Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

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

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 then 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. $KE_{trans}+KE_{rot} = PE$ $\frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2 = PE$ $\frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2 = PE$ $\frac{1}{2}Mv^2+\frac{1}{5}Mv^2 = Mgh$ $\frac{7}{10}v^2 = gh$ $v = \sqrt{\frac{10~gh}{7}}$ $v = \sqrt{\frac{(10)(9.80~m/s^2)(0.89~m)}{7}}$ $v = 3.53~m/s$ Next, we 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).
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.