# A solid disk, spinning counter-clockwise, has a mass of 4 kg and a radius of 3/2 m. If a point on the edge of the disk is moving at 6/7 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

The moment of inertia of the disk =$I = \frac{1}{2} \times m {r}^{2} = \frac{1}{2} \times 4 \times {\left(\frac{3}{2}\right)}^{2} = 4.5 k g {m}^{2}$
Angular velocity $\omega = \frac{v}{r}$, where v = linear velocity of the poit at the age of the disk and r= radius of the disk,
Here $v = \frac{6}{7} \frac{m}{s} \mathmr{and} r = 1.5 m$
Angular velocity $\omega = \frac{v}{r} = \frac{6}{7} \times \frac{1}{1.5} = \frac{4}{7} r a {\mathrm{ds}}^{-} 1$,
Angular momentum $= I \times \omega = 4.5 \times \frac{4}{7} = \frac{18}{7} k g {m}^{2} {s}^{-} 1$