A solid disk, spinning counter-clockwise, has a mass of #12 kg# and a radius of #5/2 m#. If a point on the edge of the disk is moving at #4/3 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Jul 22, 2017

The angular momentum is #=20kgm^2s^-1# and the angular velocity is #=0.053rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=4/3ms^(-1)#

#r=5/2m#

So,

#omega=(4/3)/(5/2)=8/15=0.053rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=12*(5/2)^2/2=75/2kgm^2#

The angular momentum is

#L=75/2*8/15=20kgm^2s^-1#