A solid disk, spinning counter-clockwise, has a mass of #9 kg# and a radius of #3/2 m#. If a point on the edge of the disk is moving at #9/2 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Aug 23, 2017

Answer:

The angular momentum is #=30.38kgm^2s^-1#. The angular velocity is #=3rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

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

#r=3/2m#

So,

#omega=(9/2)/(3/2)=3rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=9*(3/2)^2/2=81/8kgm^2#

#L=81/8*3=30.38kgm^2s^-1#