# A solid disk, spinning counter-clockwise, has a mass of 16 kg and a radius of 4/7 m. If a point on the edge of the disk is moving at 8/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Mar 26, 2017

The angular momentum is $= 7.31 k g {m}^{2} {s}^{-} 1$
The angular velocity is $= 2.8 r a {\mathrm{ds}}^{-} 1$

#### 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,

$v = \frac{8}{5} m {s}^{- 1}$

$r = \frac{4}{7} m$

So,

The angular velocity is $\omega = \frac{\frac{8}{5}}{\frac{4}{7}} = \frac{56}{20} = 2.8 r a {\mathrm{ds}}^{-} 1$

The angular momentum is $L = I \omega$

where $I$ is the moment of inertia

mass, $m = 16$

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

So, $I = 16 \cdot {\left(\frac{4}{7}\right)}^{2} / 2 = 2.61 k g {m}^{2}$

$L = I \cdot \omega = 2.61 \cdot 2.8 = 7.31 k g {m}^{2} {s}^{-} 1$