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: 73

Answer

The sphere rolls a distance of 4.3 meters up the incline before reversing direction.

Work Step by Step

We can find an expression for the kinetic energy of the sphere while it is rolling at a speed of 5.0 m/s. Note that the kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy. $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{3}MR^2)(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{3}Mv^2$ $KE = \frac{5}{6}Mv^2$ We can use conservation of energy to find the maximum height reached by the sphere. The potential energy at the maximum height will be equal to the kinetic energy at the bottom of the incline. $PE = KE$ $Mgh = \frac{5}{6}Mv^2$ $h = \frac{5v^2}{6g}$ $h = \frac{(5)(5.0~m/s)^2}{(6)(9.80~m/s^2)}$ $h = 2.126~m$ We can use the angle to find the distance $d$ the sphere rolls up the incline. $\frac{h}{d} = sin(\theta)$ $d = \frac{h}{sin(\theta)}$ $d = \frac{2.126~m}{sin(30^{\circ})}$ $d = 4.3~m$ The sphere rolls a distance of 4.3 meters up the incline before reversing direction.
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.