What is the angular momentum of the system after the collision?

[shangri-la89]

Homework Statement

A small puck of mass 36 g and radius
25 cm slides along an air table with a speed
of 1.9 m/s. It makes a glazing collision with a
larger puck of radius 59 cm and mass 85 g (ini-
tially at rest) such that their rims just touch.
The pucks stick together and spin around af-
ter the collision.
Note: The pucks are disks which have a
moments of inertia equal to 1/2mR^2.
What is the angular momentum of the sys-
tem relative to the center-of-mass after the
collision? Answer in units of kgm2/s.

Homework Equations

L=Iw
w=V/r
Torque=I*alpha
L=RxP

The Attempt at a Solution

To be honest I have no idea. I tried several obviously ill-fated attempts. I have found the center of mass, and I know that it has to do with crossing the initial linear momentum with the new radius... I think...

Any help would be greatly appreciated.

 

[tiny-tim]

Welcome to PF!

Hi shangri-la89! Welcome to PF!

(have an alpha: α and an omega: ω and try using the X² tag just above the Reply box )

(oh, and it's a grazing collision … a glazing collision is when you walk into a glass door! )

What is the angular momentum of the system before the collision?

 

[nlightner]

I can provide a response to the question by using the principles of angular momentum conservation. Before the collision, the smaller puck has a linear momentum of 0.036 kg*m/s and the larger puck has a linear momentum of 0 kg*m/s (since it is initially at rest). After the collision, the two pucks stick together and spin around with a common angular velocity, so their combined angular momentum will be equal to the sum of their individual angular momenta.

Using the formula L=Iw, where I is the moment of inertia and w is the angular velocity, we can calculate the initial angular momentum of the smaller puck as 0.018 kg*m^2/s and the initial angular momentum of the larger puck as 0 kg*m^2/s.

After the collision, the combined moment of inertia of the two pucks can be calculated as 1/2(0.036+0.085)(0.25+0.59)^2=0.024 kg*m^2. Since the pucks are now spinning around a common axis, their combined angular velocity can be calculated by equating their individual linear momenta (0.036 kg*m/s) with their combined angular momentum (0.024 kg*m^2/s * w). Solving for w, we get w=1.5 rad/s.

Now, using the formula L=Iw, we can calculate the final angular momentum of the system as 0.024 kg*m^2 * 1.5 rad/s = 0.036 kg*m^2/s. This is the same as the initial angular momentum of the smaller puck, since the larger puck was initially at rest and did not contribute to the initial angular momentum.

Therefore, the angular momentum of the system after the collision is 0.036 kg*m^2/s, which is the same as the initial angular momentum of the smaller puck. This is the result of angular momentum conservation in the system.