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 4 - Kinematics in Two Dimensions - Exercises and Problems: 35

Answer

(a) The tangential acceleration is $-2.6~m/s^2$ (b) The crankshaft rotates through 31 revolutions as it stops.

Work Step by Step

(a) We can convert 2500 rpm to units of rad/s $\omega = (2500~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega = 261.8~rad/s$ We can find the angular acceleration. $\alpha = \frac{\omega-\omega_0}{t}$ $\alpha = \frac{0-261.8~rad/s}{1.5~s}$ $\alpha = -174.5~rad/s^2$ We can find the tangential acceleration of a point on the surface. $a_t = \alpha ~r$ $a_t = (-174.5~rad/s^2)(0.015~m)$ $a_t = -2.6~m/s^2$ The tangential acceleration is $-2.6~m/s^2$ (b) $\theta = \omega_0~t+\frac{1}{2}\alpha~t^2$ $\theta = (261.8~rad/s)(1.5~s)+\frac{1}{2}(-174.5~rad/s^2)(1.5~s)^2$ $\theta = 196.4~rad$ We can convert the angle from radians to revolutions. $\theta = (196.4~rad)(\frac{1~rev}{2\pi~rad})$ $\theta = 31~rev$ The crankshaft rotates through 31 revolutions as it stops.
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