A solid disk, spinning counter-clockwise, has a mass of #6 kg# and a radius of #3/2 m#. If a point on the edge of the disk is moving at #3 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Feb 27, 2017

Answer:

The angular momentum is #=84.8kgm^2s^-1#
The angular velocity is #=12.57rads^-1#

Explanation:

The angular velocity is

#omega=v/r#

where,

#v=3ms^(-1)#

#r=3/2m#

So,

#omega=(3)/(3/2)*2pi=4pi=12.57rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=6*(3/2)^2/2=27/4kgm^2#

#L=12.57*27/4=84.8kgm^2s^-1#