# A solid disk, spinning counter-clockwise, has a mass of 5 kg and a radius of 8 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?

#### Answer:

$120 \setminus \setminus {\textrm{k g m}}^{2} \setminus \textrm{/ s}$ & $0.75 \setminus \setminus \textrm{r a \frac{d}{s}}$

#### Explanation:

Mas of solid disk $m = 5 \setminus k g$

Radius of solid disk $r = 8 \setminus m$

Peripheral speed of disk $v = r \setminus \omega = 6 \setminus \frac{m}{s}$

The angular velocity $\setminus \omega$ of the disk

$\setminus \omega = \frac{v}{r} = \frac{6}{8} = 0.75 \setminus \setminus \textrm{r a \frac{d}{s}}$

Mass moment of inertia ($I$) of solid disk

$I = \frac{1}{2} m {r}^{2} = \frac{1}{2} \left(5\right) {\left(8\right)}^{2} = 160 \setminus k g {m}^{2}$

Now, the angular momentum $J$ of the solid disk is given as

$J = I \setminus \omega$

$= 160 \setminus \cdot 0.75$

$= 120 \setminus \setminus {\textrm{k g m}}^{2} \setminus \textrm{/ s}$