Moment of Inertia about the CoM

[Redfire66]

Homework Statement

A puck of mass 82.0g and radius 4.09cm slides along an air table at a speed of v = 1.59m/s. It makes a glancing collision with a second puck of radius 6.97cm and mass 130g (initially at rest) such that their rims just touch. The pucks stick together and spin after the collision (b). What is the angular momentum of the system relative to the center of mass?

Homework Equations

Angular Momentum (Conserved)
Center of Mass

The Attempt at a Solution

I got the first part, using center of mass relative of the larger mass and speed of the smaller mass giving me an initial value of angular momentum during the first part of collision of 8.84×10-3 kg*m^2/s. (This is confirmed to be correct)
However the second part I'm having trouble figuring out the moment of inertia. I thought that since angular momentum is conserved, I would have L/I = ω
And I = Icm (relative to smaller mass) + m1r1^2 + Icm(relative to larger mass) + m2r2^2
Where Icm = 1/2 mr^2 (r is the distance from the center of mass relative to each mass)
According to the textbook the answer should be about 8.85rad/s however I keep getting 9.00rad/s
Not really sure what I'm doing wrong, is my perspective on center of mass wrong? Since I can't think of another way the center of mass can be set a distance of. Also I'm not really sure about the parallel axis theorem if you could also explain how it comes into play when creating a formula for moments of inertia
Thanks

 

[haruspex]

I = Icm (relative to smaller mass) + m1r1^2 + Icm(relative to larger mass) + m2r2^2
Where Icm = 1/2 mr^2 (r is the distance from the center of mass relative to each mass)

Sorry, but I'm not at all sure what the above is saying. What do you mean by "Icm (relative to smaller mass)"? Do you mean the Icm of the smaller mass about its own centre? And what does "the distance from the center of mass relative to each mass" mean? Are you saying r1 is the distance from the centre of m1 to the CoM of the combination, likewise r2, m2?

 

[ehild]

The Attempt at a Solution

I got the first part, using center of mass relative of the larger mass and speed of the smaller mass giving me an initial value of angular momentum during the first part of collision of 8.84×10-3 kg*m^2/s. (This is confirmed to be correct)
However the second part I'm having trouble figuring out the moment of inertia. I thought that since angular momentum is conserved, I would have L/I = ω
And I = Icm (relative to smaller mass) + m1r1^2 + Icm(relative to larger mass) + m2r2^2
Where Icm = 1/2 mr^2 (r is the distance from the center of mass relative to each mass)

It should be the moment of inertia of the puck with respect of its centre of mass, which is 1/2 m R², with R the radius of the puck.

 

[Redfire66]

Sorry, but I'm not at all sure what the above is saying. What do you mean by "Icm (relative to smaller mass)"? Do you mean the Icm of the smaller mass about its own centre? And what does "the distance from the center of mass relative to each mass" mean? Are you saying r1 is the distance from the centre of m1 to the CoM of the combination, likewise r2, m2?

For example
Mass1 --------------------------------CM---------Mass2
Obviously mass 1 is further away from the center of mass compared to mass 2.
Kind of confused on the concept of parallel axis theorem which I stated so I assumed that was how I was supposed to do it. I used the radius of the distance between the first mass and the com, and the second mass and the com, so I had, let's say d is the distance between center of mass and r would be radius of the circles
I = .5m1r1^2 +m1d1^2 + .5m2r2^2 + m2d2^2

 

[haruspex]

let's say d is the distance between center of mass and r would be radius of the circles
I = .5m1r1^2 +m1d1^2 + .5m2r2^2 + m2d2^2

That's right.