# A solid disk, spinning counter-clockwise, has a mass of 6 kg and a radius of 7/5 m. If a point on the edge of the disk is moving at 4/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Apr 19, 2017

The angular momentum is $= 5.6 k g {m}^{2} {s}^{-} 1$
The angular velocity is $= 0.95 r a {\mathrm{ds}}^{-} 1$

#### 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,

$v = \frac{4}{3} m {s}^{- 1}$

$r = \frac{7}{5} m$

So,

$\omega = \frac{\frac{4}{3}}{\frac{7}{5}} = \frac{20}{21} = 0.95 r a {\mathrm{ds}}^{-} 1$

The angular momentum is $L = I \omega$

where $I$ is the moment of inertia

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

So, $I = 6 \cdot {\left(\frac{7}{5}\right)}^{2} / 2 = \frac{147}{25} k g {m}^{2}$

The angular momentum is

$L = \frac{147}{25} \cdot \frac{20}{21} = 5.6 k g {m}^{2} {s}^{-} 1$