A solid disk, spinning counter-clockwise, has a mass of #8 kg# and a radius of #5/2 m#. If a point on the edge of the disk is moving at #9/8 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Apr 25, 2017

Answer:

The angular momentum is #=11.25kgm^2s^-1#
The angular velocity is #=0.45rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=9/8ms^(-1)#

#r=5/2m#

So,

#omega=(9/8)/(5/2)=0.45rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=8*(5/2)^2/2=25kgm^2#

The angular velocity is

#L=25*0.45=11.25kgm^2s^-1#