with an error that increases monotonically for smaller amplitudes. Cromer (1995) proposed the following formula: T ~ (2/π)*T0* ln(4/a).
Although it hasn’t been formulated yet, a simple approximate formula for the pendulum period valid for all possible amplitudes is not too far away as physicists are still working on it. Lima (2008) derived a weighted-average formula with this characteristic:
T ~ (r*a2 * TLima* + k2 * TCramer*) / (r*a2 + k2)
— r = 7.17, which presents a maximum error of only 0.6% (at θ0 = 95o).
What is arbitrary-amplitude angular displacement Fourier series?
The Fourier series expansion of θ(t) is expressed as:
T ~ -T0*[ln(a)/(1 - a)], where a = cos(θ0/2).
— ‘Very large-angle formulae, i.e. those which approximate the exact period asymptotically for amplitudes near to π radians, with an error that increases monotonically for smaller amplitudes. Cromer (1995) proposed the following formula: T ~ (2/π)*T0* ln(4/a).
Although it hasn’t been formulated yet, a simple approximate formula for the pendulum period valid for all possible amplitudes is not too far away as physicists are still working on it. Lima (2008) derived a weighted-average formula with this characteristic:
T ~ (r*a2 * TLima* + k2 * TCramer*) / (r*a2 + k2)
— r = 7.17, which presents a maximum error of only 0.6% (at θ0 = 95o).
What is arbitrary-amplitude angular displacement Fourier series?
The Fourier series expansion of θ(t) is expressed as:
— q = exp(-π*K’/K) : Elliptic nome
— ω = 2π/T : Angular velocity
If ε is defined above, q can be approximated using the expansion:
q = ε + 2*ε5 + 15*ε9 + 150*ε13 + 1707*ε17 + 20910*ε21 + …
Note that for θ0* < π, we have ε < 0.5, thus the approximation is applicable even for large amplitudes.
Examples
(Left to right, from top to bottom)
a. Initial angle of 0o , a stable equilibrium
b. Initial angle of 45o .
c. Initial angle of 90o .
d. Initial angle of 135o
e. Initial angle of 170o
f. Initial angle of 180o
g. Pendulum with just barely enough energy for a full swing
h. Pendulum with enough energy for a full swing
iii. Compound pendulum
Also known as a physical pendulum, a compound pendulum is a swinging rigid body that is free to rotate about a fixed horizontal axis. This setup implies the rod has mass and can stretch in size, i.e. an arbitrarily shaped rigid body swinging by a pivot. This means the pendulum’s period correlates with its moment of inertia (I) around the pivot point.
The equation of torque is:
τ = I*α
— α = Angular acceleration
— τ = Torque
The torque generated by gravity is:
τ = -m*g*L*sin(θ)
— m = Mass of the body
— L = Distance from the pivot to the object's centre of mass
— θ = Angle from the vertical
The formula for a takes the same form as the conventional simple pendulum. This means the period of oscillation can be evaluated as:
T = 2*π*[1/(m*g*L)]0.5
And frequency can be expressed as:
f = 1/T = (1/2π)*[(m*g*L)/I]0.5
We take the initial angle into consideration (for large amplitudes), then the formula for α becomes:
α = θn = -(m*g*L*sin(θ))/I
And the period becomes:
T = (4*K)*[sin2(θ0/2)]*[1/(m*g*L)]0.5
— θ0 = Maximum angle of oscillation (with respect to the vertical
— K(k) = Complete elliptic integral of the first kind
The radius of oscillation or equivalent length (L) of any physical pendulum can be calculated as:
L = I/(m*R)
— I = Moment of inertia of the pendulum about the pivot point
— m = Mass of the pendulum
— R = Distance between the pivot point and the centre of mass.
If we substitute this equation into the formula (T ~ 2*π*(L/g)0.5, then the period (T) of a compound pendulum for sufficiently small oscillations can be calculated as:
T = 2*π*(1/(m*g*R))0.5
T = 2*π*[(m*L2/3) / (m*g*(L/2))]0.5
T = 2*π*[(2*L)/(3*g)]0.5
In 1673, Huygens proved that the pivot point and the centre of oscillation are interchangeable. If we flipped the pendulum upside down and swung it from a pivot located at its previous centre of oscillation, it demonstrates the same period as before and the old pivot point becomes the new centre of oscillation. In 1817, Kater used this principle to produce a type of reversible pendulum, namely Kater pendulum, in order to detect accurate measurements of the acceleration due to gravity.
Describe the physical interpretation of the imaginary period
The Jacobian elliptic function expresses the position of a pendulum as a function of time. It is a doubly periodic function with a real period and an imaginary period. The real period is defined as the time it takes the pendulum to go through one full cycle. Appell (1978) highlighted the physical interpretation of the imaginary period: if θ0 is the maximum angle of one pendulum and (180 - θ0) is the maximum angle of another, then the real period of each is the magnitude of the imaginary period of the other.
What are coupled pendulums?
When pendulums are coupled together, they affect each other’s motion, either through a direct connection (e.g. bobs connected by a spring), or through motions in a supporting structure (such as a tabletop). The equations of motion for 2 identical simple pendulums coupled by a spring connecting the bobs are evaluated using Lagrangian Mechanics. The kinetic energy of this system can be worked out using the formula:
EK = 0.5*m*L2 (*θ12 + *θ22)
— m = Mass of the bobs
— L = Length of the strings
— θ1, θ2 = Angular displacements of the 2 bobs from equilibrium.
The potential energy of this system is given by the following formula:
Ep = m*g*L*[2 - cos(θ1) - cos(θ2)] + 0.5*k*L2*(θ2 - θ1)2
— g = Gravitational acceleration
— k = Spring constant
— L(θ2 - θ1) = Displacement of the spring from its equilibrium position, which assumes the small angle approximation.
The Lagrangian can be calculated to be:
L = 0.5*m*L2*(*θ12 + *θ22) - m*g*L*[2 - cos(θ1) - cos(θ2)] + 0.5*k*L2*(θ2 - θ1)2
This leads to the following set of coupled differential equations:
**θ1 + (g/L)*sin(θ1) + (k/m)*(θ1 - θ2) = 0
**θ2 + (g/L)*sin(θ2) - (k/m)*(θ1 - θ2) = 0
When we add and subtract these 2 equations in turn, and then apply the small angle approximation, this yields 2 harmonic oscillator equations in the variables (θ1 + θ2) and (θ1 - θ2):
**θ1 + **θ2 + (g/L)*(θ1 + θ2) = 0
**θ1 - **θ2 + (g/L + 2*(k/m))*(θ1 - θ2) = 0
along with the corresponding solutions:
θ1 + θ2 = A*cos(ω1*t + α)
θ1 - θ2 = Β*cos(ω2*t + β)
— ω1 = (g/L)0.5
— ω2 = (g/L + 2*k/m)0.5
— A, B, α, β = Constants of integration
If we express the solutions in terms of θ1 & θ2 alone, then:
θ1 = 0.5*A*cos(ω1*t + α) + 0.5* Β*cos(ω2*t + β)
θ2 = 0.5*A*cos(ω1*t + α) - 0.5* Β*cos(ω2*t + β)
If the bobs weren’t provided an initial push, then condition *θ1(0) = *θ2(0) = 0 requires α = β = 0, which yields:
A = θ1(0) + θ2(0)
B = θ1(0) - θ2(0)
Describe the history of the pendulum
One of the earliest known uses of a pendulum was a 1st-century seismometer device invented by Han Dynasty Chinese scientists Zhang Heng. Needham described its function as swaying and activating a series of levers after the disturbance by the tremor of an earthquake. Then a small ball would be released by a lever and fall out f the urn-shaped device into 1 of 8 metal toad's mouths below. Each of the 8 points of the compass signifies a direction of the earthquake’s original location.
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This is a replica of Zhang Heng’s seismometer, with the pendulum inside. |
- Around 1602, Italian scientist Galileo Galilei was the first person to study the properties of pendulums with with his paper contained in a letter to Guido Ubaldo dal Monte, from Padua, dated November 29, 1602.
- His biographer and student, Vincenzo Viviani, claimed Galileo's interest in pendulums began around 1582 by the swinging motion of a chandelier in Pisa Cathedral.
- Galileo was thought to have discovered the property of isochromism, an important property of pendulums that make them useful timekeepers. It refers to the period of the pendulum being approximately independent of the amplitude or width of the swing.
- Moreover, he discovered the period is independent of the mass of the bob, and proportional to the square root of the length of the pendulum.
II. 1656: The Pendulum clock
- Although Galileo's son, Vincenzo, began construction of a pendulum clock in 1641, he failed to complete it prior to his death in 1649. Therefore, in 1656, Dutch scientist Christiaan Huygens had the honour of being the first person to build the first ever pendulum clock.
- His invention improved timekeeping accuracy from within 15 mins to within 15 seconds a day, which lead to their widespread usage across Europe with existing clocks retrofitted with pendulum clocks.
- In 1666, English scientist Robert Hooke studied the conical pendulum, which comprised a pendulum that can swing freely in 2 dimensions, with the bob rotating in a circle or ellipse. The motions of this pendulum allowed him to model the orbital movements of the planets in the solar system.
- In 1679, Hooke provided Isaac Newton a conjecture that the components of orbital motion consisted of an inertial motion along a tangent direction plus an attractive motion in the radial direction. This helped contribute to Newton's formulation of the law of universal gravitation. Furthermore, Robert Hooke made important contributions as early as 1666 that the pendulum was a reliable tool to measure the force of gravity.
- In 1671, Jean Richer's trip to Cayenne, French Guiana coincided with his discovery of a pendulum clock being 2.5 mins per day slower at Cayenne than at Paris. This lead to the suggestion that the force of gravity was lower at Cayenne.
- In 1687, Isaac Newton's Principia Mathematica demonstrated that the Earth was not a true sphere but slightly oblate (flattened at the poles) from the effect of centrifugal force due to its rotation, which leads to the direct proportional relationship between gravity and latitude.
- Poynting & Thompson (1907) stated portable precision gravimeters were used on voyages to measure the acceleration of gravity at different points on Earth, eventually resulting in accurate models of the shape of the Earth.
III. 1673: Huygen’s Horologium Oscillatorium
- In 1673, Christiaan Huygens published his theory of the pendulum, Horologium Oscillatorium sive de motu pendulorum.
- In 1636, Marin Mersenne and René Descartes discovered the pendulum did not demonstrate isochronous motion, meaning its period slightly increased with its amplitude.
- Huygens used a tautochrone curve to determine the curved path an object follows to descend by gravity towards the same point in the same time interval, no matter the starting point.
- He demonstrated this tautochrone curve was a cycloid, rather than the circular arc of a pendulum. This verified the pendulum wasn't isochronous, which meant Galileo's observation of isochronism was accurate only for small swings not big swings.
- Huygens' discovery of the centre of oscillation, and its interchangeability with the pivot point helped solve the problem of calculating the period of an arbitrarily shaped pendulum (called a compound pendulum).
- Headrick (2002) stated the existing clock movement, called the verge escapement, provided pendulums swing arcs of about 100 degrees. Huygens demonstrated this movement lead to inaccuracies, which would cause amplitude changes (produced by unavoidable variations in the clock's drive force) to influence the period.
- Huygens installed cycloidal-shaped metal 'chops' adjacent to the the pivots in his clock, which confined the suspension cord and forced the pendulum to follow a cycloid arc in order to achieve isochronous period. However, this setup was impractical since the pendulum's swing was restricted to small angles of a few degrees (around 4 - 6 degrees).
IV. 1721: Temperature compensated pendulums
- During the 18th century, scientists identified the pendulum clock's major source of error was the rod's expansion and contraction with changes in ambient temperature, impacting the swing's period.
- The inventions of the temperature compensated pendulums such as the mercury pendulum in 1721 and gridiron pendulum in 1726 helped minimise errors in precision pendulum clocks to a few seconds per week.
- In 1673, Hugyens' discoveries include a pendulum hanging from its centre of oscillation has the same period as another pendulum hanging from its pivot, and the distance between the two points being equal to the length of a simple gravity pendulum of the same period.
- In 1818, British Captain Henry Kater invented the reversible Kater's pendulum that make accurate measurements of gravity.
V. Foucault Pendulum
In 1851, the Foucault pendulum was the first device to demonstrate the Earth's rotation without the involvement of celestial observations, which generated a "pendulum mania". The rate of precession is exaggerated in this figure.
- In 1851, Jean Bernard Léon Foucault demonstrated that the plane of oscillation of a pendulum had the tendency to remain constant regardless of the motion of the pivot, which is reminiscent of a gyroscope. This helped show the actual rotation of the Earth.
- He suspended a 67 m (220 ft) long cord attached to a pendulum free to swing in 2 dimensions from the dome of the Panthéon in Paris. He observed the plane of swing precessed or rotated 360 degrees in about 32 hours.
- Tobin (2003) described the "pendulum mania" following this discovery since it was the first device to demonstrate Earth's rotation without celestial observations.
VI. 1930: Decline in use
- Around the early 1900s, materials with low-thermal-expansion were first used to make the pendulum rods in the highest precision clocks and other instruments. Examples of such materials include invar (FeNi36, a nickel steel alloy), and fused quartz (silica). This eliminated the need for the temperature compensation materials.
- Marrison (1948) stated precision pendulums were stored in low pressure tanks to keep the air pressure constant, which eliminates changes in the period due to changes in buoyancy of the pendulum due to changing atmospheric pressure. This yielded accuracy of around a second per year.
- The quartz crystal oscillator (1921) and quartz clock (1927) replaced pendulum clocks because they demonstrated improved timekeeping accuracy. Although the pendulum clocks decreased in usage as time standards around WW2, the French Time Service continued using them in their official time standard ensemble until 1954.
- In 1950s, pendulum gravimeters were superseded by "free fall" gravimeters, however pendulum instruments were still used into the 1970s.
How was the pendulum used to measure time?
From its discovery around 1582 until development of the quartz clock in the 1930s, the pendulum was the world's standard for accurate timekeeping for roughly 3 centuries. In the 17th and 18th centuries, free-swinging seconds pendulums were extensively used as precision timers in scientific experiments.
i. Clock pendulums
https://en.wikipedia.org/wiki/Pendulum_clock
ii. Temperature compensation
iii. Atmospheric pressure
iv. Gravity
The motion of pendulums are influenced by changes in gravitational acceleration, which varies by as much as 0.5% at different locations on Earth. This requires recalibration of precision pendulum clocks after any movement.
How accurate are pendulums as timekeepers?
All clocks contain timekeeping elements called harmonic oscillators, which include pendulums, balance wheels, quartz crystals, and vibrating atoms. They are selected for timekeeping because they vibrate or oscillate at a specific resonant frequency or period and resist oscillating at other rates. Nevertheless, the resonant frequency is not infinitely distinct. Jespersen, Fitz-Randolph & Robb (1999) found a narrow natural band of frequencies (or periods) around the resonant frequency where the harmonic oscillator will oscillate, known as the resonance width or bandwidth.
— Q factor
— Escapement
— The Airy condition
How was gravity measured?
The seconds pendulum has a period of 2 seconds, thus each swing takes 1 second to complete.
Early observations:
How was the pendulum involved in the standard of length?
What are other uses of a pendulum?
— Schular tuning
— Coupled pendulums
— Religious practice
— Education
— Torture device
- Pendulums in clocks are typically composed of a weight or bob (b) suspended by a rod of wood or metal (a). To minimise air resistance, the bob is a smooth disk with a lens-shaped cross section. A common weight for seconds pendulum bobs is around 6.8 kg (15 lb). It is made as heavy as the suspension can support and the movement can drive in order to enhance the clock's regulation.
- Instead of hanging from a pivot, clock pendulums are supported by a short straight spring (d) of flexible metal ribbon. This helps prevent the friction and 'play' caused by a pivot, and the spring's slight bending force supplements the pendulum's restoring force.
- An arm hanging behind the pendulum called the crutch (e) helps maintain the pendulum's swing. It ends in a fork (f), whose prongs embrace the pendulum rod. The clock's escapement pushes the clutch back and forth (g,h).
- Each time the pendulum swings through its centre position, it releases 1 tooth of the escape wheel (g). The clock's gear turn generates the force of the mainspring or a driving weight hanging from a pulley.
- This force transmits through the clock's gear train, driving the wheel to rotate, and a tooth presses against one of the pallets (h), which pushes the pendulum a little.
- Geared to the escape wheel, the clock's wheels move forward a fixed amount with each pendulum swing, which turns the clock's hands at a steady rate.
- Milham (1945) described the pendulum's period can be adjusted by moving an adjustment nut (c) under the bob vertically.
- Beckett (1874) discussed some tower clocks and precision clocks contain a tray near to the midpoint of the pendulum rod, where small weights can be added or removed in order to effectively adjust the centre of oscillation without the necessity of halting the clock.
- Beckett (1875) highlighted the importance of a rigid support to mount the pendulum since any elasticity would create minuscule undetectable swaying motions of the support, which in turn influences the clock's period, leading to timekeeping errors.
- Commonly used in grandfather clocks, the most common pendulum length in quality clocks is about 1 metre (39 inches) long. Mantel clocks have half-second pendulums that are 25 cm (9.8 in) long, or shorter. Large tower clocks such as Big Ben use the longer 1.5 second pendulum, which are 2.25 m (7.4 ft) long or the 2-second pendulum, which are 4 m (13 ft) long.
ii. Temperature compensation
- According to Matthys (2004), the largest source of error in early pendulums was thermal expansion and contraction of the pendulum rod with changes in ambient temperature, which affected the length of pendulums. In 1658, Huygens reported the discovery of pendulum clocks running slower in summer, by as much as a minute per week.
- In 1669, Jean Picard found a pendulum with a steel rod expands by about 11.3 parts per million (ppm) with each degree Celsius increase in temperature, which leads to a loss of about 0.27 seconds per day, or 9 seconds per day for a 33 °C (59 °F) change.
- Since wood does not expand as much, losing about 6 seconds per day for a 33 °C (59 °F) change, quality clocks often had wooden pendulum rods. However, the wood requires varnish to prevent water vapour from entering due to humidity impacting the pendulum length.
- In 1721, George Graham invented the mercury pendulum, the first device to compensate for errors due to temperature changes. As the temperature increases, the pendulum rod filled with mercury lengthens, the mercury expands and its surface level elevates slightly in the container, which shifts its centre of mass closer to the pendulum pivot.
- Theoretically, these effects cancel each other out, which leaves the pendulum's centre of mass, and its period, unchanged with temperature. Matthys (2004) pointed out that the mass of mercury would take more time than the rod to adapt to any temperature changes, which would be a source of rate deviation.
iii. Atmospheric pressure
- According to Archimedes' principle, the effective weight of the bob is decreased by the buoyancy of the air it displaces. As the mass (inertia) remains constant, it decreases the pendulum's acceleration during its swing, as well as increases the period. This depends on the air pressure and the density of the pendulum, rather than its shape.
- The pendulum carries a certain amount of air during its swing, and the mass of this air increases the inertia of the pendulum, which decreases the acceleration and increases the period.
- Viscous air resistance poses a retarding effect on the pendulum's velocity. Although its effect on the period is negligible, it still dissipates energy, which decreases the amplitude. This decreases the pendulum's Q factor, which means a stronger drive force from the clock's mechanism is necessary to maintain the pendulum's motion, which causes increased disturbance to the period.
- The Encyclopædia Britannica stated that increases in barometric pressure would increase a pendulum's period by about 0.11 seconds per day per kilopascal (0.37 seconds per day per inch of mercury or 0.015 seconds per day per torr).
- In 1865, Friedrich Tiede first operated a pendulum clock in a constant-pressure tank at the Berlin Observatory.
- By 1900, the highest precision clocks were installed in tanks with constant pressure in order to remove any changes in atmospheric pressure.
iv. Gravity
The motion of pendulums are influenced by changes in gravitational acceleration, which varies by as much as 0.5% at different locations on Earth. This requires recalibration of precision pendulum clocks after any movement.
How accurate are pendulums as timekeepers?
All clocks contain timekeeping elements called harmonic oscillators, which include pendulums, balance wheels, quartz crystals, and vibrating atoms. They are selected for timekeeping because they vibrate or oscillate at a specific resonant frequency or period and resist oscillating at other rates. Nevertheless, the resonant frequency is not infinitely distinct. Jespersen, Fitz-Randolph & Robb (1999) found a narrow natural band of frequencies (or periods) around the resonant frequency where the harmonic oscillator will oscillate, known as the resonance width or bandwidth.
— Q factor
- The measure of a harmonic oscillator's resistance to disturbances to its oscillation period is a dimensionless parameter called the Q factor, which is equal to the resonant frequency divided by the resonance width.
- This means a higher Q factor leads to a smaller resonance width, and a more constant frequency or period of the oscillator for a given disturbance. Matthys (2004) stated the reciprocal of the Q is approximately proportional to the limiting accuracy achievable by a harmonic oscillator as a time standard.
Q = (2*π*Ep) / Ef
— Ep = Energy stored in the pendulum
— Ef = Energy lost to friction during each oscillation period.
— Ef = Energy lost to friction during each oscillation period.
- In clocks, the pendulum is driven by the clock's movement to keep it swinging in order compensate the energy loses to friction. The force is provided by a mechanism called the escapement.
- Therefore, as the fraction of the pendulum's energy that is lost to friction decreases, energy input decreases, the disturbance exerted from the escapement decreases, the pendulum's independence of the clock's mechanism increases, and its period becomes more constant.
The Q of a pendulum can also be calculated as:
Q = M*ω/Γ
- Μ = Mass of the bob, limited by the load capacity and rigidity of the suspension.
- ω = 2π/T = Pendulum's radian frequency of oscillation, which is fixed by the pendulum's period.
- Γ = Frictional damping force on the pendulum per unit velocity
- Matthys (2004) calculated around 99% of the energy loss in a free-swinging pendulum is attributed to air friction, therefore he stated that mounting a pendulum in a vacuum tank increases the Q factor, hence the accuracy, by a factor of 100.
- The Q of pendulums ranges from several 1000 in an ordinary clock to several hundred thousand for precision regulator pendulums swinging in vacuum. Numerous studies stated that the most accurate commercially produced pendulum clock was the Shortt-Synchronome free pendulum clock.
- Jones (2000) found the Invar master pendulum swinging in a vacuum tank had a Q of 110,000, and an error rate of around a second per year.
— Escapement
- Although the escapement theoretically can provide identical impulses per pendulum swing, hence identical pendulum response and constant period, it is found to be impossible to achieve.
- This is due to unavoidable factors such as random fluctuations in the force exerted by friction of the clock's pallets, lubrication variations, and changes in the torque provided by the clock's power source mid-swing.
— The Airy condition
- Marrison (1948) described the effect of the escapement's drive force on the period of a pendulum is minimised if exerted as a short impulse as the pendulum passes through its bottom equilibrium position.
- If the impulse occurs before the pendulum reaches the bottom of its downward swing, then it shortens the pendulum's natural period. If it occurs after the pendulum reaches the bottom of its upswing, the period would lengthen.
- In 1826 British astronomer George Airy proved the pendulum's period is unaffected by changes in the drive force that is symmetrical about its bottom equilibrium position.
How was gravity measured?
- Since the acceleration of gravity g is in the periodicity equation (1) for a pendulum, the local gravitational acceleration of the Earth can, therefore, be calculated from the period of a pendulum. Pendulums were used as gravimeters to measure the local gravity, which varies by over 0.5% across the surface of the Earth.
- Since the pendulum in a clock is easily influenced by external forces from the clock movement, free-swinging pendulums were the standard instruments of gravimetry up to the 1930s.
Early observations:
- 1620 = British scientist Francis Bacon was one of the first researchers to propose using a pendulum to measure gravity. He even suggested using the pendulum on a mountain to determine gravity's relationship with altitude.
- 1644 = French priest Marin Mersenne was the first to compare pendulum's swing with the weight's time to fall a measured distance, and measured the length of the seconds pendulum to be 0.99 m (39.1 in).
- 1669 = Jean Picard used a 25 mm (1 in) copper ball suspended by an aloe fibre and computer the length of the seconds pendulum at Paris to be 0.993 m (39.09 in). Furthermore, he performed the first experiments on thermal expansion and contraction of pendulum rods with temperature.
- 1672 = Jean Richer first observed variability of gravity at different points on Earth by using a pendulum clock in Cayenne, French Guiana and discovered it lost 2.5 minutes per day. Moreover, in order to correct time, he had to shorten the pendulum by 2.6 mm (1.25 lignes) shorter than at Paris.
- 1687 = In his Principia Mathematica, Isaac Newton demonstrated Earth's shape is slightly oblate (flattened at the poles) due to the centrifugal force of its rotation. This supported the observation that gravity was directly proportional with altitude. Newton observed equal length pendulums with bobs made of different materials had the same period, which proved that the gravitational force on different substances was proportional to their mass (inertia). This is known as the equivalence principle, which became the foundation of Albert Einstein's general theory of relativity.
- 1737 = French mathematician Pierre Bouguer brought a copper pendulum bob shaped like a double pointed cone dangled by a thread to the Andes mountains in Peru. He made corrections to his calculations due to thermal expansion of the measuring rod and barometric pressure, and the additional gravity attributed to the gravitational field of the mountainous Peruvian plateau. From the density of rock samples, he estimated the effect of the altiplano on the pendulum, and compared this with Earth's gravity. This lead to make the first approximate estimate of the density of Earth.
- 1743 = Alexis Claude Clairaut proposed the first hydrostatic model of the Earth, namely Clairaut's theorem, which lead to the calculation of Earth's ellipticity using gravity measurements.
- 1747 = Daniel Bernoulli demonstrated the method of correcting for the increasing period of the pendulum due to the finite angle of swing θ0 by using the first order correction θ02/16.
- 1792 = Jean-Charles de Borda and Jean-Dominique Cassini used a seconds pendulum at Paris to define a pendulum standard of length for use with the new metric system. The difference between the periods of the pendulums, T1 and T2, can be calculated as:
1/Δt = 1/T1 - 1/T2
- 1821 = Francesco Carlini described a number of pendulum observations on top of Mount Cenis, Italy with the aim of calculating the density of the Earth. He modelled the mountain as a segment of a sphere 18 km (11 mi) in diameter and 1.69 km (1 mi) high. Then he used rock samples to calculate Earth's gravitational field, and estimate Earth's density to be 4.39 times that of water.
How was the pendulum involved in the standard of length?
- Since gravity was relatively constant over Earth's surface, scientists were interested in the pendulum length standard due to the pendulum length being the same at any point on Earth.
- By the mid-19th century, Edward Sabine and Thomas Young discovered that gravity, and thus the length of any pendulum standard, varied with local geologic features such as mountains and dense subsurface rocks. Therefore, a pendulum length standard was defined at a single point on Earth and measurement could only occur there specifically.
- In 1791, the seconds pendulum at 45° North latitude was the leading candidate for the definition of the new unit of length known as the metre, advocated by a group led by French politician Talleyrand and mathematician Antoine Nicolas Caritat de Condorcet.
- On March 19, 1791, the committee decided to base the metre on the length of the meridian through Paris, which ultimately rejected pendulum definition on a number of reasons. They include its variability at different locations and its definition of length based on a unit of time.
- However, Britain and Denmark were the only countries that based their units of length on the pendulum temporarily.
- In 1821, the Danish inch was defined as 1/38 of the length of the mean solar seconds pendulum at 45° latitude at the meridian of Skagen, at sea level, in a perfect vacuum.
- In 1824, the British parliament passed the Imperial Weights and Measures Act, which was a reform of the British standard system. It declared that destruction of the prototype standard yard would lead to the definition of the inch in order for the length of the solar seconds pendulum at London, at sea level, in a vacuum, at 62 °F was 39.1393 inches.
- This lasted until the 1834 Houses of Parliament fire that destroyed the prototype yard, which made it impossible to recreate it accurately from the pendulum definition. Therefore, in 1855, Britain repealed the pendulum standard and returned to prototype standards.
What are other uses of a pendulum?
— Schular tuning
- In his 1923 paper, Maximilian Schuler described a pendulum whose period exactly equals the orbital period of a hypothetical satellite orbiting just above the surface of the Earth (about 84 minutes). It tends to always point at the Earth's centre upon the sudden displacement of its support. Used in inertial guidance systems onboard ships and aircraft that operate on the Earth's surface, this physics principle is known as 'Schuler tuning'.
— Coupled pendulums
- In 1665, Huygens positioned 2 clocks on his mantlepiece and observed their exhibition of an opposing motion. This refers to the pendulums oscillating in unison but in the opposite direction i.e. 180° out of phase. No matter how the 2 clocks began ticking, they eventually returned to this state, therefore being the first to observe coupled oscillation.
- This behaviour was due to the 2 pendulums influencing each other through slight motions of the supporting mantlepiece called 'entrainment' or 'mode locking'.
— Religious practice
- An example of a pendulum used in religious ceremonies is the swinging incense burner called a censer or a thurible.
— Education
- In science education, pendulums are commonly used to demonstrate a harmonic oscillator in classes that teach dynamics and oscillatory motion.
- Lecturers also use a bowling ball or wrecking ball attached to a string to demonstrate the law of conservation of energy.
— Torture device
- Scott (2009) stated there was a claim that the pendulum was an instrument of torture and execution by the Spanish Inquisition in the 18th century, which was referenced in the 1826 book The history of the Inquisition of Spain by the Spanish priest, historian and liberal activist Juan Antonio Llorente.
- Abbott (2006) described a swinging pendulum containing a knife blade on the end that gradually descends toward a bound prisoner until it penetrates the body.
- However a number of sources were skeptical of the actual using of such torture device due to the lack of credible evidence.
Known as a harmonograph, it is a mechanical apparatus that uses pendulums to create a geometric image, which illustrates simple harmonic motion. The device is named after Hugh Blackburn, who described it in 1844. However, it was first discussed by James Dean and analysed mathematically by Nathaniel Bowditch in 1815. It involves a bob being suspended from a string that in turn hangs from a V-shaped pair of strings. When the pendulum oscillates simultaneously in 2 perpendicular directions with different periods, the bob consequently follows a path resembling a Lissajous curve.
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| A picture of a harmonograph. |
Video of a harmonograph in action:
https://www.youtube.com/watch?v=Y9W-XsIQFE0
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| This is a collection of Lissajous curves. |
The movements of damped pendulums allows the harmonograph to delineate its figures. Its motion is described by the following equation:
x(t) = A*sin(t*f + p)*e-d*t
— f = Frequency
— p = Phase
— A = Amplitude
— d = damping
— t = time
If a pendulum can move about 2 axes (in a circular or elliptical shape), due to the principle of superposition, the motion of a rod attached to the bottom of the pendulum along 1 axis can be described by the following equation:
x(t) = A1*sin(t*f1 + p1)*e-d1*t + A2*sin(t*f2 + p2)*e-d2*t
A typical harmonograph with 2 pendulums move the pen by 2 perpendicular rods connected to these pendulums. Therefore, the path of the harmonograph figure drawn is described by the following equations:
x(t) = A1*sin(t*f1 + p1)*e-d1*t + A2*sin(t*f2 + p2)*e-d2*t
y(t) = A3*sin(t*f3 + p3)*e-d3*t + A4*sin(t*f4 + p4)*e-d4*t
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This is a figure depicted by a simple lateral harmonograph |
2. Ballistic pendulum
https://en.wikipedia.org/wiki/Ballistic_pendulum
https://www.youtube.com/watch?v=dBAi74A8Td8
This device measures a bullet’s momentum, from which it can calculate its velocity and kinetic energy. Before the invention of modern chronographs, ballistic pendulums were used to directly measure the projectile velocity.
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| This is a photo of a green ballistic pendulum. |
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| This is an animation of a ballistic pendulum. |
The first ballistic pendulum was invented in 1742 by English mathematician Benjamin Robins (1707-1751), and its function was published in his book New Principles of Gunnery. This revolutionised the science of ballistics because it was the first method to accurately measure the velocity of a bullet.
Most physics textbooks provide a simplified method of calculation of the bullet's velocity that uses its mass and pendulum and the height of the pendulum’s deflection to calculate the amount of energy and momentum in the pendulum and bullet system. Nonetheless, Robins’ calculations used a measure of the period of oscillation to determine the rotational inertia of the system.
Consider the motion of the bullet-pendulum system from the instant the pendulum is struck by the bullet. We can calculate the initial velocity of the bullet-pendulum system using conservation of mechanical energy (kinetic energy + potential energy), if the acceleration due to gravity (g) and the final height of the pendulum (h) are both known. Let the initial velocity be v1, the masses of the bullet and pendulum be mb & mp respectively.
If we take the initial height of the pendulum as the potential energy reference (Uinitial = 0), then the final potential energy when the bullet-pendulum system comes to a stop (Kfinal = 0) can be evaluated as Ufinal = (mb + mp)*g*h
Applying the conservation of mechanical energy gives us:
Kinitial = Ufinal
0.5*(mb + mp)*v12 = (mb + mp)*g*h
Solve for velocity to obtain: v1 = (2*g*h)0.5
Next we apply the conservation of momentum for the bullet-pendulum system to calculate the speed of the bullet (v0) before it hits the pendulum. If we equate the momentum of the bullet before it is fired to that of the bullet-pendulum system as soon as the bullet strikes the pendulum, then we can obtain:
mb*v0 = (mb + mp)*(2*g*h)0.5
To solve for v0:
v0 = (mb + mp)*(2*g*h)0.5 / mb = (1 + mp/mb )*(2*g*h)0.5
Robin’s formula:
In the 1786 edition of his book, the formula is:
v = 614.58*g*v*(p + b)/(b*i*r*n)
— v = Velocity of the ball in units per second
— b = Mass of the ball
— p = Mass of the pendulum
— g = Distance from pivot to the centre of gravity
— i = Distance from pivot to the point of the ball’s impact
— c = Chord, measured by the ribbon described in Robin’s apparatus
— r = Radius, or distance from the pivot the attachment of the ribbon
— n = Number of oscillations made by the pendulum in 1 minute
Poisson’s formula:
A rotational inertia based formula similar to Robins' was derived by French mathematician Siméon Denis Poisson, who published it in The Mécanique Physique. It was used to measure the bullet velocity by using the recoil of the gun:
m*v*c*f = M*b*k’ * (g*h)0.5
— m = Mass of the bullet
— v = Velocity of the bullet
— c = Distance from pivot to the ribbon
— f = Distance from bore axis to pivot point
— M = Combined mass of gun and pendulum
— b = Chord measured by the ribbon
— k’ = Radius from pivot to the centre of mass of gun and pendulum (measured by oscillation, as per Robins).
— g = Gravitational acceleration
— h = Distance from the centre of mass of the pendulum to the pivot
Kinitial = Ufinal
0.5*(mb + mp)*v12 = (mb + mp)*g*h
Solve for velocity to obtain: v1 = (2*g*h)0.5
Next we apply the conservation of momentum for the bullet-pendulum system to calculate the speed of the bullet (v0) before it hits the pendulum. If we equate the momentum of the bullet before it is fired to that of the bullet-pendulum system as soon as the bullet strikes the pendulum, then we can obtain:
mb*v0 = (mb + mp)*(2*g*h)0.5
To solve for v0:
v0 = (mb + mp)*(2*g*h)0.5 / mb = (1 + mp/mb )*(2*g*h)0.5
Robin’s formula:
In the 1786 edition of his book, the formula is:
v = 614.58*g*v*(p + b)/(b*i*r*n)
— v = Velocity of the ball in units per second
— b = Mass of the ball
— p = Mass of the pendulum
— g = Distance from pivot to the centre of gravity
— i = Distance from pivot to the point of the ball’s impact
— c = Chord, measured by the ribbon described in Robin’s apparatus
— r = Radius, or distance from the pivot the attachment of the ribbon
— n = Number of oscillations made by the pendulum in 1 minute
Poisson’s formula:
A rotational inertia based formula similar to Robins' was derived by French mathematician Siméon Denis Poisson, who published it in The Mécanique Physique. It was used to measure the bullet velocity by using the recoil of the gun:
m*v*c*f = M*b*k’ * (g*h)0.5
— m = Mass of the bullet
— v = Velocity of the bullet
— c = Distance from pivot to the ribbon
— f = Distance from bore axis to pivot point
— M = Combined mass of gun and pendulum
— b = Chord measured by the ribbon
— k’ = Radius from pivot to the centre of mass of gun and pendulum (measured by oscillation, as per Robins).
— g = Gravitational acceleration
— h = Distance from the centre of mass of the pendulum to the pivot
k’ = (T/π)*(g*h)0.5
— T = Half the period of oscillation
Ackley’s ballistic pendulum:
In 1962, P.O. Ackley described the method of constructing and using a ballistic pendulum, which consisted of a parallelogram linkage in standardised size. This simplified the calculation of the ballistic’s velocity. Given the horizontal swing of the pendulum, the velocity of the bullet can be calculated as:
V = (Mp/Mb)*0.2018*D
— V = Velocity of the bullet (ft / s)
— Mp = Mass of the pendulum (grains)
— Mb = Mass of the bullet (grains)
— D = Horizontal travel of the pendulum (inches)
For more accurate measurements, a number changes are required. Construction changes include the addition of a small box on top of the pendulum. Prior to the pendulum being weighed, the box is filled with bullets of the type being measured. For every shot executed, a bullet is removed from the box in order to keep the mass of the pendulum constant. The measurement change included measurement of the period of the pendulum. As the pendulum swings, the number of complete oscillations is measured over a period of time, about 5-10 minutes. That time is then divided the number of oscillations to obtain the period. Then, the formula C = π/(T*12) is applied to yield more precise constant to replace the value 0.2018 in the above equation.
3. Barton’s pendulums
https://en.wikipedia.org/wiki/Barton%27s_pendulums
The Barton’s pendulums experiment, first performed by Professor Edwin Henry Barton, demonstrated the physical phenomenon of resonance and the response of pendulums to vibration at, below and above their resonant frequencies. He hung about 10 different pendulums from 1 common system. When the system vibrates at the resonance frequency of a driver pendulum, the target pendulum swings with the maximum amplitude.
4. Conical pendulum
https://en.wikipedia.org/wiki/Conical_pendulum
This device consists of a weight (or bob) fixed on the end of a string or rod suspended from a pivot. The bob moves at a constant speed in a circle with the string (or rod) tracing out a cone. Around 1660, English scientist Robert Hooke was the first to study the conical pendulum as a model for the orbital motion of planets. In 1673 Dutch scientist Christiaan Huygens used his new concept of centrifugal force in his book Horologium Oscillatorium to calculate its period, which was then used as the timekeeping element in a few mechanical clocks and other clockwork timing devices.
During the 1800s, conical pendulums were used as the timekeeping element in a few clockwork timing mechanisms because their motion were smooth. Beckett (1874) used a conical pendulum to turn the lenses of lighthouses to sweep their beams across the sea. Furthermore, he used the location drives of equatorial mount telescopes to allow allow the telescope to follow a star smoothly across the sky as the Earth turns.
In 1788, James Watt invented the centrifugal governor, which used a conical pendulum. This invention helped regulate the speed of steam engines during the Steam Age in the 1800s. Another example of a conical pendulum is the playground game tetherball, which uses a ball attached to a pole by a cord.
Imagine a conical pendulum with a bob of mass (m) revolving without friction in a circle at a constant speed (v) on a string of length (L) at an angle (θ) from the vertical.
There are 2 forces acting on the bob:
— The tension (T) in the string, exerted along the line of the string and and acting towards the point of suspension.
— The downward bob weight (m*g), where m is the mass of the bob and g is the local gravitational acceleration.
To evaluate the force exerted by the string, it has to be resolved into a horizontal component and a vertical component. The horizontal component is T*sin(θ) toward the centre of the circle, while the vertical component is T*cos(θ) in the upward direction. From Newton’s 2nd law, the horizontal component of the tension in the string gives the bob a centripetal acceleration toward the centre of the circle:
T*sin(θ) = m*v2/r
Since there is no acceleration in the vertical direction, the vertical component of the tension in the string is equal and opposite to the weight of the bob:
T*cos(θ) = m*g
Evaluating T/m can solve and equate the above 2 equations, thereby eliminating T and m:
g/cos(θ) = v2/(r*sin(θ))
Since the speed of the pendulum bob is constant, it can be expressed as the circumference 2*π*r divided by the time (t) required for 1 revolution of the bob:
v = 2*π*r/t
If we substitute the right side of this equation for v in the previous equation, we obtain:
g/cos(θ) = [2*π*r/t]2/[r*sin(θ)] = [r*(2*π)2]/[t2 * sin(θ)]
If we use the trigonometric identity tan(θ) = sin(θ) / cos(θ) and solve for t, the time required for the bob to travel one revolution is:
t = 2*π*[r/(g*tan(θ))]2
In a practical experiment, r varies and is as difficult to measure as the constant string length (L). If r, h and L form a right-angle triangle and θ being the angle between the length (h) and the hypotenuse (L), then r can be eliminated. This means:
r = L*sin(θ)
If we substitute this value for r, this yields a formula whose only varying parameter is the suspension angle (θ),
t = 2*π*[(L*cos(θ))/g]0.5
For small angles θ, cos(θ) ≈ 1:
t ~ 2*π*(L/g)0.5
— This means the period t of a conical pendulum is equal to the period of an ordinary pendulum of the same length. Furthermore, the period for small angles is approximately independent of changes in the angle θ. Hence the period of rotation is approximately independent of the force applied to keep it rotating. This property is shared with ordinary pendulums and makes both types of pendulums useful for timekeeping, known as isochronism.
5. Cycloidal pendulum
https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Book%3A_Classical_Mechanics_(Tatum)/19%3A_The_Cycloid/19.09%3A_The_Cycloidal_Pendulum
https://en.wikipedia.org/wiki/Cycloid#Cycloidal_pendulum
If a pendulum is suspended from the cusp of an inverted cycloid, and the pendulum's length (L) is equal to that of half the arc length of the cycloid (i.e. L = 4*r), then the bob of the pendulum also traces a cycloid path. Such a cycloidal pendulum is isochronous, regardless of amplitude. If we introduce a coordinate system centred in the position of the cusp, the equation of motion would be:
x = r*[2θ(t) + sin(2θ(t))]
y = r*[-3 - cos(2θ(t))]
— T = Half the period of oscillation
Ackley’s ballistic pendulum:
In 1962, P.O. Ackley described the method of constructing and using a ballistic pendulum, which consisted of a parallelogram linkage in standardised size. This simplified the calculation of the ballistic’s velocity. Given the horizontal swing of the pendulum, the velocity of the bullet can be calculated as:
V = (Mp/Mb)*0.2018*D
— V = Velocity of the bullet (ft / s)
— Mp = Mass of the pendulum (grains)
— Mb = Mass of the bullet (grains)
— D = Horizontal travel of the pendulum (inches)
For more accurate measurements, a number changes are required. Construction changes include the addition of a small box on top of the pendulum. Prior to the pendulum being weighed, the box is filled with bullets of the type being measured. For every shot executed, a bullet is removed from the box in order to keep the mass of the pendulum constant. The measurement change included measurement of the period of the pendulum. As the pendulum swings, the number of complete oscillations is measured over a period of time, about 5-10 minutes. That time is then divided the number of oscillations to obtain the period. Then, the formula C = π/(T*12) is applied to yield more precise constant to replace the value 0.2018 in the above equation.
3. Barton’s pendulums
https://en.wikipedia.org/wiki/Barton%27s_pendulums
 |
| This is a schematic diagram of the Barton’s pendulums experiment. |
4. Conical pendulum
https://en.wikipedia.org/wiki/Conical_pendulum
This device consists of a weight (or bob) fixed on the end of a string or rod suspended from a pivot. The bob moves at a constant speed in a circle with the string (or rod) tracing out a cone. Around 1660, English scientist Robert Hooke was the first to study the conical pendulum as a model for the orbital motion of planets. In 1673 Dutch scientist Christiaan Huygens used his new concept of centrifugal force in his book Horologium Oscillatorium to calculate its period, which was then used as the timekeeping element in a few mechanical clocks and other clockwork timing devices.
 |
A photo of the monumental conical pendulum clock by Farcot, 1878 |
In 1788, James Watt invented the centrifugal governor, which used a conical pendulum. This invention helped regulate the speed of steam engines during the Steam Age in the 1800s. Another example of a conical pendulum is the playground game tetherball, which uses a ball attached to a pole by a cord.
 |
| This is a diagram of a conical pendulum whose bob travels in a horizontal circle of radius (r). The bob has mass m and is suspended by a string of length (L). The tension force of the string acting on the bob is the vector (T), and the bob's weight is the vector (m*g). |
There are 2 forces acting on the bob:
— The tension (T) in the string, exerted along the line of the string and and acting towards the point of suspension.
— The downward bob weight (m*g), where m is the mass of the bob and g is the local gravitational acceleration.
To evaluate the force exerted by the string, it has to be resolved into a horizontal component and a vertical component. The horizontal component is T*sin(θ) toward the centre of the circle, while the vertical component is T*cos(θ) in the upward direction. From Newton’s 2nd law, the horizontal component of the tension in the string gives the bob a centripetal acceleration toward the centre of the circle:
T*sin(θ) = m*v2/r
Since there is no acceleration in the vertical direction, the vertical component of the tension in the string is equal and opposite to the weight of the bob:
T*cos(θ) = m*g
Evaluating T/m can solve and equate the above 2 equations, thereby eliminating T and m:
g/cos(θ) = v2/(r*sin(θ))
Since the speed of the pendulum bob is constant, it can be expressed as the circumference 2*π*r divided by the time (t) required for 1 revolution of the bob:
v = 2*π*r/t
If we substitute the right side of this equation for v in the previous equation, we obtain:
g/cos(θ) = [2*π*r/t]2/[r*sin(θ)] = [r*(2*π)2]/[t2 * sin(θ)]
If we use the trigonometric identity tan(θ) = sin(θ) / cos(θ) and solve for t, the time required for the bob to travel one revolution is:
t = 2*π*[r/(g*tan(θ))]2
In a practical experiment, r varies and is as difficult to measure as the constant string length (L). If r, h and L form a right-angle triangle and θ being the angle between the length (h) and the hypotenuse (L), then r can be eliminated. This means:
r = L*sin(θ)
If we substitute this value for r, this yields a formula whose only varying parameter is the suspension angle (θ),
t = 2*π*[(L*cos(θ))/g]0.5
For small angles θ, cos(θ) ≈ 1:
t ~ 2*π*(L/g)0.5
— This means the period t of a conical pendulum is equal to the period of an ordinary pendulum of the same length. Furthermore, the period for small angles is approximately independent of changes in the angle θ. Hence the period of rotation is approximately independent of the force applied to keep it rotating. This property is shared with ordinary pendulums and makes both types of pendulums useful for timekeeping, known as isochronism.
https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Book%3A_Classical_Mechanics_(Tatum)/19%3A_The_Cycloid/19.09%3A_The_Cycloidal_Pendulum
https://en.wikipedia.org/wiki/Cycloid#Cycloidal_pendulum
 |
| Schematic of a cycloidal pendulum |
If a pendulum is suspended from the cusp of an inverted cycloid, and the pendulum's length (L) is equal to that of half the arc length of the cycloid (i.e. L = 4*r), then the bob of the pendulum also traces a cycloid path. Such a cycloidal pendulum is isochronous, regardless of amplitude. If we introduce a coordinate system centred in the position of the cusp, the equation of motion would be:
x = r*[2θ(t) + sin(2θ(t))]
y = r*[-3 - cos(2θ(t))]
— θ = Angle of the straight part of the string with respect to the vertical axis. This is expressed as:
sin(θ(t)) = A*cos(ω*t), ω2 = g/L = g/4r
— A < 1 : Amplitude
— ω = Radian frequency of the pendulum
— g = Gravitational acceleration
The 17th-century Dutch mathematician Christiaan Huygens discovered and proved these properties of the cycloid during his search for more accurate pendulum clock designs for navigation purposes.
6. Double pendulum
https://en.wikipedia.org/wiki/Double_pendulum
Motion of single double pendulum:
https://www.youtube.com/watch?v=U39RMUzCjiU
Animation of 2 double pendulums under near identical initial conditions:
https://www.youtube.com/watch?v=_i3WqnejOQk
This setup involves a pendulum attached to the end of another pendulum. This physical system exhibits rich dynamic behaviour with a strong sensitivity to initial conditions. The motion of a double pendulum is governed by a set of coupled ordinary differential equations, leading to chaotic motion.
Variants of the double pendulum include 2 limbs being equal or unequal lengths and masses, which may be simple pendulums or compound pendulums. The motion it produces may in 3 dimensions or restricted to the vertical plane. The following analysis interprets the limbs as identical compound pendulums of length (l) and mass (m), and their motion is restricted to 2 dimensions.
The mass is usually distributed along the length of a compound pendulum. When the mass is evenly distributed, the centre of mass of each limb is at its midpoint, and the limb has a moment of inertia of I = (m*L2)/12 about that point.
The angles between each limb and the vertical is used as the generalised coordinates to define the configuration of the system, denoted as θ1 & θ2. The position of the centre of mass of each rod is written in terms of these 2 coordinates. When the origin of the Cartesian coordinate system is interpreted at the point of suspension of the 1st pendulum, then the centre of mass of this pendulum is at:
x1 = 0.5*l*sin(θ1)
y1 = -0.5*l*sin(θ1)
and the centre of mass of the 2nd pendulum is at:
x2 = l*[sin(θ1) + 0.5*sin(θ2)]
y2 = -l*[cos(θ1) + 0.5*cos(θ2)]
The Lagrangian is:
L = Ek (Kinetic energy) - Ep (Potential energy)
= 0.5*m*(v12 + v22) + 0.5*I*(θ*12+ θ*22) - m*g*(y1 + y2)
= 0.5*m*(x*12 + y*12 + x*22+ y*22 + 0.5*I*(θ*12+ θ*22) - m*g*(y1 + y2)
— The 1st term is the linear kinetic energy of the centre of mass of the bodies.
— The 2nd term is the rotational kinetic energy around the centre of mass of each rod.
— The 3rd term is the potential energy of the bodes in a uniform gravitational field.
— The dot-notation indicates the time derivative of the variable in question.
If we substitute the coordinates above and rearrange the equation, it yields:
L = (m*l2/6)*[θ*22 + 4*θ*12 + 3*(θ*1)(θ*2)*cos(θ1 - θ2)] + 0.5*m*g*l*[3*cos(θ1) + cos(θ2)]
sin(θ(t)) = A*cos(ω*t), ω2 = g/L = g/4r
— A < 1 : Amplitude
— ω = Radian frequency of the pendulum
— g = Gravitational acceleration
The 17th-century Dutch mathematician Christiaan Huygens discovered and proved these properties of the cycloid during his search for more accurate pendulum clock designs for navigation purposes.
 |
5 isochronous cycloidal pendula with different amplitudes. |
https://en.wikipedia.org/wiki/Double_pendulum
Motion of single double pendulum:
https://www.youtube.com/watch?v=U39RMUzCjiU
Animation of 2 double pendulums under near identical initial conditions:
https://www.youtube.com/watch?v=_i3WqnejOQk
This setup involves a pendulum attached to the end of another pendulum. This physical system exhibits rich dynamic behaviour with a strong sensitivity to initial conditions. The motion of a double pendulum is governed by a set of coupled ordinary differential equations, leading to chaotic motion.
 |
This diagram illustrates a double pendulum consists of 2 pendulums attached end to end. |
The mass is usually distributed along the length of a compound pendulum. When the mass is evenly distributed, the centre of mass of each limb is at its midpoint, and the limb has a moment of inertia of I = (m*L2)/12 about that point.
The angles between each limb and the vertical is used as the generalised coordinates to define the configuration of the system, denoted as θ1 & θ2. The position of the centre of mass of each rod is written in terms of these 2 coordinates. When the origin of the Cartesian coordinate system is interpreted at the point of suspension of the 1st pendulum, then the centre of mass of this pendulum is at:
x1 = 0.5*l*sin(θ1)
y1 = -0.5*l*sin(θ1)
and the centre of mass of the 2nd pendulum is at:
x2 = l*[sin(θ1) + 0.5*sin(θ2)]
y2 = -l*[cos(θ1) + 0.5*cos(θ2)]
Top: Diagram of a double compound pendulum.
Bottom: Motion of the double compound pendulum (from numerical integration of the equations of motion).
L = Ek (Kinetic energy) - Ep (Potential energy)
= 0.5*m*(v12 + v22) + 0.5*I*(θ*12+ θ*22) - m*g*(y1 + y2)
= 0.5*m*(x*12 + y*12 + x*22+ y*22 + 0.5*I*(θ*12+ θ*22) - m*g*(y1 + y2)
— The 1st term is the linear kinetic energy of the centre of mass of the bodies.
— The 2nd term is the rotational kinetic energy around the centre of mass of each rod.
— The 3rd term is the potential energy of the bodes in a uniform gravitational field.
— The dot-notation indicates the time derivative of the variable in question.
If we substitute the coordinates above and rearrange the equation, it yields:
L = (m*l2/6)*[θ*22 + 4*θ*12 + 3*(θ*1)(θ*2)*cos(θ1 - θ2)] + 0.5*m*g*l*[3*cos(θ1) + cos(θ2)]
There is only 1 conserved quantity (the energy), and no conserved momenta. Therefore, the 2 generalised momenta can be expressed as:
These last 4 equations are explicit formulas for the time evolution of the system given its current state. Therefore, it isn’t possible to advance and integrate these equations analytically in order to yield formulas for θ1 and θ2 as functions of time. However the Runge Kutta method of similar techniques may make it possible to perform this integration numerically.
Animations show the double pendulum undergoes chaotic motion, demonstrating its sensitivity to initial conditions.
 |
This image shows the graph of the time for the pendulum to flip over as a function of initial conditions. It illustrates the amount of elapsed time before the pendulum flips over, as a function of initial position when released at rest. The initial value of θ1 ranges along the x-direction from -3 to 3. The initial value of θ2 ranges along the y-direction, from -3 to 3. The colour of each pixel indicates whether either pendulum flips within: — Green = 10*(l/g)0.5 — Red = 100*(l/g)0.5 — Purple = 1000*(l/g)0.5 — Blue = 10000*(l/g)0.5 — Initial conditions that don’t lead to a flip within 10000*(l/g)0.5 are depicted as white. |
The boundary of the central white region is defined in part by energy conservation with the following curve:
3*cos(θ1) + cos(θ2) = 2
Within the region defined by this curve, it is defined by:
3*cos(θ1) + cos(θ2) > 2
This means it is energetically impossible for either pendulum to flip, whereas the pendulum can flip outside this region. However, it is difficult to determine when it will flip. Alex Small (2013) observed similar behaviour for a double pendulum composed of 2 point masses.
The lack of a natural excitation frequency meant the double pendulum systems in seismic resistance designs can be applied to buildings. The building itself is the primary inverted pendulum, and a secondary mass is connected to complete the double pendulum.
7. Double inverted pendulum
https://en.wikipedia.org/wiki/Double_inverted_pendulum
This device is a combination of the inverted pendulum and the double pendulum. The double inverted pendulum is unstable, which leads to it falling down unless it is controlled in some way. 2 main methods of controlling a double inverted pendulum include movement of the base, or application of a torque at the pivot point between the 2 pendulums.
8. Doubochinski’s pendulum
https://en.wikipedia.org/wiki/Doubochinski%27s_pendulum
This is a classical oscillator that interacts with a high-frequency field, which allows it to develop a discrete set of stable regimes of oscillation. Each regime oscillates at a frequency near to the proper frequency of the oscillator, but each with a distinct, "quantised" amplitude. he phenomenon of amplitude quantisation in this coupled system was first discovered by the Doubochinski brothers, Danil and Yakov, in 1968–69
This figure is a schematic of the Doubochinski’s pendulum. The system is composed of 2 interacting oscillatory processes:
— A pendulum arm with a natural frequency on the order of 0.5–1 Hz.
— A stationary electromagnet (solenoid) positioned under the equilibrium point of the pendulum's trajectory and supplied with alternating current of fixed frequency, typically in the range of 10–1000 Hz.
The mechanical pendulum arm and solenoid are configured to allow the pendulum arm to interact with the oscillating magnetic field of the solenoid only over a limited portion of its trajectory. This portion is called the “zone of interaction”, which is located outside of which the strength of the magnetic field drops off rapidly to zero.This spatial inhomogeneity of the interaction is key to the quantised behaviour and other unusual properties of the system.
The table below shows the values of the quantised amplitudes, and the corresponding energies of the quantised modes. They are essentially independent of the strength of the alternating current supplied to the electromagnet, over a long range. This means the applied frequency is directly proportional to the array of stable modes that become accessible to the pendulum.
https://en.wikipedia.org/wiki/Elastic_pendulum
Also known as a spring pendulum, this physical system consists of a mass connected to a spring o produce a motion that pertains both a simple pendulum and a one-dimensional spring-mass system.
When the spring compresses, the shorter radius causes the spring to move faster due to the conservation of angular momentum. The figure shows 2 DOF elastic pendulum with polar coordinate plots.
The Lagrangian (L) is:
L = T - V
— T = Kinetic energy
— V = Potential energy
Hooke’s law is the potential energy of the spring itself:
Vk = 0.5*k*x2
— k = Spring constant
On the other hand, the height of the mass determines the potential energy from gravity. For a given angle and displacement, the potential energy is calculated as:
Vg = -g*m*(l0 + x)*cos(θ)
— g = Gravitational acceleration
The kinetic energy is calculated by:
T = 0.5*m*v2
— v = Velocity of the mass
Relating v to the other variables requires writing v as a combination of a movement along and perpendicular to the spring:
T = 0.5*m*(x*2+ (l0 + x)2 * θ*2)
So the Lagrangian becomes:
L = T - Vk - Vg
L[x,x*,θ,θ*] = 0.5*m*(x*2 + (l0 + x)2 * θ*2) - 0.5*k*x2 + g*m*(l0 + x)*cos(θ)
With 2 degrees of freedom, for x and θ, the equations of motion can be found using 2 Euler-Lagrange equations:
For x:
m*(l0 + x)*θ*2 - k*x + g*m*cos(θ) - m*x** = 0
Isolating x** leads to:
x** = (l0 + x)*θ*2 - k*x/m + g*cos(θ)
And for θ:
-g*m*(l0 + x)*sin(θ) - m*(l0 + x)2 *(θ**) - 2*m*(l0 + x)*(x*)*(θ*) = 0
Isolating θ** gives:
θ** = -[g/(l0 + x)]*sin(θ) - [(2*x*)/(l0 + x)]*(θ*)
The motion of the elastic pendulum is depicted by 2 coupled ordinary different equations, which can be solved numerically. Furthermore, analytical methods can be used to study the intriguing phenomenon of order-chaos-order in this system.
10. Foucault pendulum
https://en.wikipedia.org/wiki/Foucault_pendulum
https://en.wikipedia.org/wiki/List_of_Foucault_pendulums
Invented by French physicist Léon Foucault in 1851, this device was conceived as an experiment to demonstrate the Earth's rotation.
Foucault Pendulum at the Chicago Museum of Science and Industry:
https://www.youtube.com/watch?v=iqpV1236_Q0
Foucault Pendulum at the Panthéon, Paris:
https://www.youtube.com/watch?v=59phxpjaefA
The first public exhibition of a Foucault pendulum took place in February 1851 in the Meridian of the Paris Observatory. A few weeks later, Foucault suspended a 28-kilogram (62 lb) brass-coated lead bob with a 67-metre long (220 ft) wire from the dome of the Panthéon, Paris. The period of the pendulum was 2*π*(l/g)0.5 = 16.5 seconds. Because the latitude of its location was φ = 48°52’, the plane of the pendulum's swing traverses a full circle in approximately 24h 56’/ sin(φ) = 31.8 hours (31 hours 50 mins), rotating clockwise approximately 11.3° per hour.
The original bob used in 1851 at the Panthéon was moved in 1855 to the Conservatoire des Arts et Métiers in Paris. In 1902, a 2nd temporary bob was installed at the same Paris conservatory for the 50th anniversary.
During the museum reconstruction in the 1990s, the original pendulum was temporarily displayed at the Panthéon (1995), but was later returned to the Musée des Arts et Métiers before its reopening in 2000. On April 6, 2010, the cable suspending the bob in the Musée des Arts et Métiers fractured, causing irreparable damage to the pendulum bob and to the marble flooring of the museum. Since this incident, the original, yet-to-be-repaired pendulum bob is now displayed in a separate case adjacent to the current pendulum display. Since 1995, an exact copy of the original pendulum has been operating under the dome of the Panthéon, Paris since 1995.
Discuss the mechanics of the Foucault Pendulum
At either geographic pole, the plane of oscillation of a pendulum remains fixed relative to the distance masses of the universe while Earth rotates underneath it, which takes 1 sidereal day to complete a full rotation. Therefore, the plane of oscillation of a pendulum at the North Pole relative to earth rotates clockwise during a full day and vice versa at the South Pole.
If we suspend a Foucault pendulum at the equator, the plane of oscillation remains fixed relative to Earth. At other latitudes, the plane of oscillation precesses relative to Earth, but more slowly than at the pole. Therefore the angular speed (ω, clockwise degrees per sidereal day) is proportional to the sine of the latitude (Ψ):
ω = 360° *sin(Ψ) / day
— Latitudes north and south of the equator are defined as positive and negative, respectively.
e.g. A Foucault pendulum at 30° south latitude, viewed from above by an earthbound observer, rotates counterclockwise 360° in 2 days.
This diagram shows a Foucault pendulum at the North Pole, where it swings in the same plane as the Earth rotates beneath it.
Nobel laureate Heike Kamerlingh Onnes developed a fuller theory of Foucault pendulum for his doctoral thesis (1879). It concluded geometrical imperfection of the system or elasticity of the support wire would cause interference between two horizontal modes of oscillation. This may explain Onnes' pendulum going over from linear to elliptic oscillation in an hour.
Since air resistance damps the oscillation, some Foucault pendulums in museums incorporate an electromagnetic or other drive to keep the bob swinging. A 2009 report highlighted another main engineering problem is the preferred direction of swing, which can be alleviated with a 1-meter Foucault pendulum.
11. Furuta pendulum
https://en.wikipedia.org/wiki/Furuta_pendulum
https://www.hindawi.com/journals/jcse/2011/528341/
https://www.hindawi.com/journals/mpe/2010/742894/
Also known as the rotational inverted pendulum, the Furuta pendulum, consists of a driven arm rotating around a horizontal plane and a pendulum attached to that arm rotating in the vertical plane. It was invented in 1992 at Tokyo Institute of Technology by Katsuhisa Furuta and co. It is an example of a complex non-linear oscillator of interest in control system theory.
This is a schematic of the single rotary inverted pendulum system.
12. Gridiron pendulum
https://en.wikipedia.org/wiki/Gridiron_pendulum
- The initial tensor only has a single non-zero element (or none), and the remaining 2 diagonal terms are zero.
- It is possible to find a pendulum system where the moment of inertia in 1 of the 3 principal axes is approximately zero, but not 2.
- Studies describe slender symmetric pendulums that feature moments of inertia for 2 of the principal axes being equal and the remaining of moment of inertia is zero.
- The University of Adelaide published the equations of motion on the Furuta pendulum dynamics:
Imagine a rotational inverted pendulum is mounted to a DC motor illustrated in the figure.
- The DC motor applies a torque τ1 to Arm 1.
- The link between Arm 1 and Arm 2 is not actuated but free to rotate.
- The 2 arms have lengths L1 and L2 , with masses m1 and m2 located at l1 and l2 respectively, which are the lengths from the point of rotation of the arm to its centre of mass.
- The arms have inertia tensors J1 and J2 (about the centre of mass of the arms respectively).
- Each rotational joint is viscously damped with damping coefficients b1 and b2.
- b1 is the damping provided by the motor bearings, and b2 is the damping arising from the pin coupling between Arm 1 and Arm 2.
- A right hand coordinate system is used to define the input states and the Cartesian coordinate systems 1 and 2.
- The coordinate axes of Arm 1 and Arm 2 are the principal axes such that the inertia tensors are diagonal.
- The angular rotation of Arm 1 (θ1) is measured in the horizontal plane where a counter-clockwise direction (when viewed from above) is positive.
- The angular rotation of Arm 2 (θ2) is measured in the vertical plane where a counter-clockwise direction (when viewed from front) is positive.
- When the Arm is hanging down in the stable equilibrium position θ2 = 0.
- The torque the servo-motor applies to Arm 1 (τ1) is positive in the counter-clockwise direction (when viewed from above).
- A disturbance torque (τ2) is experienced by Arm 2, where a counter-clockwise direction (when viewed from the front) is positive.
Before we derive the dynamics of the system, we need to make a number of assumptions. They include:
- The motor shaft and Arm 1 are rigidly coupled and infinitely stiff.
- Arm 2 is infinitely stiff.
- The coordinate axes of Arm 1 and Arm 2 are the principal axes such that the inertia tensors are diagonal.
- The motor rotor inertia is negligible. However, this term can be added to the moment of inertia of Arm 1.
- Only viscous damping is considered, while all other forms of damping (such as Coulomb) are ignored. However, they can be added to the final governing DE.
The non-linear equations of motion are:
Most Furuta pendulums tend to feature long slender arms, such that the moment of inertia along the axis is negligible. Furthermore, most arms have rotational symmetry such that the moments of inertia in 2 of the principal axes are equal. Therefore, the inertia tensors may be approximated to:
We can further simplify it by making a number of substitutions.
The total moment of inertia of Arm 1 about the pivot point (using the parallel axis theorem) is:
The total moment of inertia of Arm 2 about its pivot point is:
Finally, we define the total moment of inertia the motor rotor experiences when the pendulum (Arm 2) is in its equilibrium position (hanging vertically down) as:
If we substitute the previous definitions into the governing DEs, this yields the following:
12. Gridiron pendulum
https://en.wikipedia.org/wiki/Gridiron_pendulum
A. Exterior schematicB. Normal temperatureC. Higher temperature
- Invented by British clockmaker John Harrison around 1726 and later modified by John Ellicott, this pendulum was a clock pendulum that compensated for temperature.
- The period of the pendulum's swing depends on its length, thus a pendulum clock's rate varied with changes in ambient temperature, leading to inaccurate timekeeping.
- This pendulum is composed of alternating parallel rods of 2 metals with different thermal expansion coefficients, such as steel and brass.
- The rods are connected by a frame in such a way that their different thermal expansions (or contractions) compensate for each other.
- Therefore, the overall length of the pendulum, and thus its period, remains constant with temperature.
- It was first used during the Industrial Revolution period in regulator clocks, precision clocks as they were used as time standards in workplaces to schedule work timetables and set other clocks.
- In the 19th century, scientists declared the gridiron pendulum unsuitable for the highest precision clocks.
- The friction of the rods sliding in the holes in the frame forced the rods to adjust to temperature changes in a series of small jolts, rather than smooth motion.
13. Grotta Giganta horizontal pendulums
https://www.researchgate.net/publication/236655991_The_Grotta_Gigante_horizontal_pendulums_-_Instrumentation_and_observations
https://en.wikipedia.org/wiki/Grotta_Gigante_horizontal_pendulums
This is a schematic drawing of the horizontal pendulum of grotta gigante. The rod with the mass rotates in the horizontal plane about the virtual rotation axis. The angle φ of the virtual rotation axis with the plumb line is essential for the amplification factor of the pendulum.
- They are a pair of tiltmeters mounted in the Grotta Gigante in Italy that help monitor Earth movements. Installed by the geodesist Antonio Marussi in 1959, this pendulum is sensitive to vertical deviations, and rotations and shearing motions of the cave.
- The pendulum beam is suspended horizontally by 2 steel wires, with an upper wire fixed to the cave roof and the lower wire fixed to the floor.
- The pendulum beam rotates in the horizontal plane around a virtual near-to-vertical axis that passes through the upper and lower mounting points.
14. Inertia wheel pendulum
This pendulum consists of an inertia wheel that is used as a pedagogical problem in control theory, which deals with the control of dynamical systems in engineered processes and machines.
Equations of motion
Euler – Lagrange formalism was used to derive the equations of motion of the inertia wheel pendulum.
-- q = [q1 q2]T = Generalised coordinates of the system
-- L = T - V = Lagrangian
-- T = Kinetic energy
-- V = Potential energy
-- τ = τ1 - τ2 = Vector of generalised forces acting on the system
The potential energy of the system is:
-- m1 and I1 = Mass and the moment of inertia of the pendulum
-- m2 and I2 = Mass and the moment of inertia of the disk
-- l = Distance from the pivot point to the disk shaft
-- c = Distance from the pivot point to the centre of mass of the arm
-- g = Acceleration due to gravity
The Lagrangian is:
A classical model of the torque developed by the DC motor is:
-- kT = Torque constant
-- R = Rotor coil resistance
-- Vi = Control voltage
-- kF = Back-emf constant
-- ωm = Angular velocity of the motor shaft
Since the electric time constant of the motor is significantly smaller than the mechanical time constant, it is negligible.
The angular velocity of the disk is:
-- N = The ratio between the diameter of the inertia disk and the diameter of the motor pulley, known as the reduction ratio.
Thus, the torque applied to the disk is:
If we combine derivative and the Lagrangian, this yields:
-- τf1 & τf2 = Friction torques at the joints of the pendulum and the disk, respectively.
15. Inverted pendulum
https://en.wikipedia.org/wiki/Inverted_pendulum
Lecture about inverted pendulum:
Video demonstration:
- This pendulum involves its centre of mass above its pivot point, which is unstable and can fall over without any support. Stabilising the pendulum in this inverted position requires a feedback control system that continuously monitors the pendulum's angle. Any detection of displacement from the stable position would trigger movement of the pivot point's position sideways in order to restore balance.
- In dynamics and control theory, this pendulum is widely used as a benchmark for testing control algorithms (PID controllers, state space representation, neural networks, fuzzy control, genetic algorithms, etc.).
- Example applications of the inverted pendulum include rocket or missile guidance, self-balancing personal transporters such as the Segway PT, the self-balancing hoverboard and the self-balancing unicycle, as well as Kapitza's pendulum.
Describe the equations of motion
i. Stationary pivot point
When the pivot point of the pendulum is fixed in space, the equation of motion is similar to that for an uninverted pendulum. The equation of motion below assumes there is no friction or any other resistance to movement, a rigid massless rod, and the restriction to 2-dimensional movement.
- θ** = Αngular acceleration of the pendulum
- g = Standard gravity on Earth's surface
- l = Pendulum length
- θ = Angular displacement measured from the equilibrium position
When added to both sides, it will have the same sign as the angular acceleration term:
The pendulum is assumed to consist of a point mass (m) attached to the end of a massless rigid rod of length (l) attached to a pivot point at the end opposite the point mass.
The net torque of the system equals the moment of inertia times the angular acceleration:
The torque due to gravity providing the net torque is given as:
τnet = m*g*l*sin(θ)
- θ = Angle measured from the inverted equilibrium position.
The resulting equation:
The moment of inertia for a point mass:
I = m*R2
When the radius is the length of the rod (l) in the inverted pendulum, we can substitute I = m*l2.
If we divide mass and l2 from each side, this yields:
ii. Essentials of stabilisation
- If the tilt angle (θ) is to the right, the cart must accelerate to the right and vice versa.
- The position of the cart (x) relative to the centre of the track is stabilised by slightly modulating the null angle (the angle error that the control system tries to null) by the position of the cart. Note that the null angle = θ + k*x, where k is a small value. This causes the pole to lean slightly toward the centre of the track and stabilise there where the tilt angle is exactly vertical. Any adjustment by the tilt sensor or track slope that would otherwise destabilise the pendulum would lead to a stable position offset.
- A normal pendulum influenced by a moving pivot point such as a load lifted by a crane, has a peaked response at the pendulum radian frequency of ωp = √(g/l). The frequency spectrum of the pivot motion would need to be limited near ωp in order to prevent uncontrolled swinging.
Due to the effects of the null angle modulation strategy, the position feedback is positive. This means a sudden command to move right generates an initial cart motion to the left followed by a movement to the right in order to rebalance the pendulum.
iii. From Lagrangian's equations
Refer to the illustration to the right where θ(t) is the angle of the pendulum of length l with respect to the vertical direction and the acting forces are gravity and an external force (F) in the x-direction. Let's define x(t) to be the cart's position.
The kinetic T of the system is:
- v1 = Velocity of the cart
- v2 = Velocity of the point mass m.
- v1 & v2 can be expressed in terms of x and θ by writing the velocity as the first derivative of the position.
If we simplify the equation for v2:
The kinetic energy equation would be:
The generalised coordinates of the system are θ and x, each having a generalised force. On the x axis, the generalised force Qx can be calculated through its virtual work:
Qxδx = Fδx, Qx = F,
On the θ axis, the generalised force Qθ can be also calculated through its virtual work
Qθδθ = m*g*l*sin(θ)δθ, Qθ = m*g*l*sin(θ)
According to the Lagrange's equations, the equations of motion are:
If we substitute T in these equations and simplify these equations, this would yield equations that accurately describe the motion of the inverted pendulum.
Although these equations are non-linear, however, since the aim of a control system is maintaining the pendulum's uprightness, the equations can be linearised around θ ~ 0.
iv. From Euler-Lagrange equations
According to the D'Alembert's principle, there is a relationship between generalised forces and potential energy:
However, under certain circumstances, the potential energy is not accessible, meaning only generalised forces are available.
After we denote the Lagrangian L = T - V, we can also use Euler–Lagrange equation to solve for equations of motion:
v. From Newton's second law
Newton's 2nd law describes the reaction forces at the joint between the pendulum and the cart. This yields 2 equations for each body; 1 in the x-direction and 1 in the y-direction. The equations of motion of the cart are shown below; the LHS is the sum of the forces on the body and the RHS is the acceleration.
- FN = Normal force applied to the cart.
In order to complete the equations of motion, we need to compute the acceleration of the point mass attached to the pendulum. The position of the point mass can be defined in inertial coordinates:
If we take 2 derivatives, it yields the acceleration vector in the inertial reference frame.
Next, we use Newton's 2nd law to write the 2 equations in the x-direction and the y-direction. Note that the reaction forces are positive as applied to the pendulum and negative when applied to the cart, due to the application of Newton's 3rd law.
The first equation allows us to compute the horizontal reaction force in case the applied force (F) is unknown, whereas the second equation solves for the vertical reaction force.
If we substitute:
into:
This yields:
We need to dot the equation of motion with a unit vector that runs perpendicular to the pendulum at all times and is denoted by the x-coordinate of the body frame in order to obtain the second equation.
In inertial coordinates this vector can be written using a 2-D coordinate transformation:
The pendulum equation of motion written in vector form is:
If we dot x^B with both sides, this obtains the following on the LHS:
The above equation describes the relationship between both body frame components and inertial frame components of reaction forces. If we assume the bar connecting the point mass to the cart is massless, this suggests the bar cannot transfer any load perpendicular to the bar. Therefore, the inertial frame components of the reaction forces indicates the bar can only transfer loads along the axis of the bar itself. This yields another equation to solve for the tension in the rod itself:
The above equation describes the relationship between both body frame components and inertial frame components of reaction forces. If we assume the bar connecting the point mass to the cart is massless, this suggests the bar cannot transfer any load perpendicular to the bar. Therefore, the inertial frame components of the reaction forces indicates the bar can only transfer loads along the axis of the bar itself. This yields another equation to solve for the tension in the rod itself:
If we dot x^B with the acceleration of the pendulum on the RHS of the equation, then this yields:
If we combine a the LHS with the RHS and divide through by m, this yields:
Example real world applications of the inverted pendulum include:
- Human being
- Balancing brooms or meter sticks by hand
- Early seismometers
- Personal transporters, such as the two-wheeled self-balancing scooters and single-wheeled electric unicycles.
16. Invariable pendulums
An invariable pendulum measuring gravity in Madras, India, in 1821.
- In the early 19th century, Kater introduced the idea of relative gravity measurements to augment the absolute measurements made by Kater's pendulum. It involved comparing the level of gravity at 2 different points, which helped time the period of an ordinary (single pivot) pendulum at the first point, then transport the pendulum to the second point and time its period there.
- Since the pendulum's length remains constant, the ratio of the gravitational accelerations was equal to the inverse of the ratio of the periods squared.
- Kater constructed a set of "invariable" pendulums, with only one knife edge pivot, which were transported to a number of countries after it first swung at a central station at Kew Observatory, UK.
17. Kapitza’s pendulum
https://en.wikipedia.org/wiki/Kapitza%27s_pendulum
This is a drawing of a Kapitza pendulum. It consists of a motor that rotates a crank at a high speed, a crank that vibrates a lever arm up and down, which the pendulum is attached to with a pivot.
- Named from Russian Nobel laureate physicist Pyotr Kapitza, who published a theory that explains some of its unusual properties in 1951. It consists of a vibrating suspension balancing the pendulum in an inverted position, with the bob above the suspension point.
- This pendulum was first described by A. Stephenson in 1908, who discovered a faster driving frequency stabilises the upper vertical position of the pendulum.
- Kapitza's experimental studies on this pendulum provided an analytical insight into the reasons of stability by dissecting the motion into "fast" and "slow" variables and by introducing an effective potential. This gave birth to a new branch in physics called "vibrational mechanics".
The following notation will be used:
- v = Frequency of the vertical oscillations of the suspension
- a = Amplitude of the oscillations of the suspension
- ω0 = √(g/l) = Proper frequency of the mathematical pendulum
- g = Free fall acceleration
- l = Length of rigid and light pendulum
- m = Mass
- φ = Angle between pendulum and downward direction
Time dependence of the position of pendulum is denoted as:
Energy equations
The potential energy of the pendulum is due to gravity and is defined by of the vertical position:
Ep = -m*g*[l*cos(φ) + a*cos(v*t)]The kinetic energy of the pendulum describes the velocity of a mathematical pendulum, which also accounts for the vibrations of the suspension: 
The total energy is defined by the sum of the kinetic and potential energies E = Ek + Ep and the Lagrangian by their difference L = Ek - Ep.
Equations of motion
The motion of pendulum satisfies Euler–Lagrange equations, which is satisfied by the dependence of the phase φ of the pendulum on its position.
The Lagrangian L is denoted as:
up to irrelevant total time derivative terms. The differential equation becomes:
which describes the movement of the pendulum is nonlinear due to the sin(φ) factor.
Equilibrium positions
Time-averaging over the rapid v-oscillation yields to leading order:
The "slow" equation of motion becomes: 
The total energy is defined by the sum of the kinetic and potential energies E = Ek + Ep and the Lagrangian by their difference L = Ek - Ep.
- In the mathematical pendulum, the total energy is conserved, therefore the time (t) dependence of the potential and kinetic energies is symmetric with respect to the horizontal line.
- According to the virial theorem, the mean kinetic and potential energies in harmonic oscillator are equal. This suggests that the line of symmetry corresponds to half of the total energy.
- If the suspension is vibrating, the system is no longer a closed one and the total energy is no longer conserved. Note the kinetic energy is more sensitive to vibration compared to the potential one.
- The potential energy is bound from below and above -m*g*(l + a) < Ep < m*g*(l + a) while the kinetic energy is bound only from below Ek > 0. For high frequency of vibrations (v), the kinetic energy can be large compared to the potential energy.
Equations of motion
The motion of pendulum satisfies Euler–Lagrange equations, which is satisfied by the dependence of the phase φ of the pendulum on its position.
The Lagrangian L is denoted as:
up to irrelevant total time derivative terms. The differential equation becomes:
which describes the movement of the pendulum is nonlinear due to the sin(φ) factor.
Equilibrium positions
- The Kapitza model simplifies to the simple pendulum in the limit a = 0. In that limit, the tip of the pendulum describes a circle: x2+ y2 = l2 = constant
- If the energy in the initial moment is larger than the maximum of the potential energy (E > m*g*l) then the trajectory will be closed and cyclic.
- If the initial energy is smaller (E < m*g*l) then the pendulum will oscillate close to the only stable point (x,y) = (0, -l).
- When the suspension is vibrating with a small amplitude a << 1 and with a frequency v >> ω0 significantly higher than the proper frequency ω0, the angle φ is interpreted as a superposition φ = φ0 + ζ of a "slow" component (φ0) and a rapid oscillation (ζ) with small amplitude due to the small but rapid vibrations of the suspension.
- Technically, we perform a perturbative expansion in the "coupling constants" (a/l), (ω0/v) << 1 while interpreting the ratio (a/l)*(v/ω0) as fixed.
- The perturbative treatment becomes exact in the double scaling limit a --> 0, v --> ∞. More precisely, the rapid oscillation (ζ) is defined as: ζ = (a/l)*sin(φ0)*cos(v*t)
- The equation of motion for the "slow" component (φ0) becomes:
By introducing an effective potential:
This means the effective potential has two minima if (a*v)2 > 2*g*l, or equivalently, (a/l)*(v/ω0) > √2. The first minimum is in the same position (x,y) = (0, -l) as the mathematical pendulum and the other minimum is in the upper vertical position (x,y) = (0, l). Although the upper vertical position is unstable in a mathematical pendulum, it can be stable in Kapitza's pendulum.
Rotating solutions
When the pendulum rotates around the pivot point at the same frequency that the pivot point is driven, we can derive 2 rotating solutions in this pendulum (1 clockwise and 1 counterclockwise). We can shift to the rotating reference frame using φ --> φ' + v*t to yield equation for φ:
If we consider the limit in which v is significantly higher the proper frequency ω0, we can work out the rapid-v slow-φ0' limit leads to the equation:
There is a stable equilibrium at φ0' = 0 and an unstable equilibrium at φ0' = π.
18. Kater’s pendulum
https://en.wikipedia.org/wiki/Kater%27s_pendulum
- This reversible free swinging pendulum was invented by British physicist and army captain Henry Kater in 1817 for the purposes as a gravimeter instrument to measure the local acceleration of gravity.
- Kater's pendulum consists of a rigid metal bar with 2 pivot points, one at each end of the bar, that can swing from each either pivot it is mounted to. Moreover, it either has an adjustable weight that can be shifted vertically along the bar, or one adjustable pivot, to adjust the periods of swing.
- When the periods of the swings upon each pivot are equal, this means T is equal to the period of an 'ideal' simple pendulum of length equal to the distance between the pivots. This leads to precise calculations of the acceleration of gravity based on the equation T = 2π*(√(L/g)).
- T1 and T2 = Time periods of oscillations when it is suspended from K1 and K2 respectively
- l1 and l2 = Distances of knife edges K1 and K2 from the centre of gravity respectively
Drawing of Kater's pendulum:
(a) Opposing knife edge pivots from which pendulum is suspended
(b) Fine adjustment weight moved by adjusting screw
(c) Coarse adjustment weight clamped to rod by setscrew
(d) Bob
(e) Pointers for reading
Kater used this pendulum to accurately determine the period by using the clock pendulum by the method of coincidences, which involves timing the interval between the coincidences when the 2 pendulums were swinging in synchronism. Furthermore, he determined the mean length of the solar seconds pendulum at London, at sea level, at 17 °C (62 °F), swinging at vacuum, was 0.9941 metres (39.1386 inches). This is equivalent to a gravitational acceleration of 9.81158 m/s2. Since the biggest margin of error from the mean was 7.1 μm (0.00028 inches), it translated to a precision of gravity measurement of 0.7×10-5 (7 milligals). Therefore, in 1824, the British Parliament officially made Kater's measurement of the seconds pendulum a backup standard of length for defining the yard if the yard prototype was destroyed.
19. Metronome
https://en.wikipedia.org/wiki/Metronome
- This device produces an audible click or other sound at a regular interval that can be adjusted by the user typically in beats per minute (BPM). The term 'metronome' comes from ancient Greek μέτρον (métron) meaning "measure") and νέμω (némo) meaning "I manage", "I lead".
- The metronome may have been first invented by Andalusian polymath Abbas ibn Firnas (810-887).
- In 1815, Johann Maelzel patented his mechanical, wind-up metronome as a tool for musicians, under the title "Instrument/Machine for the Improvement of all Musical Performance, called Metronome". Around the same year, Ludwig van Beethoven was the first notable composer to indicate specific metronome markings in his musical pieces, which was found on the score of the Cantata op. 112.
What are the different types of metronomes?
- Mechanical = This consists of a sliding weight on the end of an inverted pendulum rod to control tempo. If the weight slides up, the pendulum rod's tempo slows down, and vice verse if the weight slides down. Hidden inside the metronome case is a second fixed weight on the other side of the pendulum pivot, making this device a double-weighted pendulum. The pendulum swings back and forth in tempo, while a mechanism inside the metronome produces a clicking sound with each oscillation. This apparatus doesn't require any electricity, since it operates from a spring-wound clockwork escapement.
- Electromechanical = Invented by Franz Frederick, this type of metronome consisted of an electric motor to generate power the mechanism.
- Electronic = A majority of modern metronomes are electronic, which contains a quartz crystal to maintain accuracy, comparable to those used in wristwatches. They usually contain a dial or buttons to control the tempo, as well as a note tuner around the range of A440 (440 hertz).
- Software = These operate as either as standalone applications on computers and smartphones, or in music sequencing and audio multitrack software packages. Recording studio applications, such as film scoring, involve a software metronome providing a click track to synchronise musicians. iPods and other portable MP3 players have prerecorded MP3 metronome click tracks, which play different sounds and samples instead of the usual metronome beep. The Google search engine includes an interactive metronome that can play between 40 and 218 BPM.
20. Paraconical pendulum
https://www.researchgate.net/figure/The-left-panel-is-adapted-from-1-on-paraconical-pendulum-experiment-and-the-right-is_fig2_310899774
- Invented around the 1950s by French scientist Maurice Allais, this device allowed the investigation of asymmetries of inertial space. The pendulum is carried upon a ball that rests upon a flat, which allows it to rotate about its vertical axis and swing in 2 perpendicular directions. This gives this pendulum 3 degrees of freedom.
- If you account for the Earth's rotation, it complicates the formal equations in classical mechanics of the behaviour of this dynamical system. Since the rolling friction of the ball upon the flat is minimal, this gives a paraconical pendulum high Q factor, which is determined by air resistance.
21. Persoz pendulum
https://en.wikipedia.org/wiki/Persoz_pendulum
- This device is used to measure the hardness of materials. It consists of a pendulum that freely swings on 2 balls resting on a coated test panel. This is based on the principle that the amplitude of the pendulum's oscillation decreases more significantly when supported on a softer surface.
- The hardness of any coating is determined by the number of oscillations performed by the pendulum within the specified limits of amplitude determined by accurately positioned photo sensors.
22. Quantum pendulum
https://en.wikipedia.org/wiki/Quantum_pendulum
https://phys.org/news/2016-05-quantum-swinga-pendulum.html
- In the quantum world of atoms and molecules, oscillators behave differently to a classical pendulum. A time-independent wave function describes the position of an atom in a single quantum state, which suggests there are no oscillations.
- Oscillations in the quantum world need a superposition of different quantum states, sometimes known as coherence or wave-packet. When 2 quantum states superpose, it leads to an atomic motion close to the classical pendulum, or a one-phonon coherence.
- A two-phonon coherence involves an atom being at 2 different positions simultaneously. Its velocity is non-classical, which indicates the atom moves simultaneously towards both the right and the left.
- These motions exist briefly since the well-defined superposition of quantum states decays by so-called decoherence within a few picoseconds.
This is a screenshot of the non-classical quantum coherences in matter.
- The two parabolas (black curves) show the potential energy surfaces of harmonic oscillators representing the oscillations of atoms in a crystalline solid around their equilibrium positions, i.e., the so called phonons.
- The blue curves show the probability of presence of atoms at different spatial positions in thermal equilibrium. The quantum mechanical uncertainty principle requires a finite width of such distribution functions.
- The red curves show time-dependent probability distributions of coherent oscillating states in matter.
- On the left panel, it illustrates the one-phonon coherence, which is defined as the quantum mechanical motion of atoms resembles the classical motion of a pendulum (cyan ball). The latter moves during the oscillation either from left to right or vice versa.
- On the right panel, it illustrates the two-phonon coherence, which illustrates quantum mechanics initiating a nonclassical state with the quantum-mechanical property that the atom can be at two positions simultaneously.
- The velocity of the atoms also moves in a non-classical manner, i.e., the atom moves at the same time both to the right and to the left. In the case of a perfect harmonic oscillator, the currents of the two parts of the atom cancel each other perfectly. Therefore, a small anharmonicity is required to observe the emission of a coherent electric field transient.
Describe the Schrödinger equation
Using Lagrangian mechanics from classical mechanics, we can develop a Hamiltonian for the system. A simple pendulum has 1 generalised coordinate (the angular displacement, φ) and 2 constraints (the length of the string and the plane of motion). The kinetic and potential energies of the system can be computed as:
This results in the Hamiltonian
The time-dependent Schrödinger equation for the system is:
To solve the time-independent Schrödinger equation to find the energy levels and corresponding eigenstates, we need to change the independent variable as follows:
This is simply Mathieu's differential equation:
The solutions to this equation are Mathieu equations.
What are the solutions?
Provided that q are countably many special values of a, known as characteristic values, the Mathieu equation accepts solutions that are periodic with period 2π. The characteristic values of the Mathieu cosine, sine functions respectively are written an(q), bn(q), where n is a natural number. The periodic special cases of the Mathieu cosine and sine functions are often written CE(n,q,x), SE(n,q,x) respectively:
The boundary conditions in the quantum pendulum suggest that an(q), bn(q) are as follows for a given q:
The energies of the system (E) for even/odd solutions respectively, are quantised based on the characteristic values found by solving the Mathieu equation.
The effective potential depth is defined as:
The general solution of the above differential equation for a given value of a and q is a set of linearly independent Mathieu cosines and Mathieu sines, which are even and odd solutions respectively. In general, the Mathieu functions are aperiodic. However, for characteristic values of an(q), bn(q), the Mathieu cosine and sine become periodic with a period of 2π.
For positive values of q, the following is true:
Below are first few periodic Mathieu cosine functions for q = 1:
Note that the green line CE(1,1,x) resembles a cosine function, but with flatter hills and shallower valleys.
23. Rayleigh-Lorentz pendulum
https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Lorentz_pendulum
https://www.youtube.com/watch?v=VuX_UExHa0M&ab_channel=LecturesbyWalterLewin.Theywillmakeyou%E2%99%A5Physics.
https://www.youtube.com/watch?v=mvCmeUwAHpM&ab_channel=28Annie28
If the frequency is constant, the solution is x = A*cos (ω*t + φ). However, if the frequency changes slowly with time ω = ω(t), or precisely, if the characteristic time scale for the frequency variation is much smaller than the time period of oscillation, i.e.
https://www.youtube.com/watch?v=mvCmeUwAHpM&ab_channel=28Annie28
- Named after Lord Rayleigh and Hendrik Lorentz, this pendulum has a frequency that gradually varies due to an external action. It helped form the basis for the concept of adiabatic invariants in mechanics.
- This pendulum problem was formulated in 1902 by Lord Rayleigh, with arguments for Léon Lecornu's contribution back in 1895.
- At the first Solvay conference in 1991, oblivious to Rayleigh's work at the time, Hendrik Lorentz proposed a question to clarify the quantum theory, How does a simple pendulum behave when the length of the suspending thread is gradually shortened?
- The next day, Albert Einstein responded by stating that both energy and frequency of the quantum pendulum changes such that their ratio is constant, therefore the pendulum is in the same quantum state as the initial state. These works formed the basis for the concept of adiabatic invariant, which are applied in various fields and old quantum theory.
The equation of the simple harmonic motion with frequency ω for the displacement x(t) is:
d2x/dt2 + x*ω2 = 0
If the frequency is constant, the solution is x = A*cos (ω*t + φ). However, if the frequency changes slowly with time ω = ω(t), or precisely, if the characteristic time scale for the frequency variation is much smaller than the time period of oscillation, i.e.
|1/ω dω/dt| << ω
Then it can be shown that,
Then it can be shown that,
-- E = Average energy averaged over an oscillation
Since the frequency varies with time due to external action, conservation of energy no longer holds and the energy over a single oscillation is not constant. During an oscillation, the frequency varies, as does its energy. Hence, we can define average energy per unit mass for a given potential V(x;ω) as:
-- The closed integral indicates it is taken over a complete oscillation.
24. Repsoid-Bessel pendulum
25. Seconds pendulum
https://en.wikipedia.org/wiki/Seconds_pendulum
This type of pendulum has a period of exactly 2 seconds; it takes 1 second to swing and 1 second to return swing, with a frequency of 0.5 Hz. It was first used in the pendulum clock invented by Dutch scientist and inventor Christiaan Huygens in 1656.
This is a schematic of the 2nd pendulum clock built around 1673 by Christiaan Huygens, inventor of the pendulum clock. It is from his treatise Horologium Oscillatorium, published 1673, Paris, in which it records improvements to the mechanism that Huygens had illustrated in the 1658 publication of his invention, titled Horologium. It is a weight-driven clock with a verge escapement (K,L), with the 1 second pendulum (X) suspended on a cord (V). The large metal plate (T) in front of the pendulum cord is the first illustration of Huygens' 'cycloidal cheeks', which aims to improve accuracy by forcing the pendulum to follow a cycloidal path and produce isochronous swing. Huygens claimed it achieved an accuracy of 10 seconds per day.
26. Spherical pendulum
https://en.wikipedia.org/wiki/Spherical_pendulum
-- The closed integral indicates it is taken over a complete oscillation.
For a simple harmonic oscillator, it simplifies to:
- In 1835, Friedrich Bessel demonstrated that there was no need to repeatedly swing the Kater's pendulum and adjust the weights until the periods were equal. Poynting & Thompson (1907) stated the the gravity could be calculated from the 2 periods and the pendulum's centre of gravity.
- Bessel demonstrated the errors due to air drag would cancel out if the pendulum was symmetrical in form about its centre, as long as it was weighted internally at one end.
- Although Bessel wasn't involved in the construction of the Repsold-Bessel pendulum, Adolf Repsold invented one under contract by the Swiss Geodetic Commission.
25. Seconds pendulum
https://en.wikipedia.org/wiki/Seconds_pendulum
This type of pendulum has a period of exactly 2 seconds; it takes 1 second to swing and 1 second to return swing, with a frequency of 0.5 Hz. It was first used in the pendulum clock invented by Dutch scientist and inventor Christiaan Huygens in 1656.
- In 1644, Marin Mersenne determined the length of a seconds pendulum in toises.
- In 1660, the Royal Society proposed that it be the standard unit of length.
- In 1671, Jean Picard measured this length at the Paris observatory and determined it to be 440.5 lines of the Toise of Châtelet. Moreover, he suggested a universal toise that was twice the length of the seconds pendulum.
- However, the length of a seconds pendulum was discovered to vary from place to place. French astronomer Jean Richer measured the 0.3% difference in length between Cayenne (in what is now French Guiana) and Paris.
This is a schematic of the 2nd pendulum clock built around 1673 by Christiaan Huygens, inventor of the pendulum clock. It is from his treatise Horologium Oscillatorium, published 1673, Paris, in which it records improvements to the mechanism that Huygens had illustrated in the 1658 publication of his invention, titled Horologium. It is a weight-driven clock with a verge escapement (K,L), with the 1 second pendulum (X) suspended on a cord (V). The large metal plate (T) in front of the pendulum cord is the first illustration of Huygens' 'cycloidal cheeks', which aims to improve accuracy by forcing the pendulum to follow a cycloidal path and produce isochronous swing. Huygens claimed it achieved an accuracy of 10 seconds per day.
26. Spherical pendulum
https://en.wikipedia.org/wiki/Spherical_pendulum
This device consists of a mass m moving without friction on the surface of a sphere. Note the only forces acting on the mass are the reaction from the sphere and gravity.
Spherical coordinates are used to describe the position of the mass in terms of (r, θ, φ), where r is fixed, r = l.
Lagrangian mechanics
In order to write down the kinetic T = 0.5*m*v^2 and potential V parts of the Lagrangian L = T - V in arbitrary generalised coordinates the position of the mass is expressed along Cartesian axes. According to the diagram, we can outline the following equations:
x = l*sin(θ)*cos(φ)
y = l*sin(θ)*sin(φ)
z = l*(1 - cos(θ))
Next, we define the time derivatives to obtain velocities along the axes:
Yielding the following equations:
The Lagrangian, with constant parts removed, is:
The Euler–Lagrange equation involving the polar angle θ:
This yields:
When φ* = 0 the equation simplifies to the differential equation for the motion of a simple gravity pendulum.
Yielding the following equations:
The Lagrangian, with constant parts removed, is:
The Euler–Lagrange equation involving the polar angle θ:
This yields:
When φ* = 0 the equation simplifies to the differential equation for the motion of a simple gravity pendulum.
Similarly, the Euler–Lagrange equation involving the azimuth φ,
This leads to:
The last equation illustrates that angular momentum around the vertical axis, |Lz| = l*sin(θ) x m*l*sin(θ)φ* is conserved. The factor m*l2 * sin2(θ) will contribute in the Hamiltonian formulation below.
The second order differential equation determining the evolution of φ is therefore:
Hamiltonian mechanics
The Hamiltonian is:
where conjugate momenta are:
The Hamiltonian is:
where conjugate momenta are:
In terms of coordinates and momenta, it becomes:
This allows Hamilton's equations to provide time evolution of coordinates and momenta in 4 first-order differential equations.
Describe the trajectory
The trajectory of the mass on the sphere can be computed from the expression for the total energy.
Note that the component of angular momentum Lz = m*l2 * sin2(θ) is a constant of motion, independent of time.
Therefore,
This leads to an elliptic integral of the first kind for θ.
as well as an elliptic integral of the third kind for φ.
The angle θ lies between two circles of latitude, where:
27. Torsion pendulum
https://en.wikipedia.org/wiki/Torsion_pendulum_clock
27. Torsion pendulum
https://en.wikipedia.org/wiki/Torsion_pendulum_clock
https://www.youtube.com/watch?v=0GAdMAm1-3o&ab_channel=AmritaVlab
- Also known as the torsion balance, the torsion pendulum is a scientific apparatus that measures very weak forces. It was thought to have been first invented by Charles-Augustin de Coulomb in 1777, but independently invented by John Michell before 1783.
- Coulomb used it to measure the electrostatic force between charges to establish Coulomb's Law. Henry Cavendish used it in his famous 1798 Cavendish experiment to measure the gravitational force between two masses to calculate the density of the Earth, which lead to the gravitational constant.
This is a drawing of Coulomb's torsion balance. From Plate 13 of his 1785 memoir.
- The torsion balance consists of a bar suspended from its middle by a thin fibre, which acts as a weak torsion spring.
- If an unknown force is applied perpendicularly to the ends of the bar, this causes the bar to rotate and twist, until it reaches an equilibrium between the twisting force or torque of the fibre and the applied force. Subsequently, the magnitude of the force is proportional to the angle of the bar.
- Note the sensitivity of the instrument originates from the fibre's weak spring constant, therefore a very weak force induces significant rotation of the bar.
According to Wolfram, a torsional pendulum is an oscillator for which the restoring force is torsion. For instance, if we dangle a bar from a thin wire and wind it by an angle θ, it would create a torsional torque.
τ = -κ*θ
-- κ = A characteristic property of the wire, known as the torsional constant
-- I = Moment of inertia
Note this is the a simple harmonic oscillator with equation of motion:
θ = θ0 cos(ω*t + φ)
-- θ0 = Initial angle
-- ω = Angular frequency
-- φ = The phase
28. Wilberforce pendulum
https://en.wikipedia.org/wiki/Wilberforce_pendulum
- Invented by British physicist Lionel Robert Wilberforce roughly 1896, this pendulum comprises of a mass suspended by a long helical spring that is free to rotate on its vertical axis, twisting the spring.
- The mass bobs both up and down on the spring, and rotate back and forth about its vertical axis with torsional vibrations. With the correct adjustments, its motion exhibits periods of rotational oscillation that gradually alternate with periods of purely up and down oscillation.
- Berg & Marshall (1990) described the gradual back and forth energy shift between the translational 'up and down' oscillation mode and the torsional 'clockwise and counterclockwise' oscillation mode, until the motion eventually calms down.
- It does not behave like a typical pendulum because it does not swing back and forth. The mass typically has opposing pairs of radial 'arms' that protrude horizontally, threaded with small weights that can be bolted in or out to adjust the moment of inertia in order to 'tune' the torsional vibration period.
- The frequency at which the 2 modes alternate is equal to the difference between the oscillation frequencies of the modes. Therefore, if the 2 pendulums' frequencies are similar, then the alternation between them are slower. This is analogous to the beats in musical instruments, in which 2 tones combine to produce a 'beat' tone at the difference between their frequencies.
- For example, if the pendulum bobs up and down at a rate of fT = 4 Hz, and rotates back and forth about its axis at a rate of fR = 4.1 Hz, the alternation rate falt will be 0.1 Hz. Furthermore Talt = 1/falt = 10 s. Thus the motion will change from rotational to translational in 5 seconds and then back to rotational in the next 5 seconds.
29. Airy’s coal pit experiments
https://www.mezzacotta.net/100proofs/archives/120
- In 1826, British astronomer George Airy attempted to measure Earth's density by taking pendulum gravity measurements at the top and bottom of a coal mine.
- According to Gauss's law, the mass of the spherical shell of crust above the subsurface point does not contribute to the gravity. Therefore the gravitational force below the surface of the Earth decreases rather than increases with depth.
- The result was the lower pendulum moved slower by 2.24 seconds per day relative to the higher pendulum. This suggested the gravitational acceleration at the bottom of the mine 381 m (1250 ft) below the surface was roughly 1/14000 less than expected from the inverse square law.
30. Von Sterneck and Mendenhall gravimeters
- In 1887, Austro-Hungarian scientist Robert von Sterneck invented a small gravimeter pendulum that is mounted in a temperature-controlled vacuum tank in order to remove he effects of temperature and air pressure. Since the pendulum was non-reversible, it was used to obtain relative gravity measurements.
- Along with the Mendenhall pendulum, invented by Thomas C. Mendenhall of the US Coast and Geodetic Survey in 1890, the Von Sterneck gravimeter was often used for surveys into the 1920s.
- Albert A. Michelson used the Mendenhall pendulum in his 1924 measurements of the speed of light on Mt. Wilson, California because he considered it to be more accurate than the highest precision clocks of the time.
- The Repsold pendulum was around 56 cm in length and had a period of about 0.75 second. It was used extensively by European geodetic agencies, and with the Kater pendulum in the Survey of India.
31. Double pendulum gravimeters
- Invented in 1877, Hervé Faye (advocated by Peirce, Cellérier and Furtwangler) mounted 2 identical pendulums on the same support that swing with the same amplitude, 180° out of phase. It was constructed to rectify errors caused by the swing of the pendulum that staggered the tripod stand used to support portable pendulums.
- The opposite motion of the pendulums would cancel out any sideways forces on the support.
32. Gulf gravimeters
- Invented in 1929 by the Gulf Research and Development Co., this device was one of the last and most accurate pendulum gravimeters.
- It has 2 pendulums made of fused quartz, each 270 mm (10.7 inches) long with a period of 0.89 seconds, that swing on pyrex knife edge pivots, 180° out of phase.
- They were mounted in a permanently sealed temperature and humidity controlled vacuum chamber. A radioactive salt was used to capture stray electrostatic charges on the quartz pendulums.
Used into the 1960s, this instrument accurate to within (0.3–0.5)×10−7 (30–50 microgals or 3–5 nm/s2).
ii. Irreversibile processes
https://en.wikipedia.org/wiki/Irreversible_process
A loss of mechanical energy in a system often eventuated in an increase of the system's temperature. Physicist James Prescott Joule was the first to experimentally demonstrate how a certain amount of work done against friction produced a definite quantity of heat, which was conceived as the random motions of the particles that comprise matter. This highlights the equivalence between mechanical energy and heat in collisions. In inelastic collisions, the decrease in mechanical energy may have been transformed into kinetic energy of the constituent particles and temperature increases. I’ll discuss thermodynamics in another post.
iii. Satellites
https://en.wikipedia.org/wiki/Vis-viva_equation
A satellite of mass (m) at a distance (r) from the centre of Earth possesses both kinetic energy (K) and gravitational potential energy (U). Thus, mechanical energy (Emechanical) of the satellite-Earth system can be calculated with:
Emechanical = U + K
Emechanical = -G*(M*m)/r + 0.5*m*v2
If the satellite is in circulate orbit, the energy conservation equation can be further simplified into:
Emechanical = -G*(M*m) / (2*r)
In circular motion, Newton's 2nd Law of motion can be expressed as:
G*(M*m)/r2 = (m*v2)/r
What is projectile motion?
This form of motion involves an object or particle (a projectile) being projected near the Earth;s surface and traverses a curved path under the action of gravity (and air resistance). This curved path was demonstrated by Galileo to be a parabola, or a straight line if the projectile is directed upwards. The study of such motions is called ballistics, and such as trajectory is known as a ballistic trajectory. Ballistics (derived from the Greek for ‘to throw’) is the science of mechanics that deals with the flight, behaviour, and effects of projectiles, such as bullets, unguided bombs, rockets, or the like.
This graph shows the trajectories of a projectile with air drag and varying initial velocities.
This photo shows parabolic water motion trajectory.
Consider a projectile being launched with an initial velocity v(0) = v0, which can be expressed as the sum of horizontal and vertical components as follows.
v0 = v0xi + v0yj
The components v0x & v0y can be evaluated if the initial launch (i.e., elevation) angle (θ) is known:
v0x = v0*cos(θ)
v0y = v0*sin(θ)
Describe the kinematic quantities of projectile motion
In projectile motion, the horizontal motion and the vertical motion occur independently of each other. This principle of compound motion was established by Galileo in 1638, which was used to prove the parabolic form of projectile motion. An example of a ballistic trajectory with constant acceleration is a space ship in the absence of other forces. In contrast, the space ship’s acceleration on Earth changes magnitude with altitude and direction with latitude/longitude, leading to an elliptic trajectory. At higher speeds the trajectory can also be circular, parabolic or hyperbolic (unless it is distorted by other celestial objects).
i. Acceleration
In projectile motion, there is only acceleration in the vertical direction (due to free fall), so the velocity in the horizontal direction is constant, being equal to v0*cos(θ). The components of the acceleration are:
— ax = 0
— ay = g
ii. Velocity
The horizontal component of the object’s velocity remains unchanged throughout the motion. Since the acceleration due to gravity is constant, the vertical component of the object’s velocity changes linearly. We can integrate accelerations in the x and y directions to solve for the components of velocity at any time t, as follows:
— vx = v0*cos(θ)
— vy = v0*sin(θ) - g*t
Using Pythagorean theorem would yield the magnitude of the velocity:
v = (vx2 + vy2)0.5
iii. Displacement
At any time t, the projectile's horizontal and vertical displacement are:
— x = v0*t*cos(θ)
— y = v0*t*sin(θ) - 0.5*g*t2
The magnitude of the displacement is:
Δr = (x2 + y2)0.5
y = tan(θ)*x - x2*g/[2*v02*cos2(θ)]
Since g, θ, v0 are constants, the above equation is of the form
y = a*x + b*x2
— a & b = Constants
The above equation is a parabola, so the path is parabolic, hence the axis of the parabola is vertical.
If the projectile’s position (x,y) and launch angle (θ or α) are known, the initial velocity can be found solving for v0 in the aforementioned parabolic equation:
v0 = [(g*x2) / (x*sin(2*θ) - 2*y*cos2(θ)]0.5
How do we calculate the time of flight or total time of the whole journey?
The total time (t) for which the projectile is airborne is called the time of flight.
y = v0*t*sin(θ) - 0.5*g*t2
After the flight, the projectile returns to the horizontal axis (x-axis), so y = 0.
0 = v0*t*sin(θ) - 0.5*g*t2
v0*t*sin(θ) = 0.5*g*t2
v0*sin(θ) = 0.5*g*t
t = 2*v0*sin(θ) / g
— Note that air resistance is negligible on the projectile.
If the starting point is at height, y0, with respect to the point of impact, the time of flight is:
t = d / (v*cos(θ)) = {v*sin(θ) + [(v*sin(θ))2 + 2*g*y0]0.5} / g
The above equation can be simplified to
t = {v*sin(θ) + [(v*sin(θ))2]0.5}/g
= [v*sin(θ) + v*sin(θ)] / g
= 2*v*sin(θ) / g
= 2*v*sin(45) / g
= [2*v*0.5*20.5] / g
t = (v*20.5)/g
— θ = 450
— y0 = 0
How do we calculate the maximum height of the projectile?
The maximum height an object can reach is known as the peak of the object's motion. The increase in height lasts until vy = 0, which is expressed as:
0 = v0*sin(θ) - g*th
The time it takes to reach the maximum height (h):
th = (v0*sin(θ))/g
The vertical displacement of the maximum height of projectile can be calculated by the following equations:
h = v0*th*sin(θ) - 0.5*g*th2
h = (v02 * sin2(θ)) / 2g
Describe the relationship between horizontal range and maximum height
The relation between the range (R) on the horizontal plane and the maximum height (h) reached at 0.5*td is:
h = (v02 * sin2(θ)) / 2g
R = (v02 * sin(2θ)) / g
h/R = [(v02 * sin2(θ)) / 2g]*[g / (v02 * sin(2θ))] = sin2(θ) / [4*sin(θ)*cos(θ)]
h = R*tan(θ) / 4
How do we calculate the maximum distance of a projectile?
https://en.wikipedia.org/wiki/Range_of_a_projectile
If we assume the Earth’s surface is fairly level, has a uniform gravitational field, then we can predict the range of a launched projectile if the initial conditions are known. If there is no air resistance and no change in gravitational acceleration, then the projectile motion is ideal.
If we launch a projectile at 45o relative to the ground, it will travel the furthest horizontally.
d = (v2 * sin(2θ))/g
The range becomes maximum when sin(2θ) is equal to 1 (i.e. 2θ = 90o , so θ = 45 o ).
i. Flat ground
When y0 equals zero, the horizontal position of the projectile is:
x(t) = v*t*cos(θ)
In the vertical direction:
y(t) = v*t*sin(θ) - 0.5*g*t2
To compute the time it takes for the projectile to return to the same height it originated, we let tg be any time when the height of the projectile is equal to its initial value.
0 = v*t*sin(θ) - 0.5*g*t2
If we factor in: t = 0 or t = (2*v*sin(θ)) / g
Since t = T = time of flight, then T = (2*v*sin(θ)) / g
The first solution corresponds to the moment the projectile first launches, while the second solution can be used to determine the range of the projectile. If we substitute the value for t into the horizontal equation, we get:
x = [2*v2 * cos(θ)*sin(θ)] / g
Next we apply the following trigonometric identity:
sin(x + y) = sin(x)*cos(y) + sin(y)*cos(x)
If x = y, then:
sin(2θ) = 2*sin(θ)*cos(θ)
That means we can simplify the solution to:
d = (v2 * sin(2θ)) / g
If θ = 45o, then
dmax = v2/g
ii. Rough ground
If y0 is not zero, then the equations of projectile motion becomes:
x(t) = v*t*cos(θ)
y(t) = y0 + v*t*sin(θ) - 0.5*g*t2
To solve for t, the position (y) of the projectile needs to be zero.
0 = y0 + v*t*sin(θ) - 0.5*g*t2
Next we apply the quadratic formula to yield 2 solutions. After several steps of algebraic manipulation, we yield:
t = [v*sin(θ) / g] + [(v2 * sin2(θ) + 2*g*y0)0.5 / g]
Since the square root needs to be a positive number, and the velocity and the sine of the launch angle is assumed to be positive, then the solution with the greatest time can be obtained when the positive of the plus or minus sign is used.
t = [v*sin(θ) / g] + [(v2 * sin2(θ) + 2*g*y0)0.5 / g]
The projectile range would be:
d = (v*cos(θ)/g) * [v*sin(θ) + (v2 *sin2(θ) + 2*g*y0)0.5]
or, equivalently
d = (v2/2g)*{1+ [1 + (1 + (2*g*y0)/(v2 * sin2(θ))]0.5}*sin(2θ)
The range can be maximised at any height when the angle:
θ = arccos[(2*g*y0 + v2) / (2*g*y0 + 2*v2)]0.5
If we want to check the limit as y0 approaches 0, it can be written as: How do we calculate the maximum distance of a projectile?
iii. Angle of impact
The angle (ψ) at which the projectile lands on the ground is expressed as:
tan(ψ) = -vy(td) / vx(td) = (v2 * sin2(θ) + 2*g*y0)0.5 / (v*cos(θ))
To achieve maximum range, the above equation becomes:
tan2(ψ) = (2*g*y0 + v2) / v2 = C + 1
We can rewrite the original solution for θ to yield:
tan2(θ) = (1 - cos2θ) / cos2θ = v2 / (2*g*y0 + v2) = 1 / (C + 1)
If we multiply the above equation with tan2(ψ), we get:
tan2(ψ)*tan2(θ) = [(2*g*y0 + v2) / v2] * [v2 / (2*g*y0 + v2)] = 1
Because of the trigonometric identity:
tan(θ + ψ) = [tan(θ) + tan(ψ)] / [1 - tan(θ)*tan(ψ)]
— This means that (θ + ψ) must be equal to 90 degrees.
How is work energy theorem applied to projectile motion?
According to the work-energy theorem, the vertical component of velocity is:
vy2 = (v0*sin(θ))2 - 2*g*y
— Note it ignores aerodynamic drag and assumes that the landing area is at uniform height 0.
How we calculate the angle of reach?
The “angle of reach” is the angle (θ) at which a projectile must be launched in order to go a distance (d) given the initial velocity (v).
sin(2θ) = g*d/v2
There are 2 solutions to this equation:
(1) θ = 0.5*arcsin(g*d/v2)
(2) θ = 90o - 0.5*arcsin(g*d/v2)
How we calculate the angle required to hit a certain coordinate (x,y)?
If we launch a projectile from the origin (0,0) with an initial speed (v), the angle(s) of launch (θ) required to reach a target at range x and altitude y are:
tan (θ) = {[v2 + (v4 - g*(g*x2 + 2*y*v2))0.5] / (g*x)}
The 2 roots of the equation correspond to the 2 possible launch angles, assuming they aren’t imaginary, in which case the initial speed is insufficient in reaching the point (x,y) selected. Nevertheless, we can work out the angle of launch needed without the restriction of y = 0.
We can also evaluate the launch angle required to achieve the lowest possible launch velocity. If the 2 solutions above equal one another, this suggests that the quantity under the square root sign is zero. So, solving the quadratic equation for v2 yields:
v2 /g = y + (y2 + x2)0.5
This gives out the angle:
θ = arctan[y/x + (y2 /x2 + 1)0.5]
If we designate the angle whose tangent is y/x by α, then:
tan(θ) = (sin(α) + 1)/cos(α)
tan(π/2 - θ) = cos(α)/(sin(α) + 1)
cos2(π/2 - θ) = 0.5*(sin(α) + 1)
2*cos2(π/2 - θ) -1 = cos(π/2 - α)
Hence:
θ = 0.5*π - 0.5*(π/2 - α)
This means the launch should be at the angle halfway between the target and Zenith (vector opposite to Gravity).
If we know the height of launch and landing is identical and ignore air resistance, we can compute the length of the parabolic arc (L) traced by a projectile using the following formula:
L = [(v2 * cos2(θ))/2g] * [2*sec(θ)*tan(θ) - ln|(1 - sin(θ))/(1 + sin(θ))|
— v = Initial velocity
— θ = Launch angle
— g = Acceleration due to gravity as a positive value.
The above equation is obtained through evaluation of the arc length integral for height-distance parabola between the bounds initial and final displacements (i.e. between 0 and the horizontal range of the projectile).
Describe the trajectory of a projectile with air resistance
If air resistance is accounted for, it directly affects the velocity of the particle (Fa ∝ v→), which is only valid at Reynolds Number below about 1000. Since air has a kinematic viscosity around 0.15 cm2/s , the product of speed and diameter must be less than about 150 cm2/s.
At higher values of Reynolds number, the force of air resistance is proportional to the square of the particle's velocity.
The free body diagram illustrates a projectile under the effects of air resistance and gravity. It assumes air resistance to be in the direction opposite of the projectile's velocity. It’s more realistic to use the relationship Fair = -k*v2 to achieve an analytic solution. Nevertheless, Fair = -k*v is used because of the initial assumption of direct proportionality, which indicates the difference between the air resistance and the velocity being a constant arbitrary factor with units of N*s/m.
This graph shows the trajectories of a mass thrown at an angle of 70°:
— Black = Without drag
— Blue = With Stokes drag
— Green = With Newton drag
— x direction: ΣF = -k*vx = m*ax
— y direction: ΣF = -k*vy - m*g = m*ay
This implies that:
[1] ax = -k*vx /m = d*vx /dt
&
[2] ay = (-k*vy - m*g)/m = -k*vy /m - g = dvy /dt
The solution to equation [1] is an elementary differential equation, therefore the steps leading to a unique solution for vx and, subsequently, x will not be enumerated. Given the initial conditions vx = vxo (vxo is the x component of the initial velocity) and sx = 0 for t = 0:
[1a] vx = vxo*e-k*t/m
[1b] sx = (m*k)*vxo*(1 - e*-k*t/m)
Extensively solving [2] requires the initial conditions vy = vyo & sy = 0 when t = 0:
[2a] - g = dvy/dt + k*vy/m
We can solve the first order, linear, non-homogeneous differential equation via an integrating factor e ∫k/m dt .
[2b] et*k/m *(dvy/dt + vy*k/m) = et*k/m *(-g)
[2c] (vy*et*k/m)’ = et*k/m *(-g)
[2d] ∫(vy*et*k/m)’ dt = vy*et*k/m = ∫et*k/m *(-g) dt
[2e] vy*et*k/m = (m*k)* et*k/m *(-g) + C
[2f] vy = -m*g/k + C*et*k/m
Integration of [2f] yields the following equation:
[3] sy = -m*g/k + (m/k)*(vyo + m*g/k)*et*k/m + C
Solving for our initial conditions gets us the following functions:
[2g] vy(t) = -m*g/k + (vyo + m*g/k)*et*k/m
[3a] sy(t) = -m*g/k - (m/k)*(vyo + m*g/k)*et*k/m- (m/k)*(vyo + m*g/k)
Equation [3a] can be simplified into:
[3b] sy(t) = -(m*g/k)*t + (m/k)*(vyo + m*g/k)*(1 - et*k/m)
m = 0.145 kg (5.1 oz)
v0 = 44.7 m/s (100 mph)
g = -9.81 m/s2 (-32.2 ft/s2)
vt = -33.0 m/s (-73.8 mph)
k = m*g/vt = 0.145*(-9.81)/(-33) = 0.0431 kg/s
θ = 45o
The more realistic trajectory Fair = -k*|v|*v can not be calculated analytically, but only by numerical simulations.
ax = -k*vx2/m = dvx/dt
vx = dsx/dt
— If vxo is not 0, vx = 1/(1/vxo + k*t/m).
— If vxo is 0, vx = 0
sx = m*ln(1 + vxo*k*t)/k
Since acceleration is negative while the velocity is positive and vice versa, a projectile fired upwards requires the absolute value to be taken of the vertical velocity. This would make an analytical solution for vertical position more complex.
Where g0 is gravitational acceleration set to some constant, such as standard gravity:
ay = (-k*|vy|*vy)/m - g0 = dvy/dt for constant gravity
A more complex solution:
ay = (-k*|vy|*vy)/m - g0*(r /(r+sy))2 = dvy/dt for gravity as a function of height above a planet's surface.
— g0 = Planet’s gravity at the surface
— r = Planet’s radius
How do we calculate the time of flight with air resistance?
The total time of the journey in the presence of air resistance can be determined by solving the equation sy(t) = 0. The Lambert W function is required to solve sy(t) = 0 in the case of zero air resistance. Since the equation [3b] is of the form c1*t + c2 + c3 *ec4*t = 0, it can be transformed into an equation solvable by the W equation. In closed form, the total time of flight can be evaluated algebraically by the following equation:
t = (m/k) + vyo/g + (m/k)*W*{-[1 + k*vyo/(m*g)*e-(1 + k*vyo(m*g))]
Solving the problem of motion of a projectile with air resistance modelled as Fair = -k*v2 follows. Assumptions include constant gravitational acceleration and air resistance dictated by the following drag equation: FD = 0.5*c*ρ*A*v2
— FD = Drag force
— c = Drag coefficient
— ρ = Air density
— A = Cross sectional area of the projectile
Consider a projectile of mass (m) launching from a point (x0, y0), with an initial velocity (V0) in an initial direction that makes an angle (ψ0) with the horizontal. The air resistance experienced by the projectile is dictated by the formula Fair = -k*v2, which acts tangentially to the path of travel at any point. If we apply Newton’s 2nd of motion in the x-direction, we get:
[1] fx = -k*V2 *cos(ψ) = m*(d2x/dt2)
— V = (u2 + v2)0.5
— u & v = Horizontal and vertical components of the velocity V respectively.
If μ = k/m, du/dt = d2x/dt2, and cos(ψ) = u/V, then equation [1] becomes:
[1a] du/dt = -μ*u*(u2 + v2)0.5
In the y-direction:
[2] fy = -k*V*sin(ψ) - m*g = m*(d2y/dt2)
If μ = k/m, dv/dt = d2y/dt2, and sin(ψ) = u/V, then equation [2] becomes:
[2a] dv/dt = -μ*u*(u2 + v2)0.5 - g
Since we know that dv/du = (dv/dt) / (du/dt), we can divide equation [2a] by equation [1a] to obtain:
[2b] dv/dt = v/u + g/[μ*u*(u2 + v2)0.5]
If we introduce a quantity q such that v = q*u, then we get:
[2c] dv/du = d(q*u)/du = q + u*(dq/du)
Notice that if equation [2b] = equation [2c], then it can be rewritten as: u*(dq/du) = g/[μ*u*(u2 + v2)0.5]. If we separate the variables and integrate them, we yield:
[2d] (g/μ)* ∫(u-3) du = ∫ (1+q2)0.5 dq
The left hand side of equation [2d] is ∫(u-3) du = -1/(2*u2). Meanwhile, for the right-hand side of equation [2d], let q = sinh(Q), such that 1 + u2 = 1 + sinh2(Q) = cosh2(Q) and, dq = cosh(Q) dQ. Hence, ∫(1+q2)0.5 dq = ∫cosh2(Q) dQ. Also cosh2(Q) = 0.5*(1+cosh(2Q))
This yields ∫cosh2(Q) dQ = 0.5*[Q + 0.5*sinh(2Q)].
Equation [2d] becomes -g/(d*μ*u2) = 0.5*[Q + 0.5*sinh(2Q)] - Ã
For which;
u = (g/μ)0.5 * 1/[A - (Q + 0.5*sinh(2Q))0.5
Since v = u*q = u*sinh(Q)
v = (g/μ)0.5 *sinh(Q) / [A - (Q + 0.5*sinh(2Q))]0.5
If we denote λ = Α - (Q + 0.5*sinh(2Q), this means:
[2e] u = (g/μ)0.5 * (1/λ0.5)
[2f] v = (g/μ)0.5 * [sinh(Q)]/λ0.5
At the start of the motion, t = 0 and V02 = u02 + v02 .
Thus, tan(ψ0) = v0/u0 = q0 = sinh(Q0), such that A = g/(μ*u02) + [Q0 + 0.5*sinh(2*Q0)]
As the motion proceeds, u —> 0 and v —> (a negative constant), i.e. q —> ∞ & Q —> -∞.
This means that, λ —> ∞ & 1/λ0.5 —> 0. Hence:
When a state of dynamic equilibrium is achieved under vertical free fall, the opposing forces of gravity and drag are equalised, i.e. k*vt2 = m*g.
[2g] vt = (g/μ)0.5
In equation [1a], if we substitute the values for u and v from equations [2e] and [2f], this leads to:
du/dt = (-g/λ)*cosh(Q)
Moreover, du/dQ = [1/(2*λ1.5)]*(g/μ)*(1 + cosh(2Q))
Since dQ/dt = (du/dt)/(du/dQ), this means:
dt/dQ = -cosh(Q) / [(g*μ)0.5 * λ0.5]
[2h]
Since dx/dQ = (dx/dt)/(dQ/dt) = (-1/μ)*(cosh(Q)/λ)
[2i]
[2j]
I. Time of flight
To determine the time of flight (T), we set y to 0 in equation [2j] and then solve for the value of the variable Q.
[2k]
Equation [2h] with QT substituted for Q yields:
[2l]
Equation [2i] gives the horizontal range R as:
[2m]
III. Maximum height
At the highest point of the projectile path ψ = 0, and Q = 0, it gives the maximum height from equation [2j].
[2n]
Describe lofted trajectory
A special case of a ballistic trajectory for a rocket is a lofted trajectory, which has an apogee greater than the minimum-energy trajectory to the same range. Generally speaking, the rocket travels higher by it expends more energy to get to the same landing point. A lofted trajectory is preferred for various reasons such as increasing distance to the horizon to give greater viewing/communication range or for changing the angle with which a missile will impact on landing.
Eddie Woo’s videos on projectile motion
Part 2:
https://www.youtube.com/watch?v=-50vg_970Ik
Part 3:
https://www.youtube.com/watch?v=dK39bPl-jOU
Part 4:
https://www.youtube.com/watch?v=wSZL-SLGl04
Part 5:
https://www.youtube.com/watch?v=xwPAXHBLvvw
Khan Academy videos on projectile motion:
— Projectile at an angle:
https://www.youtube.com/watch?v=ZZ39o1rAZWY
— Horizontally launched projectile:
https://www.youtube.com/watch?v=jmSWImPs6fQ
— Projectile motion (1D motion)
Part 1:
https://www.youtube.com/watch?v=15zliAL4llE
Part 2:
https://www.youtube.com/watch?v=emdHj6WodLw
Part 3:
https://www.youtube.com/watch?v=Y5cSGxdDHz4
Part 4:
https://www.youtube.com/watch?v=-W3RkgvLrGI
Part 5:
https://www.youtube.com/watch?v=dlpmllTx5MY
Part 6:
https://www.youtube.com/watch?v=bl2DvFn8LjM
Part 7:
https://www.youtube.com/watch?v=bl2DvFn8LjM
— Projectile motion (2D motion)
Part 1:
https://www.youtube.com/watch?v=bl2DvFn8LjM
Part 2:
https://www.youtube.com/watch?v=cEOxZWGp-8E
Part 3:
https://www.youtube.com/watch?v=o2ZrX9MbIwU
Walter Lewin’s 1999 Lecture on the Motion of Projectiles:
https://www.youtube.com/watch?v=92Em8p6Qzmo
Examples:
(1) At the MCG (Melbourne Cricket Ground, the distance from the side boundary rope to the cricket pitch in the centre of the ground is about 75 metres. The stumps at each end of the pitch is about 71.1cm high. Steve Smith is on the boundary is about throw a 200g cricket ball towards the pitch with an initial throwing speed of 100 km/h. Ignore air resistance and assume the weather is calm.
(a) At what angle does Ponting needs to throw the ball to hit the bottom of the stumps?
y = tan(θ)*x - x2*g/[2*v02*cos2(θ)]
0 = tan(θ)*75 - 752*9.81/[2*(27.78)2*cos2(θ)]
(5.52*104)/[1.54*103 *cos2(θ)] = 75*tan(θ)
35.8/cos2(θ) = 75*tan(θ)
0.477 = tan(θ) * cos2(θ)
θ = 36.3o
(b) Using the angle you calculated from (1), how long would it take for the cricket ball to reach the bottom of the stumps?
t = 2*v0*sin(θ) / g
t = 2*27.78*sin(36.3o) / 9.81 = 3.35 seconds
(c) Consider a batsman named Virat Kohli running from one crease to the other crease, which is about 17.68m. From rest, he accelerates up to 20 km/h in 0.5 seconds. If he crosses the crease before the ball hits the stumps, he will be adjudged OUT by the umpire. Consider Kohli running from the crease takes place at the same time Smith releases the ball from the boundary under the same initial conditions. Using the answers from questions (a) and (b), Ponting manages to hit the bottom of the stumps, would Kohli be OUT?
Distance elapsed during acceleration
r = r0 + 0.5*(v + v0)*t
r = 0 + 0.5*(5.56 + 0)*0.5
r = 1.39 m
t = 0.5 seconds
v = 5.56 m/s
Distance remaining under constant velocity
r = 17.68 - 1.39 = 16.29 m
https://www.youtube.com/watch?v=-50vg_970Ik
Part 3:
https://www.youtube.com/watch?v=dK39bPl-jOU
Part 4:
https://www.youtube.com/watch?v=wSZL-SLGl04
Part 5:
https://www.youtube.com/watch?v=xwPAXHBLvvw
Khan Academy videos on projectile motion:
— Projectile at an angle:
https://www.youtube.com/watch?v=ZZ39o1rAZWY
— Horizontally launched projectile:
https://www.youtube.com/watch?v=jmSWImPs6fQ
— Projectile motion (1D motion)
Part 1:
https://www.youtube.com/watch?v=15zliAL4llE
Part 2:
https://www.youtube.com/watch?v=emdHj6WodLw
Part 3:
https://www.youtube.com/watch?v=Y5cSGxdDHz4
Part 4:
https://www.youtube.com/watch?v=-W3RkgvLrGI
Part 5:
https://www.youtube.com/watch?v=dlpmllTx5MY
Part 6:
https://www.youtube.com/watch?v=bl2DvFn8LjM
Part 7:
https://www.youtube.com/watch?v=bl2DvFn8LjM
— Projectile motion (2D motion)
Part 1:
https://www.youtube.com/watch?v=bl2DvFn8LjM
Part 2:
https://www.youtube.com/watch?v=cEOxZWGp-8E
Part 3:
https://www.youtube.com/watch?v=o2ZrX9MbIwU
Walter Lewin’s 1999 Lecture on the Motion of Projectiles:
https://www.youtube.com/watch?v=92Em8p6Qzmo
(1) At the MCG (Melbourne Cricket Ground, the distance from the side boundary rope to the cricket pitch in the centre of the ground is about 75 metres. The stumps at each end of the pitch is about 71.1cm high. Steve Smith is on the boundary is about throw a 200g cricket ball towards the pitch with an initial throwing speed of 100 km/h. Ignore air resistance and assume the weather is calm.
(a) At what angle does Ponting needs to throw the ball to hit the bottom of the stumps?
y = tan(θ)*x - x2*g/[2*v02*cos2(θ)]
0 = tan(θ)*75 - 752*9.81/[2*(27.78)2*cos2(θ)]
(5.52*104)/[1.54*103 *cos2(θ)] = 75*tan(θ)
35.8/cos2(θ) = 75*tan(θ)
0.477 = tan(θ) * cos2(θ)
θ = 36.3o
(b) Using the angle you calculated from (1), how long would it take for the cricket ball to reach the bottom of the stumps?
t = 2*v0*sin(θ) / g
t = 2*27.78*sin(36.3o) / 9.81 = 3.35 seconds
(c) Consider a batsman named Virat Kohli running from one crease to the other crease, which is about 17.68m. From rest, he accelerates up to 20 km/h in 0.5 seconds. If he crosses the crease before the ball hits the stumps, he will be adjudged OUT by the umpire. Consider Kohli running from the crease takes place at the same time Smith releases the ball from the boundary under the same initial conditions. Using the answers from questions (a) and (b), Ponting manages to hit the bottom of the stumps, would Kohli be OUT?
Distance elapsed during acceleration
r = r0 + 0.5*(v + v0)*t
r = 0 + 0.5*(5.56 + 0)*0.5
r = 1.39 m
t = 0.5 seconds
v = 5.56 m/s
Distance remaining under constant velocity
r = 17.68 - 1.39 = 16.29 m
17.68 = 1.39 + 5.56*t + 0.5*0*t2
16.29 = 5.56*t + 0
t = 2.93 seconds
tT = 0.5 + 2.93 = 3.43 seconds
Since 3.43 > 3.35 seconds, Virat Kohli would be run out.
(d) What’s the maximum height the cricket ball thrown by Ponting reaches?
h = (v02 * sin2(θ)) / 2g
h = (27.782 * sin2(36.3o)) / (2*9.81)
h = 13.785 m
I’ll delve into the details of other types of energy in another post.
16.29 = 5.56*t + 0
t = 2.93 seconds
tT = 0.5 + 2.93 = 3.43 seconds
Since 3.43 > 3.35 seconds, Virat Kohli would be run out.
(d) What’s the maximum height the cricket ball thrown by Ponting reaches?
h = (v02 * sin2(θ)) / 2g
h = (27.782 * sin2(36.3o)) / (2*9.81)
h = 13.785 m
I’ll delve into the details of other types of energy in another post.
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