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Derive a formula for the maximum speed \({v_{\max }}\) of a simple pendulum bob in terms of g, the length l, and the maximum angle of swing \({\theta _{\max }}\).

Short Answer

Expert verified

The formula for the maximum speed is \({v_{\max }} = \sqrt {2gl\left( {1 - \cos {\theta _{\max }}} \right)} \).

Step by step solution

01

Understanding the simple pendulum

The simple pendulum is said to be in simple harmonic motion when it oscillates back and forth, and the effect of friction can be ignored.

The speed of the pendulum bob is maximum when it moves through its equilibrium position.

02

Given Data

The maximum angle of swing \({\theta _{\max }}\).

The maximum speed is \({v_{{\rm{max}}}}\).

The length of the pendulum is \(l\).

03

Representation of the simple pendulum and determination of the energy of pendulum bob

A simple pendulum of length l consisting of a bob of mass mat the maximum angle of swing \({\theta _{\max }}\) is shown in the figure below:

Take the lowest point of the bob as the reference level (h = 0).

The total energy of the pendulum bob at the lowest point (\(\theta = 0\)) where its speed is maximum \(\left( {{v_{\max }}} \right)\)is given as:

\(\begin{aligned}{c}{E_{{\rm{bottom}}}} &= K{E_{{\rm{bottom}}}} + P{E_{{\rm{bottom}}}}\\ &= \frac{1}{2}mv_{\max }^2 + mg(0)\\ &= \frac{1}{2}mv_{\max }^2 + 0\end{aligned}\)

Here, \(K{E_{{\rm{bottom}}}}\) and \(P{E_{{\rm{bottom}}}}\) are the kinetic and potential energy at the bottom, m is the mass and g is the gravitational acceleration.

The total energy of the pendulum bob at the highest point (at top where \(\theta \) is maximum) is given as:

\(\begin{aligned}{c}{E_{{\rm{top}}}} &= K{E_{{\rm{top}}}} + P{E_{{\rm{top}}}}\\ &= 0 + mgh\end{aligned}\)

Here, \(K{E_{{\rm{top}}}}\) and \(P{E_{{\rm{top}}}}\) are the kinetic and potential energy at the top.

04

Determination of the maximum speed of the pendulum bob

According to the law of conservation of energy, the total energy at the top \(\left( {{E_{{\rm{top}}}}} \right)\)must be equal to the total energy at the bottom \(\left( {{E_{{\rm{bottom}}}}} \right)\). Thus,

\(\begin{aligned}{c}{E_{{\rm{top}}}} &= {E_{{\rm{bottom}}}}\\(0 + mgh) &= \left( {\frac{1}{2}mv_{\max }^2 + 0} \right)\\mgh &= \frac{1}{2}mv_{\max }^2\\{v_{\max }} = \sqrt {2gh} \end{aligned}\) ….. (i)

Thus, the height of the bob at the top point is:

\(h = \left( {l - l\cos {\theta _{\max }}} \right)\)

Substitute this value of h in equation (i).

\(\begin{aligned}{c}{v_{\max }} &= \sqrt {2gh} \\ &= \sqrt {2gl\left( {1 - \cos {\theta _{\max }}} \right)} \end{aligned}\)

This is the required formula for the maximum velocity of the pendulum bob.

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