A 2.60-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so

Question

A 2.60-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. A solid sphere of mass 0.25 kg is thrown horizontally with a speed

v1 = 15.5 m/s

to hit the rod at the point A one-fifth of the way up from the bottom of the rod. The sphere bounces back horizontally with a speed

v2 = 9.50 m/s,

A 2.60-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. A solid sphere of mass 0.25 kg is thrown horizontally with a speed v1 = 15.5 m/s to hit the rod at the point A one-fifth of the way up from the bottom of the rod. The sphere bounces back horizontally with a speed v2 = 9.50 m/s, while the rod swings to the right through an angle ? before swinging back toward its original position. What is the angular velocity of the rod immediately after the collision? What is the angular momentum of an object moving with a linear velocity about a given point? Consider conservation of angular momentum. rad/s

Explanation / Answer

(Angular momentum of sphere)1 = m v1 L * 4/5
(Angular momentum of sphere)2 = -m v2 L * 4/5
Total change of angular momentum of sphere = 4 m L / 5 * (15.5 + 9.5)
Change in angular momentum = .8 * .25 * L * 25 = 5 L
Angular momentum of rod = 1/3 M L^2 * w    where w is angular speed
1/3 M L^2 w = 5 L
w = 5 * 3 / (2.6 * 2) = 2.88 / sec    initial angular speed of rod

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