A solid disk, spinning counter-clockwise, has a mass of #2 kg# and a radius of #8/9 m#. If a point on the edge of the disk is moving at #11/9 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Apr 23, 2017

Answer:

The angular momentum is #=1.09kgm^2s^-1#
The angular velocity is #=1.375rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=11/9ms^(-1)#

#r=8/9m#

So,

#omega=(11/9)/(8/9)=1.375rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=2*(8/9)^2/2=64/81kgm^2#

The angular momentum is

#L=64/81*1.375=1.09kgm^2s^-1#