# A solid disk, spinning counter-clockwise, has a mass of 2 kg and a radius of 5/2 m. If a point on the edge of the disk is moving at 15/4 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Dec 19, 2016

Angular momentum $= 9.375 k g \cdot {m}^{2} {s}^{-} 1$
Angular velocity$= 1.5 r a \mathrm{dc} {\dot{s}}^{-} 1$

#### Explanation:

For any object which is rotating about an axis, each point located on the object has the same angular velocity $\omega$. Its units are $r a \mathrm{dc} {\dot{s}}^{-} 1$. It can be found with the help of velocity $v$ and radius $r$ of a point using the relation

$\omega = \frac{v}{r}$

Inserting given values we get
$\omega = \frac{\frac{15}{4}}{\frac{5}{2}}$
$\implies \omega = \left(\frac{15}{4}\right) \times \left(\frac{2}{5}\right) = 1.5 r a \mathrm{dc} {\dot{s}}^{-} 1$

The angular momentum $L$ of a solid disc can be found using the following formula

L = Ixxω
where $I$ is the moment of inertia of a solid disc and is given as $\frac{1}{2} M {r}^{2}$. $M$ being mass of disc.

If we substitute the value of $\omega$ in terms of $v \mathmr{and} r$, the expression for angular momentum reduces to
$L = \frac{1}{2} M {r}^{2} \times \frac{v}{r}$
$\implies L = \frac{1}{2} M r v$
Inserting given values
$L = \frac{1}{2} \times 2 \times \frac{5}{2} \times \frac{15}{4}$
$\implies L = 9.375 k g \cdot {m}^{2} {s}^{-} 1$