# A solid disk, spinning counter-clockwise, has a mass of 11 kg and a radius of 4/7 m. If a point on the edge of the disk is moving at 2/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Nov 22, 2016

#### Answer:

The angular velocity is $= \frac{7}{6} H z$
The angular momentum is $= \frac{308}{147} k g {m}^{2} {s}^{- 1}$

#### Explanation:

The angular velocity is given by the formula $\omega = \frac{v}{r}$

The velocity $v = \frac{2}{3} m {s}^{- 1}$

The radius $r = \frac{4}{7} m$

So, $\omega = \frac{\frac{2}{3}}{\frac{4}{7}} = \frac{2}{3} \cdot \frac{7}{4} = \frac{7}{6} H z$

The angular momentum, $L = I \omega$

where $I =$moment of inertia

For a solid disc,

$I = \frac{m {r}^{2}}{2}$

$I = \frac{1}{2} \cdot 11 \cdot {\left(\frac{4}{7}\right)}^{2} = \frac{11 \cdot 8}{49} = \frac{88}{49} k g {m}^{2}$

So,

$L = \frac{88}{49} \cdot \frac{7}{6} = \frac{308}{147} k g {m}^{2} {s}^{- 1}$