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 #7/4 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Feb 11, 2017

Answer:

The angular momentum is #=57kgms^(-1)#
The angular veocity is #=13.7rads^-1#

Explanation:

The angular velocity is

#omega=v/r#

where,

#v=7/4ms^(-1)#

#r=4/5m#

So,

#omega=(7/4)/(4/5)*2pi=70/16pi=13.7 rads^-1#

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

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

So, #I=13*(4/5)^2/2=208/50=4.16kgm^2#

#L=13.7*4.16=57kgms^(-1)#