A solid disk, spinning counter-clockwise, has a mass of #8 kg# and a radius of #3/2 m#. If a point on the edge of the disk is moving at #5/8 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Nov 15, 2017

Answer:

The angular momentum is#=3.75kgm^2s^-1# and the angular velocity is #=0.417rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=5/8ms^(-1)#

#r=3/2m#

So,
The angular velocity is

#omega=(5/8)/(3/2)=0.417rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

The mass of the disc is #m=8kg#

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

So, #I=8*(3/2)^2/2=9kgm^2#

The angular momentum is

#L=9*0.417=3.75kgm^2s^-1#