A puck of mass 80.0 g and radius 3.90 cm slides along an air table at a speed of

Question

A puck of mass 80.0 g and radius 3.90 cm slides along an air table at a speed of 1.50 m/s as shown in the figure below. It makes a glancing collision with a second puck of radius 6.00 cm and mass 140.0 g (initially at rest) 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 position of the center of mass in the lab frame?

(b) What is the angular speed about the center of mass?

Explanation / Answer

Let's find the distance of the CM from the center of the larger puck:

d = 80.0g*(3.90 + 6.00)cm / (80.0 + 140.0)g =3.6 cm

Then the distance from the center of the smaller puck to the CM is
x = (3.70 + 6.00 - 3.6)cm = 6.1 cm
so

the initial angular momentum is
L = mvr = 0.0800kg * 1.50m/s * 0.061m = 0.0732 kg·m²/s
and this is also the final angular momentum.

Now for the tough part -- we need to determine the moment of inertia of the assembly about the CM (which lies on the axis of rotation).
small puck:
I1 = ½mr² + mx² = 80.0g * [½ * (0.0370m)² + (0.0610m)²] = 0.2036 g·m²
I2 = ½MR² + Md² = 140.0g * [½ * (0.060m)² + (0.036m)²] = 0.342 g·m²
total I = I1 + I2 = 0.545 g·m² = 5.40e-4 kg·m²

L = I*
0.0732 kg·m²/s = 5.40e-4kg·m² *
=135.55rad/s ..............b)

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