The figure shows an overhead view of a uniform 2.50-kg plastic rod of length 1.2

Question

The figure shows an overhead view of a uniform 2.50-kg plastic rod of length 1.20 m on a very slick 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 56.0 g slides without friction toward the opposite end of the rod with an initial speed of 20.5 m/s and in the direction shown. The disk strikes the rod and sticks to it. After the collision, the rod (with the disk stuck to its end) rotates about the pivot point. For the question below, treat the disk as if it were a point mass.

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

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

KEf = J

Explanation / Answer

I'll assume that 1.00 103 means 1x10^3 a) Yes. Any object which has a velocity vector which does not pass directly through the chosen axis has angular momentum about that axis. b) No, mechanical energy is not conserved. Angular momentum is conserved, and any excess energy is dissipated within the door. c) W.before = W.after Solve for both, then solve for final w. (note that W is angular momentum, w is angular velocity) W.before = I.bullet * w.bullet + I.door * w.door I.bullet = m.bullet * (.91 m)^2 w.bullet = V.bullet / (.91 m) W.before = m.bullet * V.bullet * (.91 m) + 0 W.after = I.bullet * w.bullet + I.door + w.door W.after = m.bullet * (.91 m)^2 * w.bullet + 1/3 * m.door * (1m)^2 * w.door since w.bullet = w.door, we'll just call that w W.after = w * (m.bullet * (.91 m)^2 + 1/3 * m.door * (1m)^2) Set W.before = W.after and solve for w m.bullet * V.bullet * (.91 m) + 0 = w * (m.bullet * (.91 m)^2 + 1/3 * m.door * (1m)^2) .006 kg * 1000 m/s * .91 m = w * (.006 kg * (.91 m)^2 + 1/3 * 18.2 kg * (1 m)^2) 5.46 kgm^2/s = w * 6.0716 kgm^2 w = .8992 radians/second d) Before: Ek = 1/2 * m.bullet * V.bullet^2 Ek = 1/2 * .006 kg * 1000000 m^2/s^2 Ek = 3000 joules After: Ek = I.bullet * w^2 + I.door * w^2 Ek = (.8992 rad/sec)^2 * (.006 kg * (.91 m)^2 + 1/3 * 18.2 kg * (1m)^2) Ek = 4.909 joules Note that kinetic energy after is significantly less than kinetic energy before.

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