The figure shows an overhead view of a 2.30-kg plastic rod of length 1.20 m on a

Question

The figure shows an overhead view of a 2.30-kg plastic rod of length 1.20 m on a table. One end of the rod is attached to the table, and the rod is free to pivot about this point without friction. A disk of mass 33.0 g slides toward the opposite end of the rod with an initial velocity of 34.0 m/s. The disk strikes the rod and sticks to it. After the collision, the rod rotates about the pivot point.

(a) What is the angular velocity of the two after the collision?
1 rad/s

(b) What is the kinetic energy before and after the collision?

KEi

2

KEf

KEi

2

KEf

Explanation / Answer

initial angular momentum L1 = m*v*L = 0.033*34*1.2 = 1.3464 kg m^2/s

after collision I = (1/3)*M*L^2 +m*L^2

I = ((1/3)*2.3*1.2*1.2)+(0.033*1.2*1.2) = 1.15152 kg m^2/s

final angular momentum L2 = I*w

law of conservation of angualr momentum

when the net external torque acting on the system is zero the total angualr momentum of the system is conserved(constant)

L1 = L2

w = 1.3464/1.1515 = 1.17 rad/s

KEi = 0.5*m*v^2 = 0.5*0.033*34*34 = 19.074 J

KE f = 0.5*I*w^2 = 0.5*1.15152*1.17*1.17 = 0.788 J

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