A solid disk, spinning counter-clockwise, has a mass of #9 kg# and a radius of #3/2 m#. If a point on the edge of the disk is moving at #5/2 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Feb 15, 2018

Answer:

The angular momentum is #=16.875kgm^2s^-1# and the angular velocity is #=1.67rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=5/2ms^(-1)#

#r=3/2m#

So,

The angular velocity is positive (spinning counter clockwise)

#omega=(5/2)/(3/2)=5/3=1.67rads^-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= 9kg#

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

The angular momentum is

#L=81/8*5/3=16.875kgm^2s^-1#