A puck of mass 80.0 g and radius 3.50 cm slides along an air table at a speed of

Question

A puck of mass 80.0 g and radius 3.50 cm slides along an air table at a speed of 1.60 m/s as shown in the Figure below. It makes a glancing collision with a second puck of radius 6.00 cm and mass 120.0 g (initially at rest) such that their rims just touch. Because their nims are coated with instant acting glue, pucks stick together are 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? Your response is within 10% of the correct value. The may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. (b) What is the angular speed about the center of mass?

Explanation / Answer

a)

when they touch ,the center of mass is distant from the center of the larger puck given by

Ycm=[0+120*(3.5+6)]/(120+80)=5.7 cm

angular momentum of the system

L=m1r1v1+m2r2v2=0+(0.057)(80*10-3)*1.5

L=0.00684 Kg-m2

b)

Moment of inertia about the center of mass

I=(1/2)m1r12+m1d12+(1/2)m2r22+m2d22

I=(1/2)*0.08*0.0352+0.08*0.0572+(1/2)*0.12*0.062+0.12*(0.095-0.057)2

I=6.982*10-4 kg-m2

angular momentum

Wf=L/I=0.00684/(6.982*10-4)

Wf=9.8 rad/sec

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