A puck with a mass of 0.08 kg and a radius of 0.04 m slides along an air table at a speed of 1.5 m/s. It makes a glancing collision with a second puck at rest having a radius of 0.06 m and a mass of 0.12 kg such that their rims just touch. Because their rims are coated with instant-acting glue, the pucks stick together and spin after the collision.
a) What is the angular momentum of the system relative to the center of mass?
b) What is the angular speed about the center of mass?
Center of Mass equation
Angular momentum equation
Conservation of angular momentum equation
The Attempt at a Solution
First I found the center of mass of the two puck system for the y-axis relative to the smaller pucks' center of mass.
CM ( j or y axis ) = [(0.08 kg * 0 m) + (0.12 kg * 0.1 m)] / 0.2 kg = 0.06 m
Modeling the 2 puck system as a non rigid body I plugged the center of mass into the angular momentum equation for the initial situation. The larger puck is stationary so has an angular momentum of zero.
L (total sum) = m * r * v (total sum) = 0.06 m * 0.08 kg * 1.5 m/s = 7.2 x 10^-3 kg * m/s^2
My problem is that I cannot conceptually understand how I am suppose to find the angular speed by using the moment of inertia for this situation. In other words I am having trouble finding a solution to the moment of inertia for the two puck system.
I = 1/2 M R^2 for a solid cylinder and MR^2 for a particle
Can anyone help?