3. The figure shows an overhead view of a 2.20-kg plastic rod of length 1.20 m o

Question

3. The figure shows an overhead view of a 2.20-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 50.0 g slides toward the opposite end of the rod with an initial velocity of 32.5 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?

___________rad/s

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

KEi =______________J

KEf =_____________J

Explanation / Answer

a)   Conservation of angular momentum states

L = L',

where primed quantities stand for conditions after the collision and both momenta are calculated relative to the pivot point. The initial angular momentum of the system of stick-disk is that of the disk just before it strikes the stick. That is,

L=I,

where I is the moment of inertia of the disk and is its angular velocity around the pivot point. Now, I=mr2 (taking the disk to be approximately a point mass) and =v/r, so that

L = mvr

After the collision,   L'=I.

It is ' that we wish to find. Conservation of angular momentum gives   I''=mvr. ====> ' = mvr/I'

where I' is the moment of inertia of the stick and disk stuck together, which is the sum of their individual moments of inertia about the nail. The formula for a rod rotating around one end to be I=Mr2/3. Thus,

   I' = mr^2 + Mr2/3 = (m + M/3)r^2 = (0.050 + 0.733)*1.2^2 = 1.128 kgm^2

Now   ' = mvr/I' = (0.050*32.5*1.2)/1.128 = 1.73 rad/s

b) The kinetic energy before the collision is the incoming disk’s translational kinetic energy, and after the collision, it is the rotational kinetic energy of the two stuck together.

First, we calculate the translational kinetic energy by entering given values for the mass and speed of the incoming disk.   KE=0.5mv2 = 0.5*0.050*32.5^2 = 26.4 J

After the collision, the rotational kinetic energy can be found because we now know the final angular velocity and the final moment of inertia. Thus, entering the values into the rotational kinetic energy equation gives

KE' = 0.5I''^2 = 0.5*1.128*1.73^2 = 1.69 J

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