A 3.60-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so

Question

A 3.60-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 = 13.5 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

Conserve angular momentum:
initial sphere mvr = final sphere mvr + I
where I = mL²/3 = 3.2kg * (2m)² / 3 = 4.27 kg·m²
0.25kg * (17 + 9.5)m/s * (4/5)2m = 4.27kg·m² *
= 2.48 rad/s

So for the rod, initial E = KE = ½I² = ½ * 4.27kg·m² * (2.48rad/s)²
E = 13.2 J becomes PE = mgh, so
13.2 J = 3.2kg * 9.8m/s² * h
h = 0.420 m
(That seems awfully high. You'll have to let me know if that's right.)

h = L(1 - cos) where here L is the distance to the CM
0.420m = 1m(1 - cos) = 1m - 1m*cos
= arccos((1-0.420)/1) = 54.5º

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