## University Physics with Modern Physics (14th Edition)

Published by Pearson

# Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 331: 10.29

#### Answer

(a) $\omega = 0.309~rad/s$ (b) $Work = 100~J$ (c) The average power is 6.67 watts.

#### Work Step by Step

(a) The angular impulse provided by the child will be equal in magnitude to the angular momentum of the merry-go-round. $I\omega = \tau ~t$ $\omega = \frac{R~F ~t}{I}$ $\omega = \frac{(2.40)(18.0~N)(15.0~s)}{2100~kg~m^2}$ $\omega = 0.309~rad/s$ (b) The work done by the child is equal to the rotational kinetic energy of the merry-go-round. $Work = K_{rot}$ $Work = \frac{1}{2}I\omega^2$ $Work = \frac{1}{2}(2100~kg~m^2)(0.309~rad/s)^2$ $Work = 100~J$ (c) $P_{ave} = \frac{Work}{t}$ $P_{ave} = \frac{100~J}{15.0~s}$ $P_{ave} = 6.67~W$ The average power is 6.67 watts.

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