Answered: A simple pendulum is released from rest…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

A simple pendulum is released from rest when the mass is 33.1 cm above its lowest point. What is the speed of the suspended mass at the bottom of its swing? g= 9.80 m/s2 and ignore air restistance. No mass was given.

Expert Solution

Step 1

Given data

The height from which the simple pendulum is released is calculated as h_r = 33.1 cm.

The gravitational acceleration of the g = 9.8 m/s².

Initially, the body is at rest, therefore, v_i = 0

The velocity of the pendulum is calculated by the conservation of energy as,

$\begin{array}{rcl} K E_{i} + P E_{i} & = & K E_{f} + P E_{f} \\ \frac{1}{2} m v_{i}^{2} + m g h_{r} & = & \frac{1}{2} m v_{f}^{2} + m g h_{b} \\ \frac{v_{i}^{2}}{2} + g h_{r} & = & \frac{v_{f}^{2}}{2} + g h_{b} \end{array}$

Here, h_r is the released height, h_b is the bottom height.

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