# A solid disk, spinning counter-clockwise, has a mass of 12 kg and a radius of 5/4 m. If a point on the edge of the disk is moving at 2/9 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Jun 28, 2017

The angular momentum is $= \frac{5}{3} k g {m}^{2} {s}^{-} 1$ and the angular velocity is $= \frac{8}{45} r a {\mathrm{ds}}^{-} 1$

#### 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,

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

$r = \frac{5}{4} m$

So,

$\omega = \frac{\frac{2}{9}}{\frac{5}{4}} = \frac{8}{45} r a {\mathrm{ds}}^{-} 1$

The angular momentum is $L = I \omega$

where $I$ is the moment of inertia

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

So, $I = 12 \cdot {\left(\frac{5}{4}\right)}^{2} / 2 = \frac{75}{8} k g {m}^{2}$

The angular momentum is

$L = \frac{75}{8} \cdot \frac{8}{45} = \frac{5}{3} k g {m}^{2} {s}^{-} 1$