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

Question

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 45.0 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?
  
What is the angular momentum of an object moving with a linear velocity, about a given point? rad/s

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

KEi

KEf

KEi

=

KEf

=
Initially the disk has translational motion while after collision the disk and stick have rotational motion. J

Explanation / Answer

appply law of conservation of energy

0.5*m*v^2 = 0.5*(I1+I2)*wf^2

I1 = m*r^2 = 0.045*1.2*1.2 = 0.0648 kg-m^2

I2 = (1/3)*m*l^2 = (1/3)*2.2*1.2*1.2 = 1.056 kg-m^2

I1+I2 = 1.1208

then after collision angular velocity is wf= sqrt(0.045*32.5*32.5/(1.1208)) = 42.4 rad/s = 6.5 rad/s

angular mpmentum L = m*v*r sin(theta)

if theta = 90 degrees

then L =m*v*r = 0.045*32.5*1.2 = 1.755 kg*m^2/s

B) KEi = 0.5*m*v^2 = 0.5*0.045*32.5*32.5 = 23.76 J

KEf = 0.5*(I1+I2)*wf^2 = 0.5*1.1208*6.5^2 = 23.76 J

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