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: 69

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 highest point reached by the sphere. The potential energy at the highest point on the incline 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.
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