A solid disk, spinning counter-clockwise, has a mass of #7 kg# and a radius of #5/2 m#. If a point on the edge of the disk is moving at #6/5 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Feb 22, 2018

Answer:

The angular momentum is #=10.5kgm^2s^-1# and the angular velocity is #=0.48rads^-1#

Explanation:

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=6/5ms^(-1)#

#r=5/2m#

So,

The angular velocity is positive (spinning counter clockwise)

#omega=(6/5)/(5/2)=12/25=0.48rads^-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= 7kg#

So, #I=7*(5/2)^2/2=175/8kgm^2#

The angular momentum is

#L=175/8*0.48=10.5kgm^2s^-1#