A solid disk, spinning counter-clockwise, has a mass of #12 kg# and a radius of #3/8 m#. If a point on the edge of the disk is moving at #7/5 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Jul 3, 2017

Answer:

The angular momentum is #=3.15kgm^2s^-1# and the angular velocity is #=3.73rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=7/5ms^(-1)#

#r=3/8m#

So,

#omega=(7/5)/(3/8)=56/15=3.73rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=12*(3/8)^2/2=0.84kgm^2#

The angular momentum is

#L=0.84*3.73=3.15kgm^2s^-1#