A solid disk, spinning counter-clockwise, has a mass of #3 kg# and a radius of #8/9 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
Oct 5, 2017

Answer:

The angular momentum is #=3.67kgm^2s^-1#
The angular velocity is #=3.09rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

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

#r=8/9m#

So,

The angular velocity is #omega=(11/4)/(8/9)=99/32=3.09rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

The mass of the disc is #m=3kg#

The radius of the disc is #r=8/9m#

So, #I=3*(8/9)^2/2=32/27kgm^2#

The angular momentum is

#L=32/27*99/32=3.67kgm^2s^-1#