# A solid disk, spinning counter-clockwise, has a mass of 9 kg and a radius of 3/2 m. If a point on the edge of the disk is moving at 5/2 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Feb 15, 2018

The angular momentum is $= 16.875 k g {m}^{2} {s}^{-} 1$ and the angular velocity is $= 1.67 r a {\mathrm{ds}}^{-} 1$

#### 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,

$v = \frac{5}{2} m {s}^{- 1}$

$r = \frac{3}{2} m$

So,

The angular velocity is positive (spinning counter clockwise)

$\omega = \frac{\frac{5}{2}}{\frac{3}{2}} = \frac{5}{3} = 1.67 r a {\mathrm{ds}}^{-} 1$

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 is $m = 9 k g$

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

The angular momentum is

$L = \frac{81}{8} \cdot \frac{5}{3} = 16.875 k g {m}^{2} {s}^{-} 1$