A solid disk, spinning counter-clockwise, has a mass of #4 kg# and a radius of #3/7 m#. If a point on the edge of the disk is moving at #7/9 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Apr 11, 2017

Answer:

The angular momentum is #=0.67kgm^2s^-1#
The angular velocity is #=1.81rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=7/9ms^(-1)#

#r=3/7m#

So,

#omega=(7/9)/(3/7)=49/27=1.81rads^-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*(3/7)^2/2=18/49kgm^2#

The angular momentum is

#L=18/49*1.81=0.67kgm^2s^-1#