# A solid disk, spinning counter-clockwise, has a mass of 8 kg and a radius of 5/2 m. If a point on the edge of the disk is moving at 9/4 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Nov 18, 2016

The angular velocity $= \frac{9}{40} H z$
The angular momentum $= \frac{45}{8} k g {m}^{2} {s}^{- 1}$

#### Explanation:

First, let's calculate the angular velocity, $\omega$

$\omega = \frac{v}{r}$

$v = \frac{9}{4} m {s}^{- 1}$

$r = \frac{5}{2} m$

$\omega = \frac{\frac{9}{4}}{\frac{5}{2}} = \frac{9}{4} \cdot \frac{2}{5} = \frac{9}{40} H z$

The angular momentum, $L$ is

$L = I \cdot \omega$

where, $I =$moment of inertia of the disc

Here, $I = \frac{m {r}^{2}}{2}$

$m = 8 k g$

$r = \frac{5}{2} m$

$I = 8 \cdot \frac{25}{4} \cdot \frac{1}{2} = 25 k g {m}^{2}$

So, $L = 25 \cdot \frac{9}{40} = \frac{45}{8} k g {m}^{2} {s}^{- 1}$