University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 298: 9.59


(a) $A = 0.600~m/s^3$ (b) $\alpha(t) = (2.40~rad/s^3)~t$ (c) $t = 3.54~s$ (d) $\theta = 17.7~rad$

Work Step by Step

(a) $a(t)= A~t$ $a(3)= A~(3.00~s) = 1.80~m/s^2$ $A = 0.600~m/s^3$ (b) $\alpha(t) = \frac{a(t)}{r}$ $\alpha(t) = \frac{(0.600~m/s^3)~t}{0.250~m}$ $\alpha(t) = (2.40~rad/s^3)~t$ (c) $\omega (t) = \int_{0}^{t}\alpha(t)~dt$ $\omega (t) = \int_{0}^{t}(2.40~rad/s^3)~t~dt$ $\omega (t) = (1.20~rad/s^3)~t^2$ We can find $t$ when $\omega = 15.0~rad/s$ $(1.20~rad/s^3)~t^2 = 15.0~rad/s$ $t = \sqrt{\frac{15.0~rad/s}{1.20~rad/s^3}}$ $t = 3.54~s$ (d) $\theta (t) = \int_{0}^{t}~\omega(t)~dt$ $\theta (t) = \int_{0}^{t}~(1.20~rad/s^3)~t^2~dt$ $\theta (t) = (0.400~rad/s^3)~t^3$ We can find $\theta$ when $t = 3.54~s$: $\theta = (0.400~rad/s^3)(3.54~s)^3$ $\theta = 17.7~rad$
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