A solid disk, spinning counter-clockwise, has a mass of #8 kg# and a radius of #5/2 m#. If a point on the edge of the disk is moving at #5/8 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Question

A solid disk, spinning counter-clockwise, has a mass of #8 kg# and a radius of #5/2 m#.  If a point on the edge of the disk is moving at #5/8 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Isabella Alvarez · Accepted Answer

#....#

#r=5/2m#, The circumference of the disk is #C=2\pi r#

Because a point on the outside of the disk moves along the circumference with #v=5/8 m/s#, one rotation is completed in

#T=C/v={2\pi r}/v#

The angular velocity, #w#, has units of radians per second. One rotation has #2\pi# radians so

#w={2\pi}/T=v/r= 0.25# radians per second

As a side note, you could obtain this immediately if you are familiar with the expression #v_\text{perpendicular}=w\times r#

The angular momentum of the disk is #L=Iw#, where #I# is the inertia of the disk. You could obtain the inertia by integration, or just look it up in a table.

For a uniform disk, #I=1/2mR^2= 25 \quad kg\cdot m^2#.

Consequently,

#L=Iw=6.25 \quad kg\cdot m^2/s#

Done, but you would need to integrate if you wanted to figure out the inertia on your own without consulting a reference.

Assume the disk has uniform density #\sigma={mass}/{area}=m/{\pi R^2}#

#I=\int r^2 dm=\int r^2 (\sigma dA)#

where #dm=\sigma dA#. You just need to substitute #(dA=r\quad d\phi\quad dr)# and integrate to get the answer, #I=1/2 mR^2#.

Benjamin Asiedu · Answer

undefined Angular momentum (L) of the disk:

$$ L = I \cdot \omega $$

Where:
- $ I $ is the moment of inertia of the disk.
- $ \omega $ is the angular velocity of the disk.

Moment of inertia (I) of a solid disk:

$$ I = \frac{1}{2} m r^2 $$

Given:
- $ m = 8 \, \text{kg} $
- $ r = \frac{5}{2} \, \text{m} $

$$ I = \frac{1}{2} \cdot 8 \cdot \left(\frac{5}{2}\right)^2 = \frac{1}{2} \cdot 8 \cdot \frac{25}{4} = 50 \, \text{kg} \cdot \text{m}^2 $$

Now, we need to find the angular velocity ($ \omega $) using the linear velocity ($ v $):

$$ v = r \cdot \omega $$

Given:
- $ v = \frac{5}{8} \, \text{m/s} $
- $ r = \frac{5}{2} \, \text{m} $

$$ \frac{5}{8} = \frac{5}{2} \cdot \omega $$

$$ \omega = \frac{5}{8} \cdot \frac{2}{5} = \frac{1}{8} \, \text{rad/s} $$

Now, we can calculate the angular momentum:

$$ L = 50 \cdot \frac{1}{8} = 6.25 \, \text{kg} \cdot \text{m}^2/\text{s} $$

So, the disk's angular momentum is $ 6.25 \, \text{kg} \cdot \text{m}^2/\text{s} $ and its angular velocity is $ \frac{1}{8} \, \text{rad/s} $.