# A solid disk, spinning counter-clockwise, has a mass of 7 kg and a radius of 5 m. If a point on the edge of the disk is moving at 2 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Oct 29, 2017

$\omega = 0.4 \textcolor{w h i t e}{i} {s}^{-} 1$
$L = 7 \textcolor{w h i t e}{i} k g {m}^{2} {s}^{-} 1$

#### Explanation:

Angular Velocity:

$\omega = \frac{\Delta \theta}{\Delta t}$

and $v = r \cdot \left(\frac{\Delta \theta}{\Delta t}\right) = r \omega$

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

Here,

• $v = 2 \frac{m}{s}$
• r=5m/

$\therefore \omega = \frac{2 \frac{m}{s}}{5 m} = 0.4 \textcolor{w h i t e}{i} {s}^{-} 1$

Angular Momentum:

$L = I \omega$

Here,

• $I \left(\text{moment of inertia}\right) = \frac{m {r}^{2}}{2}$
$= \frac{7 k g \cdot {\left(5 m\right)}^{2}}{2} = 17.5 \textcolor{w h i t e}{i} k g {m}^{2}$
• $\omega = 0.4 \textcolor{w h i t e}{i} {\text{rads}}^{-} 1$

$\therefore L = 17.5 \textcolor{w h i t e}{i} k g {m}^{2} \cdot 0.4 \textcolor{w h i t e}{i} {\text{rads}}^{-} 1$
$= 7 \textcolor{w h i t e}{i} k g {m}^{2} {s}^{-} 1$