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?

1 Answer
Apr 19, 2016

Answer:

#....#

Explanation:

#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#.

Therefore,

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

Done, but if you wanted to solve for the inertia on your own without looking it up, you would have to integrate.

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#.