The figure shows an overhead view of a 2.80-kg plastic rod of length 1.20 m on a

Question

The figure shows an overhead view of a 2.80-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 31.0 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) By law of conservation of momentum

Li=Lf

mvrsinq = (Irod++ Idisc )w

mvrsinq = (Irod++ Idisc )w

mvr=(ML2/3 + mL2) w

0.045*31*1.20= ( [(2.8*1.22/3) + (0.045*1.22)]w

w=1.2 rad/s

b) KEi = 1/2mv2 = 0.5*0.045*312 = 21.62 J

KEf = 1/2Iw2 =- ½*((Irod++ Idisc ) w2 = ½*( [(2.8*1.22/3) + (0.045*1.22)]*1.22 = 1.0 J

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