# A solid disk, spinning counter-clockwise, has a mass of 8 kg and a radius of 3/2 m. If a point on the edge of the disk is moving at 5/8 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Nov 15, 2017

The angular momentum is$= 3.75 k g {m}^{2} {s}^{-} 1$ and the angular velocity is $= 0.417 r a {\mathrm{ds}}^{-} 1$

#### 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,

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

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

So,
The angular velocity is

$\omega = \frac{\frac{5}{8}}{\frac{3}{2}} = 0.417 r a {\mathrm{ds}}^{-} 1$

The angular momentum is $L = I \omega$

where $I$ is the moment of inertia

The mass of the disc is $m = 8 k g$

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

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

The angular momentum is

$L = 9 \cdot 0.417 = 3.75 k g {m}^{2} {s}^{-} 1$