A solid disk, spinning counter-clockwise, has a mass of #6 kg# and a radius of #2 m#. If a point on the edge of the disk is moving at #5/3 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 · 2018-02-23T10:22:55

The angular momentum is #=10kgm^2s^-1# and the angular velocity is #=0.83rads^-1#

The angular velocity is

#omega=(Deltatheta)/(Deltat)#

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

#omega=v/r#

where,

#v=5/3ms^(-1)#

#r=2m#

So,

The angular velocity is positive (spinning counter clockwise)

#omega=(5/3)/(2)=5/6=0.83rads^-1#

The angular momentum is #L=Iomega#

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

The mass is #m= 6kg#

So, #I=6*(2)^2/2=12kgm^2#

The angular momentum is

#L=12*0.83=10kgm^2s^-1#