Solveeit Logo

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.

A

Angular velocity of rod after collision is 60rad/sec

B

Angular velocity of rod after collision is 30rad/sec

C

Velocity of midpoint of rod just after collision is 15 m/s downward

D

Velocity of midpoint of rod just after collision is 10 m/s downward

Answer

Angular velocity ≈ 30 rad/s and Midpoint velocity = 15 m/s downward

Explanation

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=2×9.8×2019.8\mboxm/s.v=\sqrt{2gh}=\sqrt{2\times9.8\times20}\approx 19.8\mbox{ m/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.

  1. The moment of inertia of a rod about one end is
IP=Icm+M(L2)2=112ML2+14ML2=ML23.I_P=I_{cm}+M\Big(\frac{L}{2}\Big)^2=\frac{1}{12}ML^2+\frac{1}{4}ML^2=\frac{ML^2}{3}\,.
  1. Just before impact, the entire rod is translating downward so that the angular momentum about P (taking the lever arm as L/2L/2) is
Li=Mv(L2)=M(19.8)12.L_i = Mv\Big(\frac{L}{2}\Big)= M\,(19.8)\frac{1}{2}\,.
  1. Let the rod acquire an angular speed ω\omega (about P) immediately after collision. Then
Lf=IPω=ML23ω.L_f = I_P \,\omega = \frac{ML^2}{3}\,\omega\,.

Conservation about P gives

Mv(L2)=ML23ωω=3v2L.M\,v\Big(\frac{L}{2}\Big)=\frac{ML^2}{3}\,\omega\quad\Longrightarrow\quad \omega=\frac{3v}{2L}\,.

With v19.8\mboxm/sv\approx19.8\,\mbox{m/s} and L=1\mboxmL=1\,\mbox{m}:

ω3×19.8229.7\mboxrad/s.\omega\approx\frac{3\times19.8}{2}\approx29.7\,\mbox{rad/s}\,.

Rounded suitably, the angular velocity is about 30 rad/s.

  1. The centre of mass, being at a distance L/2L/2 from P, has a speed due to rotation:
vcm=ω(L2)30×0.5=15\mboxm/s.v_{cm}=\omega\Big(\frac{L}{2}\Big)\approx30\times0.5=15\,\mbox{m/s}\,.

Its direction is downward (as can be checked using the geometry of the rotation).