# Errors in conservation of energy equation of a falling stick

I have been trying to write out a conservation of energy equation of a falling stick (at an angle $$\theta$$ to the horizontal) at the instant of its bounce.

Firstly, I write out the equation that relates the change in angular velocity of the rod to the contact angle as such:

$$\omega=\frac{m(u+v)l\sin \theta}{I}$$ where $$u$$ is the impact velocity, $$v$$ is the rebounding velocity, $$l$$ is the moment arm of the torque done by normal force. The collision is assumed to be instantaneous. Since the stick is hitting the ground without any rotation, friction is assumed to be negligible hence there is no torque done by kinetic or static friction here.

Substituting this into $$mu^2=mv^2+I\omega ^2$$ (conservation of energy)

I get: $$mu^2=mv^2+\frac{[m(u+v)l \sin \theta]^2}{I}$$

The equation itself seems to be correct.

However, when I tried to plot the rebounding velocity $$v$$ against the contact angle $$\theta$$, with the actual values of all the constants, we get the following graph:

As shown in the graph, where the y axis is rebounding velocity and x axis is $$\theta$$, the rebounding velocity becomes negative at a certain point (meaning the stick actually enters the ground).

I have conducted experiments to verify this graph, and the rebounding velocity is never negative.

Why is this happening? Is there something wrong with the conservation of energy equation? What is the reason behind the "negative" rebounding velocity when it is supposed to be positive.

P.S: I am not sure if anyone else faces this problem as well. Sorry if this question appears to be irrelevant. But I believe this is a legit physics question. :)

• Please consider writing more useful question titles, see How do we write good question titles? Commented Nov 29, 2020 at 12:15
• Is there something wrong with the equation? You should explain where you got your first equation from. It looks like you pulled from it thin air...
– Gert
Commented Nov 29, 2020 at 12:20
• Shouldn't that be $\cos\theta$ if angle $\theta$ is measured from the horizontal? Commented Nov 30, 2020 at 23:34

Note that your equation is quadratic in $$v$$. This means that there are two solutions. What is happening here is that you are picking the wrong solution in that range. Try plotting both solutions and see if they cross at a point such that one solution is valid at each point.