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
(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
Answer in units of kgm2/s.
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
L= (.47)*(0.881578947)*(1.06)+(ycm=0 so this equals 0)
Where did I mess up?