Question

A disk of mass 2M and radius R is freely spinning in space at a constant speed ω when a second disk of mass M and radius R with no angular speed is carefully dropped on top such that the centers of the disks align. Friction causes the two items to move together as one. What is the new angular speed of the pair?

Answer 1

2ω/3

Step-by-step explanation:

The moments of inertia of the 1st disk is:

`I_1 = (2MR^2)/2 = MR^2`

The moments of inertia of the 2nd disk is:

`I_2 = MR^2/2`

So the total moments of inertia of the system of 2 disks after the drop is:

`I = I_1 + I_2 = MR^2 + MR^2/2 = 1.5MR^2`

Using law of angular momentum conservation we have the following equation for before and after the drop

`I_1\omega = I\omega_2`

`\omega_2 = \omega(I_1)/(I) = \omega(MR^2)/(1.5MR^2) = 2\omega/3`

Answer 2

The new angular speed is 2/3omega

Step-by-step explanation:

See attached handwritten document for more details