A solid disk, spinning counter-clockwise, has a mass of #6 kg# and a radius of #1/2 m#. If a point on the edge of the disk is moving at #12/5 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Jun 3, 2017

Answer:

The angular momentum is #=3.6kgm^2s^-1#
The angular velocity is #=4.8rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

#v=r*((Deltatheta)/(Deltat))=r omega#

#omega=v/r#

where,

#v=12/5ms^(-1)#

#r=1/2m#

So,

#omega=(12/5)/(1/2)=24/5=4.8rads^-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*(1/2)^2/2=0.75kgm^2#

#L=0.75*4.8=3.6kgm^2s^-1#