A solid disk, spinning counter-clockwise, has a mass of #12 kg# and a radius of #5/4 m#. If a point on the edge of the disk is moving at #2/9 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Jun 28, 2017

Answer:

The angular momentum is #=5/3kgm^2s^-1# and the angular velocity is #=8/45rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=2/9ms^(-1)#

#r=5/4m#

So,

#omega=(2/9)/(5/4)=8/45rads^-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*(5/4)^2/2=75/8kgm^2#

The angular momentum is

#L=75/8*8/45=5/3kgm^2s^-1#