# A solid disk, spinning counter-clockwise, has a mass of 2 kg and a radius of 5 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?

Jun 28, 2018

The angular momentum is $= 30 k g {m}^{2} {s}^{-} 1$ and The angular velocity is $= 1.2 r a {\mathrm{ds}}^{-} 1$

#### 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,

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

$r = 5 m$

So,

The angular velocity is

$\omega = \frac{6}{5} = 1.2 r a {\mathrm{ds}}^{-} 1$

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 $m = 2 k g$

So, the moment of inertia is

$I = 2 \cdot {\left(5\right)}^{2} / 2 = 25 k g {m}^{2}$

The angular momentum is

$L = I \omega = 25 \cdot 1.2 = 30 k g {m}^{2} {s}^{-} 1$