The figure shows an overhead view of a 2.40-kg plastic rod of length 1.20 m on a

Question

The figure shows an overhead view of a 2.40-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 52.0 g slides toward the opposite end of the rod with an initial velocity of 38.0 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? Incorrect: Your answer is incorrect. 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 = J KEf = J

Explanation / Answer

Here ,

mass of rod , m = 2.40 Kg

lemgth , L = 1.20 m

mass of disk , m = 0.052 Kg

a) Using conservation of angular momentum

I * w = m * v *r

(m * r^2 + I ) * w = m * v * r

(2.4*1.2^2/3 + 0.052 * 1.2^2) * w = 0.052 * 38 * 1.2

solving for w

w = 1.933 rad/s

the angular speed of the rod is 1.933 rad/s

B)

kinetic energy before = 0.5 * 0.052 * 38^2

kinetic energy before = 37.544 J

the kinetic energy before is 37.544 J

kinetic energy after = 0.5 * (2.4*1.2^2/3 + 0.052 * 1.2^2) * w^2

kinetic energy after = 0.5 * (2.4*1.2^2/3 + 0.052 * 1.2^2) * 1.933^2

kinetic energy after = 2.292 J

the kinetic energy after is 2.292 J

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