# A solid disk, spinning counter-clockwise, has a mass of 3 kg and a radius of 7/5 m. If a point on the edge of the disk is moving at 8/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Nov 13, 2016

The angular momentum is $= \frac{28}{5} k g {m}^{2} {s}^{- 1}$
The angular velocity is $= \frac{40}{12} H z$

#### Explanation:

mass $m = 3 k g$

radius of solid disc $r = \frac{7}{5} m$

Velocity on the edge $v = \frac{8}{3} m {s}^{- 1}$

The angular velocity $\omega = \frac{v}{r} H z$

$\omega = \frac{\frac{8}{3}}{\frac{7}{5}} = \frac{8}{3} \cdot \frac{5}{7} = \frac{40}{21} H z$

The angular momentum $L = I \cdot \omega$

where $I =$moment of inertia

For a solid disc, $I = \frac{m {r}^{2}}{2} = 3 \cdot \frac{49}{25} \cdot \frac{1}{2} = \frac{147}{50} k g {m}^{2}$

So, angular momentum $L = I \cdot \omega = \frac{147}{50} \cdot \frac{40}{21} = \frac{28}{5} k g {m}^{2} {s}^{- 1}$