A solid disk, spinning counter-clockwise, has a mass of #12 kg# and a radius of #7/5 m#. If a point on the edge of the disk is moving at #5/3 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Apr 27, 2017

Answer:

The angular momentum is #=14kgm^2s^-1#
The angular velocity is #=1.19rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=5/3ms^(-1)#

#r=7/5m#

So,

#omega=(5/3)/(7/5)=1.19rads^-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*(7/5)^2/2=294/25kgm^2#

The angular momentum is

#L=294/25*1.19=14kgm^2s^-1#