# A solid disk, spinning counter-clockwise, has a mass of 2 kg and a radius of 7/4 m. If a point on the edge of the disk is moving at 7/9 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Nov 11, 2016

The angular momentum is $= \frac{49}{36} k g {m}^{2} {s}^{- 1}$
The angular velocity is $= \frac{4}{9} H z$

#### Explanation:

The angular velocity is $\omega = \frac{v}{r}$

Here $v = \frac{7}{9} m {s}^{- 1}$ and $r = \frac{7}{4} m$

therefore $\omega = \frac{7}{9} \cdot \frac{4}{7} = \frac{4}{9} H z$

The angular momentum is $L = I \cdot \omega$
Where $I$ is the moment of inertia
For a solid disc $I = \frac{m {r}^{2}}{2} = 2 \cdot {\left(\frac{7}{4}\right)}^{2} / 2 = \frac{49}{16} k g {m}^{2}$

So, the angular momentum is $\frac{49}{16} \cdot \frac{4}{9} = \frac{49}{36} k g {m}^{2} {s}^{- 1}$