# A solid disk, spinning counter-clockwise, has a mass of 4 kg and a radius of 3 m. If a point on the edge of the disk is moving at 8 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Jul 19, 2017

1. $\omega = \frac{8}{3} \frac{1}{s}$
2. $L = 96 \pi \frac{k g \cdot {m}^{2}}{s}$

#### Explanation:

For this equation, we need to find two things: 1. the angular velocity of the disk and 2. the angular momentum of the disk

Angular velocity, $\omega$, is the rate of change of angular displacement, and it is the angular analogue to linear velocity. Likewise, these two share a relationship of $\omega = \frac{v}{r}$
If we substitute our linear velocity of $8 \frac{m}{s}$ and our radius of $3 m$into the equation:

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

$\omega = \frac{8 \frac{\cancel{m}}{s}}{3 \cancel{m}}$

$\omega = \frac{8}{3} \frac{1}{s}$

We now have the answer to part 1.

Part 2. is asking about angular momentum. One equation used to calculate this is $L = I \omega$ where $L$ is angular momentum, $I$ is the moment of inertia, and $\omega$ is angular velocity.

The moment of inertia is an objects resistance to angular acceleration, and its calculation changes on the shape of the object. For a solid disk, the equation is
$I = \frac{1}{2} M {r}^{2}$ where $M$ is the object's mass and $r$ is the radius of the disk. If we substitute in for those values, we see that

$I = \frac{1}{2} \cdot 4 k g \cdot 3 {m}^{2}$

$I = 18 k g \cdot {m}^{2}$

If we substitute in this value into our first equation, we see that

$L = I \omega$

$L = 18 k g \cdot {m}^{2} \cdot \frac{16 \pi}{3} \frac{1}{s}$

$L = 96 \pi \frac{k g \cdot {m}^{2}}{s}$