# A solid disk, spinning counter-clockwise, has a mass of 7 kg and a radius of 3 m. If a point on the edge of the disk is moving at 6 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Nov 19, 2017

The angular momentum is $= 63 k g {m}^{2} {s}^{-} 1$ and the angular velocity is $= 2 r a {\mathrm{ds}}^{-} 1$

#### 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,

$v = 6 m {s}^{- 1}$

$r = 3 m$

So,

The angular velocity is

$\omega = \frac{6}{3} = 2 r a {\mathrm{ds}}^{-} 1$

The angular momentum is $L = I \omega$

where $I$ is the moment of inertia

The mass of the disc is $m = 7 k g$

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

So, $I = 7 \times {\left(3\right)}^{2} / 2 = 31.5 k g {m}^{2}$

The angular momentum is

$L = 31.5 \times 2 = 63 k g {m}^{2} {s}^{-} 1$