# A solid disk, spinning counter-clockwise, has a mass of 13 kg and a radius of 4/5 m. If a point on the edge of the disk is moving at 7/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Dec 28, 2017

The angular momentum is $= 12.15 k g {m}^{2} {s}^{-} 1$ and the angular velocity is $= 2.92 r a {\mathrm{ds}}^{-} 1$

#### 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,

$v = \frac{7}{3} m {s}^{- 1}$

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

So,

The angular velocity is

$\omega = \frac{\frac{7}{3}}{\frac{4}{5}} = \frac{35}{12} = 2.92 r a {\mathrm{ds}}^{-} 1$

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 $m = 13 k g$

So, $I = 13 \cdot {\left(\frac{4}{5}\right)}^{2} / 2 = 4.16 k g {m}^{2}$

The angular momentum is

$L = 4.16 \cdot 2.92 = 12.15 k g {m}^{2} {s}^{-} 1$