A solid disk, spinning counter-clockwise, has a mass of #9 kg# and a radius of #6 m#. If a point on the edge of the disk is moving at #15 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Aug 9, 2017

Answer:

The angular momentum is #=405kgm^2s^-1# and the angular velocity is #=2.5rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=15ms^(-1)#

#r=6m#

So,

#omega=(15)/(6)=5/2=2.5rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=9*(6)^2/2=162kgm^2#

The angular momentum is

#L=162*2.5=405kgm^2s^-1#