A solid disk, spinning counter-clockwise, has a mass of #9 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
May 10, 2017

Answer:

The angular momentum is #=2.36kgm^2s^-1#
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)=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=9*(3/8)^2/2=0.63kgm^2#

THe angular momentum is

#L=0.63*3.73=2.36kgm^2s^-1#