# A solid disk, spinning counter-clockwise, has a mass of 12 kg and a radius of 6 m. If a point on the edge of the disk is moving at 2 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Nov 28, 2016

The angular velocity $= \frac{1}{3} H z$
The angular momentum is $= 72 k g m {s}^{- 1}$

#### Explanation:

The angular velocity is

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

Where $v = v e l o c i t y = 2 m {s}^{- 1}$

and $r = r a \mathrm{di} u s = 6 m$

So, $\omega = \frac{2}{6} = \frac{1}{3} H z$

The angular momentum is

$L = I \omega$

Where, $I =$moment of inertia

The moment of inertia of a solid disc is $= \frac{m {r}^{2}}{2}$

where, $m =$ mass of the disc

$I = 12 \cdot 6 \cdot \frac{6}{2} = 216 k g {m}^{2}$

and $L = 216 \cdot \frac{1}{3} = 72 k g m {s}^{- 1}$