Energy: Pendulums and Tension

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Unit: Work, Energy, and Power

Prerequisites

Energy Conservation for Pendulums

Later Topics

None

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Part 1

A simple pendulum consists of a small bob of mass $m = 0.5$ kg attached to a light string of length $L = 2.4$ m. The bob is pulled aside until the string makes an angle of $\theta_0 = 33^\circ$ with the vertical and then released from rest. Neglect air resistance. Take the lowest point of the swing as the zero of gravitational potential energy. Take $g = 9.8$ m/s $^2$ .

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Find the speed $v_{\text{bottom}}$ of the bob as it passes through the lowest point.

Correct!

Solution:

Solution

We apply conservation of mechanical energy between the release point and the lowest point. The bob is released from rest at angle $\theta_0 = 33^\circ$ .

At the release point, the height above the lowest point is:

h = L - L \cos \theta_0 = L(1 - \cos \theta_0)

The total mechanical energy at release (taking the lowest point as zero potential energy):

E_{\text{initial}} = K_{\text{initial}} + U_{\text{initial}} = 0 + mgh = mgL(1 - \cos \theta_0)

At the lowest point:

E_{\text{bottom}} = K_{\text{bottom}} + U_{\text{bottom}} = \frac{1}{2}mv_{\text{bottom}}^2 + 0

By conservation of energy:

mgL(1 - \cos \theta_0) = \frac{1}{2}mv_{\text{bottom}}^2

Solving for $v_{\text{bottom}}$ :

v_{\text{bottom}} = \sqrt{2gL(1 - \cos \theta_0)}

Substituting the values $g = 9.8$ m/s $^2$ , $L = 2.4$ m, and $\theta_0 = 33^\circ$ :

v_{\text{bottom}} = \sqrt{2 \cdot 9.8 \cdot 2.4 \cdot (1 - \cos 33^\circ)} = 2.75 \text{ m/s}

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