Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 35

Answer

To have the same speed as the sphere, the hoop should be released from a height of 43 cm

Work Step by Step

We can use the formula for the speed of a rotating object to find an expression for the sphere's speed $v_s$ at the bottom. Let $h_s$ be the initial height of the sphere. $v_s = \sqrt{\frac{2gh_s}{1+c}}$ $v_s = \sqrt{\frac{2gh_s}{1+2/5}}$ $v_s = \sqrt{\frac{10gh_s}{7}}$ We can use the formula for the speed of a rotating object to find an expression for the hoop's speed $v_h$ at the bottom. Let $h_h$ be the initial height of the hoop. $v_h = \sqrt{\frac{2gh_h}{1+c}}$ $v_h = \sqrt{\frac{2gh_h}{1+1}}$ $v_h = \sqrt{gh_h}$ We can equate the speeds to find the value of $h_h$ $v_h = v_s$ $\sqrt{gh_h} = \sqrt{\frac{10gh_s}{7}}$ $gh_h = \frac{10gh_s}{7}$ $h_h = \frac{10h_s}{7}$ $h_h = \frac{(10)(0.30~m)}{7}$ $h_h = 0.43~m$ To have the same speed as the sphere, the hoop should be released from a height of 43 cm.
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