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

Jan 8, 2018

The angular momentum is $= 7.59 k g {m}^{2} {s}^{-} 1$ and the angular velocity is $= 12 r a {\mathrm{ds}}^{-} 1$

#### Explanation:

The angular velocity is

$\omega = \frac{\Delta \theta}{\Delta t}$

$v = r \cdot \left(\frac{\Delta \theta}{\Delta t}\right) = r \omega$

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

where,

$v = \frac{9}{2} m {s}^{- 1}$

$r = \frac{3}{8} m$

So,

The angular velocity is

$\omega = \frac{\frac{9}{2}}{\frac{3}{8}} = 12 r a {\mathrm{ds}}^{-} 1$

The angular momentum is $L = I \omega$

where $I$ is the moment of inertia

For a solid disc, $I = \frac{m {r}^{2}}{2}$

The mass of the disc is $m = 9 k g$

So, $I = 9 \cdot {\left(\frac{3}{8}\right)}^{2} / 2 = \frac{81}{128} k g {m}^{2}$

The angular momentum is

$L = \frac{81}{128} \cdot 12 = 7.59 k g {m}^{2} {s}^{-} 1$