# A solid disk, spinning counter-clockwise, has a mass of 8 kg and a radius of 3/2 m. If a point on the edge of the disk is moving at 9/8 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Nov 20, 2016

The angular momentum is $= \frac{27}{4} k g m {s}^{- 1}$
The angular velocity is $= \frac{3}{4} H z$

#### Explanation:

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

where
$v = v e l o c i t y = \frac{9}{8} m {s}^{- 1}$
$r = r a \mathrm{di} u s = \frac{3}{2} m$

so, $\omega = \frac{9}{8} \cdot \frac{2}{3} = \frac{3}{4} H z$

The angular momentum is

$L = I \omega$

Where $I$=moment of inertia

In the case of a solid disc, $I = \frac{m {r}^{2}}{2}$

$m =$mass of the disc, $= 8 k g$

$r =$ radius of the disc, $= \frac{3}{2} m$

Therefore, $I = \frac{1}{2} \cdot 8 \cdot \frac{9}{4} = 9 k g {m}^{2}$

The angular momentum is $L = 9 \cdot \frac{3}{4} = \frac{27}{4} k g m {s}^{- 1}$