# Chapter 11 - Exercises and Problems - Page 201: 40

a) $14.2 \ Js$ b) $57 \ Nm$

#### Work Step by Step

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: $\tau = \frac{L}{t}=\frac{14.2}{.25}=57 \ Nm$

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