A small puck of mass 46 g and radius 42 cm slides along an air table with a spee

Question

A small puck of mass 46 g and radius 42 cm slides along an air table with a speed of 1.3 m/s. It makes a glazing collision with a larger puck of radius 68 cm and mass 75 g (initially at rest) such that their rims just touch. Both are disks with a moment o f inertia equal to 1/2 m r^2. The pucks stick together and spin a round after the collision. After the collision the center of mass CM has a linear velocity v and an angular velocity omega about the CM. What is the angular omentum of the system relative to th e center o f mass after the collision? Answer in u n its o f kg · m^2/s. What is the system's angular speed about the CM after the collision? Answer in u n its of rad/s .

Explanation / Answer

a)

The distance between puck centers when they touch is r1+r2 = 1.10m.

As we all know that the center of mass is located at a point along the line segment joining their centers.

Their moments of mass are equal:

r1m1 = r2m2

Here r1 + r2 = 1.10 m

So that r2 = (1.10 m - r1).

r1*46 = r2*75 = (110 cm - r1)*75

r1 = 68.18cm

so r2 = 41.82 cm

Angular momentum Am1 = (130 cm/s)*(68.18cm)(46gm)

Am1 = 407716.4 (cm^2*gm/s) = 4.077*10^-2 (kg*m^2/s)

b)

The angular momentum will be the same after the collision as it was before it, but each mass will have only a part of this total angular momentum. The two masses will have the same angular speed around the center of mass after the collision: call this Va.

The angular speed equals the linear speed divided by the radius from the center or rotation, which is the center of mass in this case. So after the collision:

m1*Va*r1^2 + m2*Va*r2^2 = 407716.4 (cm^2*gm/s)

[46 gm*(68.18 cm)^2] + [75 gm*(41.82 cm)^2]*Va = 407716.4 (cm^2*gm/s)

[345000 (g*cm^2)]*Va = 407716.4  (cm^2*gm/s)

Va = 1.18 rad/s

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