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: 34

Answer

(a) $\omega = 88.3~rad/s$ (b) The fraction of the kinetic energy that is rotational kinetic energy is $\frac{2}{7}$ (which is equal to 0.286)

Work Step by Step

(a) We can find the height $h$ of the incline: $\frac{h}{L} = sin(\theta)$ $h = L~sin(\theta)$ $h = (2.1~m)~sin(25^{\circ})$ $h = 0.89~m$ We can use conservation of energy to solve this question. The potential energy at the top of the incline will be converted into translational kinetic energy and rotational kinetic energy at the bottom of the incline. We can find the speed at the bottom. $KE_{trans}+KE_{rot} = PE$ $\frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2 = PE$ $\frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2 = PE$ $\frac{1}{2}Mv^2+\frac{1}{5}Mv^2 = Mgh$ $\frac{7}{10}v^2 = gh$ $v = \sqrt{\frac{10~gh}{7}}$ $v = \sqrt{\frac{(10)(9.80~m/s^2)(0.89~m)}{7}}$ $v = 3.53~m/s$ We can find the angular velocity; $\omega = \frac{v}{R}$ $\omega = \frac{3.53~m/s}{0.040~m}$ $\omega = 88.3~rad/s$ (b) $KE_{rot} = \frac{1}{5}Mv^2$ $KE = \frac{7}{10}Mv^2$ We can find the fraction of the kinetic energy that is rotational kinetic energy. $\frac{KE_{rot}}{KE} = \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2}$ $\frac{KE_{rot}}{KE} = \frac{2}{7}$ The fraction of the kinetic energy that is rotational kinetic energy is $\frac{2}{7}$ (which is equal to 0.286).
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