A solid disk, spinning counter-clockwise, has a mass of #3 kg# and a radius of #8/3 m#. If a point on the edge of the disk is moving at #11/4 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
May 31, 2018

Answer:

The angular momentum is #=11kgm^2s^-1# and the angular velocity is #=1.03125rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=11/4ms^(-1)#

#r=8/3m#

So,

The angular velocity is

#omega=(11/4)/(8/3)=1.03125rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

The mass is #m=3 kg#

So, #I=3*(8/3)^2/2=10.67kgm^2#

The angular momentum is

#L=10.67*1.03125=11kgm^2s^-1#