A 2.20-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so

Question

A 2.20-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 = 12.5 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?

Explanation / Answer

Conserve angular momentum:
initial sphere mvr = final sphere mvr + I
where I = mL²/3 = 2.2kg * (2m)² / 3 = 2.933 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 2.933kg·m² *
= 3.00 rad/s

So for the rod, initial E = KE = ½I² = ½ * 2.933kg·m² * (3.00rad/s)²
E = 13.19 J becomes PE = mgh, so
13.19 J = 2.2kg * 9.8m/s² * h
h = 0.611 m

h = L(1 - cos) where here L is the distance to the CM
0.611m = 1m(1 - cos) = 1m - 1m*cos
= arccos((1-0.611)/1) = 67.10º

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