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 `8/5 m/s` in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Answer 1

The angular momentum of the disk is `~~5` `(kgm^2)/s` and the angular velocity is `14/5 (rad)/s (~~2.8 (rad)/s)`.

Explanation:

Angular momentum is given by `vecL=Iomega`, where `I` is the moment of inertia of the object, and `omega` is the angular velocity of the object.

The moment of inertia of a solid disk is given by `I=1/2mr^2`, and angular velocity is given by `v/r`, where `v` is the tangential velocity and `r` is the radius.

We are given that `m=11kg`, `r=4/7m`, and `v=8/5m/s`. We can use these values to calculate the moment of inertia and angular velocity, and ultimately the angular momentum.

`omega=v/r=(8/5m/s)/(4/7m)`

`=>omega=14/5(rad)/s`

This is the angular velocity.

`I=1/2mr^2=1/2(11kg)(4/7m)^2`

`=>I=88/49 kgm^2`

This is the moment of inertia.

We can now calculate the angular momentum:

`vecL=Iomega=(88/49kgm^2)*(14/5(rad)/s)`

`=>vecL=176/35 (kgm^2)/s`

`=>vecL~~5 (kgm^2)/s`