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

distance of center of mass from center of larger puck x = (80*9.9)/(80+140) = 3.6 cm

ditance of center of mass from smaller puck d = 3.9 + 6 - 3.6 = 6.3 cm

initial angular momentum Li = m*v*d = 0.08*1.5*0.063 = 0.00756 kg m^2/s <<<<<----answer

after coliision

moment of inertia of smaller puck = I1 = (1/2)*m*r^2 + m*d^2

I1 = ((1/2)*0.08*0.039^2) + (0.08*0.063^2) = 0.000378 kg m^2

moment of inertia of larger puck = I2 = (1/2)*M*R^2 + M*x^2

I2 = ((1/2)*0.14*0.06^2) + (0.14*0.036^2) = 0.000433 kg m^2

final angular Lf = (I1+I2)*w

Lf =Li

(0.000378 + 0.000433)*w = 0.00756

w = 9.32 rad/s <<<<<----answer

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