Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 11 - Rolling, Torque, and Angular Momentum - Problems - Page 321: 10d

Answer

$v = 1.80~m/s$

Work Step by Step

In part (c), we found that the total kinetic energy is $~~6.9~J$ The rotational inertia of a hollow sphere is $I = \frac{2}{3}MR^2$ We can find the mass of the sphere: $I = \frac{2}{3}MR^2 = 0.040~kg~m^2$ $M = \frac{(3)(0.040~kg~m^2)}{2R^2}$ $M = \frac{(3)(0.040~kg~m^2)}{(2)(0.15~m)^2}$ $M = 2.67~kg$ We can find an expression for the total kinetic energy: $K = \frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$ $K = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{3}MR^2)(\frac{v}{R})^2$ $K = \frac{1}{2}Mv^2+\frac{1}{3}Mv^2$ $K = \frac{5}{6}Mv^2$ We can find the speed of the center of mass when the sphere has moved $1.0~m$ up the incline from its initial position: $K = \frac{5}{6}Mv^2 = 6.9~J$ $v^2 = \frac{(6)(6.9~J)}{5M}$ $v = \sqrt{\frac{(6)(6.9~J)}{5M}}$ $v = \sqrt{\frac{(6)(6.9~J)}{(5)(2.67~kg)}}$ $v = 1.80~m/s$
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