Question
Question: A horizontal rigid rod of length 1m is dropped from height h = 20 m, as shown in figure. The end of ...
A horizontal rigid rod of length 1m is dropped from height h = 20 m, as shown in figure. The end of the rod collides with the table. If collision is perfectly elastic.

Angular velocity of rod after collision is 60rad/sec
Angular velocity of rod after collision is 30rad/sec
Velocity of midpoint of rod just after collision is 15 m/s downward
Velocity of midpoint of rod just after collision is 10 m/s downward
Angular velocity ≈ 30 rad/s and Midpoint velocity = 15 m/s downward
Solution
We are given a 1 m long horizontal rod dropped from 20 m so that, just before impact, every point is moving downward with
v=2gh=2×9.8×20≈19.8\mboxm/s.When one end (say the left end, P) hits the table, a very short (impulsive) force reverses its vertical velocity from downward 19.8 m/s to upward 19.8 m/s (elastic collision). During this impulse the net external torque about P is zero so we can use conservation of angular momentum about P.
- The moment of inertia of a rod about one end is
- Just before impact, the entire rod is translating downward so that the angular momentum about P (taking the lever arm as L/2) is
- Let the rod acquire an angular speed ω (about P) immediately after collision. Then
Conservation about P gives
Mv(2L)=3ML2ω⟹ω=2L3v.With v≈19.8\mboxm/s and L=1\mboxm:
ω≈23×19.8≈29.7\mboxrad/s.Rounded suitably, the angular velocity is about 30 rad/s.
- The centre of mass, being at a distance L/2 from P, has a speed due to rotation:
Its direction is downward (as can be checked using the geometry of the rotation).