What is the Angular Momentum of the System After a Collision?

AI Thread Summary
The discussion revolves around calculating the angular momentum of a system of two pucks after a collision. The smaller puck, with a mass of 47 g and a speed of 1.5 m/s, collides with a larger stationary puck of mass 67 g. Participants analyze the center-of-mass (CM) coordinates and velocities before and after the collision, applying the moment of inertia formula for disks. The user expresses confusion over discrepancies in their calculations for angular momentum, specifically regarding the values for the CM and the velocities of the pucks. Clarification and assistance are requested to resolve the calculation errors.
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Homework Statement


A small puck of mass 47 g and radius
44 cm slides along an air table with a speed
of 1.5 m/s. It makes a glazing collision with a
larger puck of radius 62 cm and mass 67 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
.5mr^2.
(a) before (b) during (c) after
After the collisions the center-of-mass has a
linear velocity V and an angular velocity ω
about the center-of-mass “+ cm”.
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





The Attempt at a Solution



I found an example of this problem online and tried to follow it. But my online homework says my answers is wrong. Here is the work of the problem I tried to follow.


(a) Before the collision, the y-coordinate of the CM is
(m1y1 + m2y2)/M = (0.08/0.2)0.1 m = 0.04 m.
The x-coordinate of the CM is (m1x1 + m2x2)/M = (0.08/0.2)x1 = (0.4)x1.
The velocity of the CM is
vCM = dxCM/dt = (0.4) dx1/dt = (0.4)1.5 m/s = 0.6 m/s in the x-direction.
In the lab frame the particle moves with velocity v = 1.5 m/si
and the CM moves with vCM =0.6 m/s i.
With respect to the CM m1 moves with velocity v1 = v - vCM = 0.9i m/s
and m2 moves with velocity v2 = 0 - vCM = -0.6i m/s.
The angular momentum of the system about the CM is
L = -(m1v1(y1 - yCM) + m2v2yCM)k = -(7.2*10-3 kgm2/s)k.

Here is my work:
ycm = (.47/1.14)*1.06=0.437017544
xcm=0.412280702(x1)
vcm=(0.412280702)*1.5=0.618421053
v1=v-vcm=0.881578947

L= (.47)*(0.881578947)*(1.06)+(ycm=0 so this equals 0)
= 0.439202631

Where did I mess up?
 
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I realized my answer was way wrong. But what I fixed didn't actually fix it.

L = (.47)(0.881578947)(1.06-0.437017544) + (.67)(0.618421053)(0.437017544)
= 0.439202632

Please help. I'm really confused.
 
Can anyone try to help me? Thanks.
 
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