A solid disk, spinning counter-clockwise, has a mass of #6 kg# and a radius of #3/5 m#. If a point on the edge of the disk is moving at #1/6 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Mar 6, 2017

Answer:

The angular momentum is #=1.88kgm^2s^-1#
The angular velocity is #=1.75rads^-1#

Explanation:

The angular velocity is

#omega=v/r#

where,

#v=1/6ms^(-1)#

#r=3/5m#

So,

#omega=(1/6)/(3/5)*2pi=1.75rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

For a solid disc, #I=(mr^2)/2#

So, #I=6*(3/5)^2/2=27/25kgm^2#

#L=1.75*27/25=1.88kgm^2s^-1#