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 5/2 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Jan 14, 2018

$\omega = 1.25 \left(r a \mathrm{di} a n \frac{s}{s}\right)$ and $L = 15 \frac{k g \cdot {m}^{2}}{s}$

Explanation:

The relationship between instantaneous velocity, v and angular velocity, $\omega$, is
$\omega = \frac{v}{r}$

So $\omega = \frac{\frac{5}{2} \frac{m}{s}}{2 m} = 1.25 \frac{r a \mathrm{di} a n s}{s}$

Angular momentum is $I \cdot \omega$ where
$I = \frac{M \cdot {R}^{2}}{2}$

So, the angular momentum, $L$, is
$L = \frac{6 k g \cdot {\left(2 m\right)}^{2}}{2} \cdot 1.25 \frac{r a \mathrm{di} a n s}{s} = 15 \frac{k g \cdot {m}^{2}}{s}$

I hope this helps,
Steve