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

a) $14.2 \ Js$ b) $57 \ Nm$
a) We use the parallel axis theorem to obtain: $I_t = .048 +mR^2 = .048+(.88)(.43)^2 = .21$ Recall that $\omega=\frac{v}{r}$, it follows: $L=I\omega=I\frac{v}{r}=(.021)\frac{50}{.74}=14.2 \ Js$ b) We obtain: $\tau = \frac{L}{t}=\frac{14.2}{.25}=57 \ Nm$