Question

A solid disk, spinning counter-clockwise, has a mass of `11 kg` and a radius of `4/7 m`. If a point on the edge of the disk is moving at `2/3 m/s` in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Answer 1

The angular velocity is `=7/6Hz`
The angular momentum is `=308/147kgm^2s^(-1)`

Explanation:

The angular velocity is given by the formula `omega=v/r`

The velocity `v=2/3ms^(-1)`

The radius `r=4/7 m`

So, `omega=(2/3)/(4/7)=2/3*7/4=7/6 Hz`

The angular momentum, `L=Iomega`

where `I=`moment of inertia

For a solid disc,

`I=(mr^2)/2`

`I=1/2*11*(4/7)^2=(11*8)/(49)=88/49kgm^2`

So,

`L=88/49*7/6=308/147kgm^2s^(-1)`