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

1 Answer
Jul 1, 2017

Answer:

The angular momentum is #=6kgm^2s^-1# and the angular velocity is #=1rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=2ms^(-1)#

#r=2m#

So,

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

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=3*(2)^2/2=6kgm^2#

The angular momentum is

#L=6*1=6kgm^2s^-1#