A solid disk, spinning counter-clockwise, has a mass of #15 kg# and a radius of #5/2 m#. If a point on the edge of the disk is moving at #4/3 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Oct 22, 2017

Answer:

The angular momentum is #=24.8kgm^2s^-1#
The angular velocity is #=0.53rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=4/3ms^(-1)#

#r=5/2m#

So,

#omega=(4/3)/(5/2)=8/15=0.53rads^-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=15kg#

So, #I=15*(5/2)^2/2=375/8kgm^2#

The angular momentum is

#L=375/8*0.53=24.8kgm^2s^-1#