Conservaion of Angular Momentum and Center of Mass

[Benkyou]

Homework Statement

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?

Homework Equations

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?

 

[Benkyou]

I found that the answer lies in using the parallel axis theorem. I am still not sure how to applies this as there are two objects glued together.

I = Icm + MD^2

Icm = Moment of Inertia for Center of Mass

What I don't understand is how I am suppose to apply the parallel axis theorem in this situation.

 

[LowlyPion]

I found that the answer lies in using the parallel axis theorem. I am still not sure how to applies this as there are two objects glued together.

I = Icm + MD^2

Icm = Moment of Inertia for Center of Mass

What I don't understand is how I am suppose to apply the parallel axis theorem in this situation.

To apply the || axis theorem, you are going to figure it about the center of mass. The Moment of inertia of each puck then 1/2mr² - offset by their distances from the center of mass - md² - looks like what is called for here. (i.e. .08*(.06)² for the smaller and similarly for the larger.)

 

[Benkyou]

Thanks I got it. Appreciate the help.


Ip = (1/2 * 0.12 kg * .06m^2) + (12 kg * .04m ^2) + (1/2 * 0.08 kg * 0.04m^2) + (0.08 * 0.06m^2)

And from the moment of inertia of the pucks I can plug in and get the angular momentum

Thanks again