he figure shows an overhead view of a uniform 2.50-kg plastic rod of length 1.20
ID: 2217272 • Letter: H
Question
he 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=11.767J
KEf= J
he 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=11.767J KEf= JExplanation / Answer
as there is no external moment actring on the system angular momentum is conserved
initial angular momentrum about pivot = 0.056 kg*20.5m/s*1.20m= 1.3776
final angular speed be
then final angular momentum =[ 0.056*l^2+(2.50*l^2/3)]
equate these to get
KEi=0.5*0.056*20.5^2 =
KEf= 0.5*[ 0.056*l^2+(2.50*l^2/3)]*^2 =
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