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,

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?
  
rad/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? rad/s

Explanation / Answer

apply m1u1 + m2u2   = m1v1 -m2v2

m2v2 = 2.6*0 + (0.25 * 15.5 ) - (0.25 * 9.5)

V2 = 1.5/0.25

v2 = 6 m/s

so

now use V = rW

W = 6/(1/5)

W = 30 rad/s

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