Pendulum (mathematics) - Simple Gravity Pendulum

Simple Gravity Pendulum

A so called "simple pendulum" is an idealization of a "real pendulum" but in an isolated system using the following assumptions:

  • The rod or cord on which the bob swings is massless, inextensible and always remains taut;
  • Motion occurs only in two dimensions, i.e. the bob does not trace an ellipse but an arc.
  • The motion does not lose energy to friction or air resistance.

The differential equation which represents the motion of a simple pendulum is

(Eq. 1)

where is acceleration due to gravity, is the length of the pendulum, and is the angular displacement.

"Force" derivation of (Eq. 1)

Please take the time to consider Figure 1 on the right, showing the forces acting on a simple pendulum. Note that the path of the pendulum sweeps out an arc of a circle. The angle is measured in radians, and this is crucial for this formula. The blue arrow is the gravitational force acting on the bob, and the violet arrows are that same force resolved 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, which is considered the tangential axis because its direction is always tangent to the circle. Consider Newton's second law,

where F is the sum of forces on the object, m is mass, and a is the acceleration. Because we are only concerned with changes in speed, and because the bob is forced to stay in a circular path, we apply Newton's equation to the tangential axis only. The short violet arrow represents the component of the gravitational force in the tangential axis, and trigonometry can be used to determine its magnitude. Thus,

where

is the acceleration due to gravity near the surface of the earth. The negative sign on the right hand side implies that and always point in opposite directions. This makes sense because when a pendulum swings further to the left, we would expect it to accelerate back toward the right.

This linear acceleration along the red axis can be related to the change in angle by the arc length formulas; is arc length:

thus:

"Torque" derivation of (Eq. 1)

Equation (1) can be obtained using two definitions for torque.

First start by defining the torque on the pendulum bob using the force due to gravity.

where is the length vector of the pendulum and is the force due to gravity.

For now just consider the magnitude of the torque on the pendulum.

where is the mass of the pendulum, is the acceleration due to gravity, is the length of the pendulum and is the angle between the length vector and the force due to gravity.

Next rewrite the angular momentum.

.

Again just consider the magnitude of the angular momentum.

.

and is time derivative

,

According to, we can get by comparing the magnitudes

,

thus:

which is the same result as obtained through force analysis.

"Energy" derivation of (Eq. 1)

It can also be obtained via the conservation of mechanical energy principle: any object falling a vertical distance would acquire kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy. Change in potential energy is given by

change in kinetic energy (body started from rest) is given by

Since no energy is lost, those two must be equal

Using the arc length formula above, this equation can be rewritten in favor of

is the vertical distance the pendulum fell. Look at Figure 2, which presents the trigonometry of a simple pendulum. If the pendulum starts its swing from some initial angle, then, the vertical distance from the screw, is given by

similarly, for, we have

then is the difference of the two

in terms of gives

(Eq. 2)

This equation is known as the first integral of motion, it gives the velocity in terms of the location and includes an integration constant related to the initial displacement . We can differentiate, by applying the chain rule, with respect to time to get the acceleration


\begin{align}
{d^2\theta\over dt^2} & = {1\over 2}{-(2g/\ell) \sin\theta\over\sqrt{(2g/\ell) \left(\cos\theta-\cos\theta_0\right)}}{d\theta\over dt} \\
& = {1\over 2}{-(2g/\ell) \sin\theta\over\sqrt{(2g/\ell) \left(\cos\theta-\cos\theta_0\right)}}\sqrt{{2g\over \ell} \left(\cos\theta-\cos\theta_0\right)} = -{g\over \ell}\sin\theta
\end{align}

which is the same result as obtained through force analysis.

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