A solid disk, spinning counter-clockwise, has a mass of #4 kg# and a radius of #2 m#. If a point on the edge of the disk is moving at #6/7 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Sep 23, 2017

Answer:

The angular momentum is #=3.43kgm^2s^-1# and the angular velocity is #=0.43rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

The velocity is #v=r*((Deltatheta)/(Deltat))=r omega#

Therefore, #omega=v/r#

where,

#v=6/7ms^(-1)#

#r=2m#

So,

#omega=(6/7)/(2)=3/7=0.43rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=4*(2)^2/2=8kgm^2#

The angular momentum is

#L=8*3/7=24/7=3.43kgm^2s^-1#