# A solid disk, spinning counter-clockwise, has a mass of #6 kg# and a radius of #3 m#. If a point on the edge of the disk is moving at #5/2 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

##### 1 Answer

The angular momentum of the disk is

#### Explanation:

Angular momentum is given by

The moment of inertia of a solid disk is given by

We are given that

#omega=v/r=(5/2m/s)/(3m)#

#=>omega=5/6(rad)/s#

This is the angular velocity.

**Note:** The radian is a *defined* unit. It's definition of a ratio of two lengths makes it a *pure number* without dimensions. The unit of angle, be it radians, degrees, or revolutions, is really just a name to remind us that we're dealing with angle. Consequently, the radian unit seems to appear or disappear without warning. Above we have velocity divided by distance, which we would expect to have units of *dimensionless* unit (radians) to give **Do not mistake this for a frequency and multiply by #2pi#.**

#I=1/2mr^2=1/2(6kg)(3m)^2#

#=>I=27 kgm^2#

This is the moment of inertia.

We can now calculate the angular momentum:

#vecL=Iomega=(27kgm^2)*(5/6(rad)/s)#

#=>vecL=135/6 (kgm^2)/s#

#=>vecL=22.5 (kgm^2)/s~~23(kgm^2)/s#