A solid disk, spinning counter-clockwise, has a mass of 2 kg and a radius of 7/4 m. If a point on the edge of the disk is moving at 5/9 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Nov 16, 2016

The angular velocity is $= \frac{20}{63} H z$
The angular momentum is $= \frac{35}{36} k g {m}^{2} {s}^{- 1}$

Explanation:

The angular velocity is $\omega = \frac{v}{r}$

where $v = v e l o c i t y = \left(\frac{5}{9}\right) m {s}^{- 1}$

and $r = r a \mathrm{di} u s = \frac{7}{4} m$

$\omega = \frac{\frac{5}{9}}{\frac{7}{4}} = \frac{5}{9} \cdot \frac{4}{7} = \frac{20}{63} H z$

The angular momentum is $L = I \omega$

Where $\omega =$angular velocity

and $I =$ moment of inertia

For a solid disc, $I = \frac{m {r}^{2}}{2}$

$I = 2 \cdot {\left(\frac{7}{4}\right)}^{2} \cdot \left(\frac{1}{2}\right) = \frac{49}{16} k g {m}^{2}$

So, $L = \left(\frac{49}{16}\right) \cdot \frac{20}{63} = \frac{35}{36} k g {m}^{2} {s}^{- 1}$