A solid disk, spinning counter-clockwise, has a mass of #4 kg# and a radius of #3/7 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?

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Answer 1 · 2017-08-09T06:11:41

The angular momentum is #=2.14kgm^2s^-1# and the angular velocity is #=5.83rads^-1#

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

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

#r=3/7m#

So,

#omega=(5/2)/(3/7)=35/6=5.83rads^-1#

The angular momentum is #L=Iomega#

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

So, #I=4*(3/7)^2/2=18/49kgm^2#

The angular momentum is

#L=18/49*35/6=2.14kgm^2s^-1#