# A solid disk, spinning counter-clockwise, has a mass of 8 kg and a radius of 5/2 m. If a point on the edge of the disk is moving at 5/8 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Apr 19, 2016

$\ldots .$

#### Explanation:

$r = \frac{5}{2} m$, The circumference of the disk is $C = 2 \setminus \pi r$

Because a point on the outside of the disk moves along the circumference with $v = \frac{5}{8} \frac{m}{s}$, one rotation is completed in

$T = \frac{C}{v} = \frac{2 \setminus \pi r}{v}$

The angular velocity, $w$, has units of radians per second. One rotation has $2 \setminus \pi$ radians so

$w = \frac{2 \setminus \pi}{T} = \frac{v}{r} = 0.25$ radians per second

As a side note, you could obtain this immediately if you are familiar with the expression ${v}_{\setminus} \textrm{p e r p e n \mathrm{di} c \underline{a} r} = w \setminus \times r$

The angular momentum of the disk is $L = I w$, where $I$ is the inertia of the disk. You could obtain the inertia by integration, or just look it up in a table.

For a uniform disk, $I = \frac{1}{2} m {R}^{2} = 25 \setminus \quad k g \setminus \cdot {m}^{2}$.

Therefore,

$L = I w = 6.25 \setminus \quad k g \setminus \cdot {m}^{2} / s$

Done, but if you wanted to solve for the inertia on your own without looking it up, you would have to integrate.

Assume the disk has uniform density $\setminus \sigma = \frac{m a s s}{a r e a} = \frac{m}{\setminus \pi {R}^{2}}$

$I = \setminus \int {r}^{2} \mathrm{dm} = \setminus \int {r}^{2} \left(\setminus \sigma \mathrm{dA}\right)$

where $\mathrm{dm} = \setminus \sigma \mathrm{dA}$. You just need to substitute $\left(\mathrm{dA} = r \setminus \quad d \setminus \phi \setminus \quad \mathrm{dr}\right)$ and integrate to get the answer, $I = \frac{1}{2} m {R}^{2}$.