## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 10 - Exercises and Problems - Page 184: 27

#### Answer

a) $1.1\times10^{-3} Nm$ b) $3.63\times10^{-3} Nm$

#### Work Step by Step

a) We use the equation for the rotational inertia of a hollow cylinder, and we add the inertia of the end disks: $I = \frac{1}{2}(.033)(.085^2)(2 \ disks) + (12)(.085^2)=1.1\times10^{-3}$ b) We find: $\tau = I\alpha = (1.1\times10^{-3})(3.3)=3.63\times10^{-3} Nm$

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