A solid disk, spinning counter-clockwise, has a mass of #13 kg# and a radius of #4/5 m#. If a point on the edge of the disk is moving at #7/3 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Dec 28, 2017

Answer:

The angular momentum is #=12.15kgm^2s^-1# and the angular velocity is #=2.92rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=7/3ms^(-1)#

#r=4/5m#

So,

The angular velocity is

#omega=(7/3)/(4/5)=35/12=2.92rads^-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=13 kg#

So, #I=13*(4/5)^2/2=4.16kgm^2#

The angular momentum is

#L=4.16*2.92=12.15kgm^2s^-1#