College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Review & Synthesis: Chapters 6-8 - Review Exercises - Page 324: 24


The speed of the solid cylinder is $~1.84~m/s$

Work Step by Step

We can find the total kinetic energy of a solid cylinder moving at a speed $v$: $KE = KE_{tr}+KE_{rot}$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR^2)~(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{4}Mv^2$ $KE = \frac{3}{4}Mv^2$ We can use conservation of energy to find the speed of the solid cylinder: $KE_f = U_0$ $\frac{3}{4}Mv^2 = Mgh$ $\frac{3}{4}Mv^2 = Mgd~sin~\theta$ $v^2 = \frac{4gd~sin~\theta}{3}$ $v = \sqrt{\frac{4gd~sin~\theta}{3}}$ $v = \sqrt{\frac{(4)(9.80~m/s^2)(0.30~m)~sin~60.0^{\circ}}{3}}$ $v = 1.84~m/s$ The speed of the solid cylinder is $~1.84~m/s$
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.