A solid disk, spinning counter-clockwise, has a mass of #4 kg# and a radius of #7/4 m#. If a point on the edge of the disk is moving at #5/9 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Apr 7, 2017

Answer:

The angular momentum is #=1.94kgm^2s^-1#
The angular velocity is #=0.32rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=5/9ms^(-1)#

#r=7/4m#

So,

#omega=(5/9)/(7/4)=20/63=0.32rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=4*(7/4)^2/2=49/8kgm^2#

The angular momentum is

#L=49/8*0.32=1.94kgm^2s^-1#