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 295: 9.7


(a) $a=\pi/4 \space rad$ $\space \space \space \space \space b=2 \space rad/s$ $\space \space \space \space \space c=-0.139 \space rad/s^3$ (b) $t=0\space s$ (c) $\theta=19.5 \space rad$ $\space \space \space \space \space \omega = 9.36 \space rad/s$

Work Step by Step

(a)Derive $\omega$ and $\alpha$ from $\theta$ and plug in the given values: $\theta(t)=a+bt-ct^3$ $\theta(0)=a+b(0s)-c(0s)^3=\frac{\pi}{4}rad$ $a=\frac{\pi}{4}rad$ $\omega(t)=\frac{d\theta}{dt}$ $\omega(t)=b-3ct^2$ $\omega(0)=b-3c(0s)^2=2 \space rad/s$ $b=2 \space rad/s$ $\alpha(t)=\frac{d\omega}{dt}$ $\alpha(t)=-6ct$ $\alpha(1.5s)=-6c(1.5s)=1.25 \space rad/s^2$ $c=\frac{1.25 \space rad/s^2}{-6(1.5s)}$ $c=-0.139 \space rad/s^3$ (b)$\theta(t)=\frac{\pi}{4}+2t+0.139t^3=\frac{\pi}{4}$ $2t+0.139t^3=0$ Factor t then solve for t: $t(2+0.139t^2)=0$ $t=0 \space$ or $t=\sqrt{\frac{-2}{0.139}}=imaginary \space number$ Since t has to be a real number: $t=0$ (c)$\alpha(t)=-6ct=3.5 \space rad/s^2$ $t=\frac{3.5 \space rad/s^2}{-6(-0.139 \space rad/s^3)}$ $t=4.2 \space s$ $\theta(t)=a+bt-ct^3$ $\theta(4.2s)=\frac{\pi}{4}rad+(2 \space rad/s)(4.2s)-(-0.139 \space rad/s^3)(4.2s)^3$ $\theta(4.2s)=19.5 \space rad$ $\omega(t)=b-3ct^2$ $\omega(4.2s)=2 \space rad/s-3(-0.139 \space rad/s^3)(4.2s)^2$ $\omega(4.2s)=9.36 \space rad/s$
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