# A solid disk, spinning counter-clockwise, has a mass of 7 kg and a radius of 5/2 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?

Dec 14, 2016

The angular momentum is $= 146.6 k g m {s}^{- 1}$

The angular velocity is $= 3.35 r a {\mathrm{ds}}^{- 1}$

#### Explanation:

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

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

$r = \frac{5}{2} m$

$\omega = \frac{4}{3} \cdot \frac{2}{5} = \frac{8}{15} H z = \frac{16}{15} \pi r a {\mathrm{ds}}^{- 1} = 3.35 r a {\mathrm{ds}}^{- 1}$

The angulur momentum is $L = I \omega$

The moment of inertia of a solid disc is $= m {r}^{2} / 2$

$I = 7 \cdot {\left(\frac{5}{2}\right)}^{2} = 7 \cdot \frac{25}{4} k g {m}^{2}$

Therefore,

$L = \frac{16}{15} \cdot \pi \cdot \frac{175}{4} k g {m}^{2} {s}^{- 1} = 146.6 k g m {s}^{- 1}$