A solid disk, spinning counter-clockwise, has a mass of 4 kg4kg and a radius of 3 m3m. If a point on the edge of the disk is moving at 6/5 m/s65ms in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Feb 22, 2017

The angular momentum is =45.2kgm^2s^-1=45.2kgm2s1
The angular velocity is =2.51rads^-1=2.51rads1

Explanation:

The angular velocity is

omega=v/rω=vr

where,

v=6/5ms^(-1)v=65ms1

r=3mr=3m

So,

omega=(6/5)/(3)*2pi=4pi/5=2.51rads^-1ω=6532π=4π5=2.51rads1

The angular momentum is L=IomegaL=Iω

Where II is the moment of inertia

For a solid disc, I=(mr^2)/2I=mr22

So, I=4*(3)^2/2=18kgm^2I=4(3)22=18kgm2

L=I* omega=2.51*18=45.2kgm^2s^-1L=Iω=2.5118=45.2kgm2s1