# A solid disk, spinning counter-clockwise, has a mass of 3 kg and a radius of 8/9 m. If a point on the edge of the disk is moving at 11/4 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Oct 5, 2017

The angular momentum is $= 3.67 k g {m}^{2} {s}^{-} 1$
The angular velocity is $= 3.09 r a {\mathrm{ds}}^{-} 1$

#### 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,

$v = \frac{11}{4} m {s}^{- 1}$

$r = \frac{8}{9} m$

So,

The angular velocity is $\omega = \frac{\frac{11}{4}}{\frac{8}{9}} = \frac{99}{32} = 3.09 r a {\mathrm{ds}}^{-} 1$

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 $m = 3 k g$

The radius of the disc is $r = \frac{8}{9} m$

So, $I = 3 \cdot {\left(\frac{8}{9}\right)}^{2} / 2 = \frac{32}{27} k g {m}^{2}$

The angular momentum is

$L = \frac{32}{27} \cdot \frac{99}{32} = 3.67 k g {m}^{2} {s}^{-} 1$