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29. Problem 12.70P (HRW) Two 2.00 kg balls are attached to the ends of a thin rod of negligible mass, 50.0 cm long. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its centre. While the rod is horizontal (see figure), a 50.0 g putty wad drops onto one of the balls with a speed of 3.0 m/s and sticks to it. What is the angular speed of the system just after the putty wad hits? What is the ratio of the kinetic energy of the entire system after the collision to that of the putty wad just before? Through what angle will the system rotate until it momentarily stops?
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Solution: Click For PDF Version (a) The key concept to be used in solving this problem is the principle of conservation of angular momentum. The angular momentum of the putty wad,
As the putty wad sticks to the ball on the right, the moment of inertia of the system comprising of two balls, a rod of negligible mass and the putty ball about its axis of rotation will be
Let
This gives
(b) The K.E. of the putty wad just before collision
The K.E. of the entire system just after the putty wad hits can be computed by considering it as a rigid body. It is given by
We thus find
(c) When the system stops momentarily its kinetic energy would have completely
changed into potential energy. Let
By equating the above value for the change in P.E. to
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, just
before it hits the ball on the right about its centre of rotation as shown in
the figure is
be the
angular speed of the system just after the putty wad hits. By applying
conservation of angular momentum, we get
be the
angle by which the system would have rotated clockwise when it momentarily
stops. The change in potential energy will be
,
we can calculate
. We find