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

a) $1.1\times10^{-3} Nm$ b) $3.63\times10^{-3} Nm$
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: $\tau = I\alpha = (1.1\times10^{-3})(3.3)=3.63\times10^{-3} Nm$