# A solid disk, spinning counter-clockwise, has a mass of 6 kg and a radius of 2 m. If a point on the edge of the disk is moving at 5/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Feb 23, 2018

The angular momentum is $= 10 k g {m}^{2} {s}^{-} 1$ and the angular velocity is $= 0.83 r a {\mathrm{ds}}^{-} 1$

#### 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,

$v = \frac{5}{3} m {s}^{- 1}$

$r = 2 m$

So,

The angular velocity is positive (spinning counter clockwise)

$\omega = \frac{\frac{5}{3}}{2} = \frac{5}{6} = 0.83 r a {\mathrm{ds}}^{-} 1$

The angular momentum is $L = I \omega$

where $I$ is the moment of inertia

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

The mass is $m = 6 k g$

So, $I = 6 \cdot {\left(2\right)}^{2} / 2 = 12 k g {m}^{2}$

The angular momentum is

$L = 12 \cdot 0.83 = 10 k g {m}^{2} {s}^{-} 1$