A solid disk, spinning counter-clockwise, has a mass of #6 kg# and a radius of #2 m#. If a point on the edge of the disk is moving at #2/3 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Mar 26, 2017

Answer:

The angular momentum is #=4kgm^2s^-1#
The angular velocity is #=0.33rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=2/3ms^(-1)#

#r=2m#

So,

angular velocity, #omega=(2/3)/(2)=0.33rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=6*(2)^2/2=12kgm^2#

#L=I omega=12*0.33=4kgm^2s^-1#