Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems - Page 223: 34


a) $1.373\times10^{-3} kg ⋅ m^{2} ≈ 1.37\times10^{-3} kg ⋅ m^{2}$ b) $=5.42\times10^{-2} m ⋅ N$

Work Step by Step

(a) The moment of inertia of a cylinder is $\frac{1}{2} MR^{2}$. $I = \frac{1}{2}MR^{2} = \frac{1}{2}(0.380 kg)(0.0850 m)^{2} = 1.373\times10^{-3} kg ⋅ m^{2} ≈ 1.37\times10^{-3} kg ⋅ m^{2}$ (b) The wheel slows down “on its own” from 1500 rpm to rest in 55.0 s. This is used to calculate the frictional torque. $τ_{fr} = I α_{fr} = I\frac{\Delta\omega}{\Delta t} =(1.373\times10^{-3} kg ⋅ m^{2}) \frac{(0-1500 rev/min)(2π rad/rev)(1min/60 s)}{55.0 s} $ $= -3.921\times10^{-3} m ⋅ N$ The net torque causing the angular acceleration is the applied torque plus the (negative) frictional torque. $Στ=τ_{applied} + τ_{fr} = Iα → τ_{applied} = Iα - τ_{fr} = I\frac{\Delta\omega}{\Delta t} - τ_{fr}$ $=(1.373\times10^{-3}kg ⋅ m^{2})\frac{(1750 rev/min)(2π rad/rev)(1min/60 s)}{5.00 s}-(-3.921\times10^{-3} m ⋅ N)$ $=5.42\times10^{-2} m ⋅ N$
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