A solid disk, spinning counter-clockwise, has a mass of #9 kg# and a radius of #3/8 m#. If a point on the edge of the disk is moving at #9/2 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Jan 8, 2018

Answer:

The angular momentum is #=7.59kgm^2s^-1# and the angular velocity is #=12rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=9/2ms^(-1)#

#r=3/8m#

So,

The angular velocity is

#omega=(9/2)/(3/8)=12rads^-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=9 kg#

So, #I=9*(3/8)^2/2=81/128kgm^2#

The angular momentum is

#L=81/128*12=7.59kgm^2s^-1#