The figure shows an overhead view of a uniform 1.50-kg plastic rod of length 1.2

Question

The figure shows an overhead view of a uniform 1.50-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 41.0 g slides without friction toward the opposite end of the rod with an initial speed of 27.0 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. Find the angular velocity of the two after the collision.

Explanation / Answer

The initial kinetic energy must be enough to increase the potential energy enough to let it reach the 12 O'clock position. The initial kinetic energy is: KE = (1/2) I w^2 = (1/2)(1/3)M L^2*(Vti/L)^2 = (1/6)M Vti^2 (I is the moment of inertia and w is angular velocity) The potential energy change (with the CM in the middle of the rod) is deltaPE = M g [L/2 - (-L/2)] = M g L Therefore the requirement is (1/6)Vti^2 = gL Vti = sqrt(6 g L)

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