# A solid disk, spinning counter-clockwise, has a mass of 11 kg and a radius of 4/7 m. If a point on the edge of the disk is moving at 8/5 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Dec 26, 2016

The angular momentum of the disk is $\approx 5$ $\frac{k g {m}^{2}}{s}$ and the angular velocity is $\frac{14}{5} \frac{r a d}{s} \left(\approx 2.8 \frac{r a d}{s}\right)$.

#### Explanation:

Angular momentum is given by $\vec{L} = I \omega$, where $I$ is the moment of inertia of the object, and $\omega$ is the angular velocity of the object.

The moment of inertia of a solid disk is given by $I = \frac{1}{2} m {r}^{2}$, and angular velocity is given by $\frac{v}{r}$, where $v$ is the tangential velocity and $r$ is the radius.

We are given that $m = 11 k g$, $r = \frac{4}{7} m$, and $v = \frac{8}{5} \frac{m}{s}$. We can use these values to calculate the moment of inertia and angular velocity, and ultimately the angular momentum.

$\omega = \frac{v}{r} = \frac{\frac{8}{5} \frac{m}{s}}{\frac{4}{7} m}$

$\implies \omega = \frac{14}{5} \frac{r a d}{s}$

This is the angular velocity.

$I = \frac{1}{2} m {r}^{2} = \frac{1}{2} \left(11 k g\right) {\left(\frac{4}{7} m\right)}^{2}$

$\implies I = \frac{88}{49} k g {m}^{2}$

This is the moment of inertia.

We can now calculate the angular momentum:

$\vec{L} = I \omega = \left(\frac{88}{49} k g {m}^{2}\right) \cdot \left(\frac{14}{5} \frac{r a d}{s}\right)$

$\implies \vec{L} = \frac{176}{35} \frac{k g {m}^{2}}{s}$

$\implies \vec{L} \approx 5 \frac{k g {m}^{2}}{s}$