College Physics (4th Edition)

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

Chapter 8 - Problems - Page 316: 61

Answer

$a = 4.02~m/s^2$

Work Step by Step

Since the sphere is rolling without slipping, the force of static friction directed up the incline provides the torque to rotate the sphere. We can find an expression for the force of friction $F_f$: $\tau = I~\alpha$ $R~F_f = I~\alpha$ $R~F_f = I~\frac{a}{R}$ $F_f = I~\frac{a}{R^2}$ $F_f = (\frac{2}{5}MR^2)(\frac{a}{R^2})$ $F_f = \frac{2}{5}Ma$ We can find the acceleration of the sphere: $Mg~sin~\theta-F_f = Ma$ $Mg~sin~\theta = Ma+\frac{2}{5}Ma$ $Mg~sin~\theta = \frac{7}{5}Ma$ $a = \frac{5}{7}~g~sin~\theta$ $a = \frac{5}{7}~(9.80~m/s^2)~sin~35^{\circ}$ $a = 4.02~m/s^2$
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