If there was a situation where you had two rods pinned in the center and the left rod having an initial angular velocity $\omega_1$, and the right rod was at rest. I am wondering what the final angular velocities would be if there was a coefficient of restitution during the impact between the rods, $e = 0.8$, i.e. can you apply the traditional coefficient of restitution relation for angular velocities?

Does $$e = \frac{\omega_2' - \omega_1'}{\omega_2 - \omega_1}$$ still apply?

Then you could use the conservation of the total angular momentum equation to find the angular velocity of each rod where

$$ H_1 + H_2 = H_1' + H_2' $$ $$ I_1\omega_1 = I_1\omega_1' + I_2\omega_2' $$

enter image description here


2 Answers 2


The problem is that angular momentum is not conserved in this scenario. In order to define angular momentum in a system, there needs to be an origin. If we put the origin at the center of the first rod, $M_1$, then it is clear that any force that the pin at $M_1$ exerts does not result in any torque in the system. However, we then have a problem because $M_2$ will exert a torque on the system; the pin of the second rod holds it in place and exerts forces so that it doesn't fly upwards; instead, it only rotates.

  • $\begingroup$ If I were to modify the last two equations and let the conservation of the total angular momentum with respect to $M_1$ to be $$ H_1 + \int_0^t{M_{21}d\tau} = H_1' + H_2' $$ where $M_{21}$ is the moment that the pin at $M_2$ exerts on $M_1$, would the system of equations then be able to be solved when substituting the relation of the angular velocities as shown with $e$? $\endgroup$ May 17, 2022 at 19:29
  • $\begingroup$ I'm not familiar with coefficient of restitutions for angular momentum cases, so I don't know if we just define $e$ as $\frac{|\omega_2'-\omega_1'|}{|\omega_2-\omega_1|}$. However, if we can say that (which we very well could), and we know how much angular momentum was lost by $M_2$, then you are correct: we can solve the system of equations to find $\omega_1'$ and $\omega_2'$ after one collision. $\endgroup$
    – Eli Yablon
    May 17, 2022 at 20:34

No the coefficient of restitution does not apply to with rotational velocity. In fact conservation of momentum will not apply either because the bodies are pinned and can transfer momentum to the earth or receive momentum from the earth as needed.

In addition the contact between the bodies is still done through a force applied over time resulting in an impulse exchanged between the bodies. As a result the coefficient of restitution applies to the relative velocity between the bodes at the point of contact

$$ v_{\rm rel}^{\rm after} = -\epsilon\; v_{\rm rel}^{\rm before} \tag{1}$$

If each rod has distance $r_1$ and $r_2$ between the pivot and the contact point then, and by convention we have positive rotation being counterclockwise you can work out the relative speed as

$$ v_{\rm rel} = \omega_1 r_2 + \omega_2 r_2 $$

For the relative velocity to be zero (as if they were gears) they would need to move in opposite sense such that $\omega_1 r_1 = - \omega_2 r_2$.

At the moment of impact, the center of mass of each body does not move (due to the pivot) and as a result an equal and opposite reaction impulse must act through each pivot on each body to cancel out the contact impulse.

For the first body this looks like this


and for the second body this looks like this


These pairs of impulses impart a moment of impulse on each rod changing their rotational speed. According to the above conventions and the diagrams each impulse pair acts in a clockwise (negative) fashion on both bodies.

$$ \begin{aligned} I_1 \omega_1 ^{\rm after} & = I_1 \omega_1^{\rm before} - r_1 J \\ I_2 \omega_2 ^{\rm after} & = I_2 \omega_2^{\rm before} - r_2 J \\ \end{aligned} \tag{2} $$

The before and after relative velocity is

$$ \begin{aligned} v_{\rm rel}^{\rm before} & = \omega_1^{\rm before} r_2 + \omega_2^{\rm before} r_2 \\ v_{\rm rel}^{\rm after} & = \omega_1^{\rm after} r_2 + \omega_2^{\rm after} r_2 \end{aligned} \tag{3} $$

Use the after rotational velocities from (2) in (3) and then plug the relative velocities in (1) to get an equation in terms of the impulse $J$.

$$ J = (1+\epsilon) \frac{ v_{\rm rel}^{\rm before} }{ \frac{r_1^2}{I_1} + \frac{r_2^2}{I_2} } \tag{4} $$

Then back substitute into the expressions for the rotational speed after to get

$$ \begin{aligned}\omega_{1}^{{\rm after}}=\left(\frac{\frac{r_{2}^{2}}{I_{2}}-\epsilon\frac{r_{1}^{2}}{I_{1}}}{\frac{r_{1}^{2}}{I_{1}}+\frac{r_{2}^{2}}{I_{2}}}\right)\omega_{1}^{{\rm before}}+\left(-\frac{(1+\epsilon)\frac{r_{1}r_{2}}{I_{1}}}{\frac{r_{1}^{2}}{I_{1}}+\frac{r_{2}^{2}}{I_{2}}}\right)\omega_{2}^{{\rm before}}\\ \omega_{1}^{{\rm after}}=\left(-\frac{(1+\epsilon)\frac{r_{1}r_{2}}{I_{2}}}{\frac{r_{1}^{2}}{I_{1}}+\frac{r_{2}^{2}}{I_{2}}}\right)\omega_{1}^{{\rm before}}+\left(\frac{\frac{r_{1}^{2}}{I_{1}}-\epsilon\frac{r_{2}^{2}}{I_{2}}}{\frac{r_{1}^{2}}{I_{1}}+\frac{r_{2}^{2}}{I_{2}}}\right)\omega_{2}^{{\rm before}} \end{aligned} \tag{5} $$

As you can see from above, the relationship between the before and after rotational speeds is rather complex in the general case.

Consider the simplified scenario of both rods having the same length and mass to arrive at $$ \begin{aligned}\omega_{1}^{{\rm after}}=\left(\frac{1-\epsilon}{2}\right)\omega_{1}^{{\rm before}}+\left(-\frac{1+\epsilon}{2}\right)\omega_{2}^{{\rm before}}\\ \omega_{1}^{{\rm after}}=\left(-\frac{1+\epsilon}{2}\right)\omega_{1}^{{\rm before}}+\left(\frac{1-\epsilon}{2}\right)\omega_{2}^{{\rm before}} \end{aligned} $$

For such as case the ratio $\lambda$ you are asking about is $$ \lambda = \frac{ \omega_{2}^{\rm after}-\omega_{1}^{\rm after}}{\omega_{2}^{\rm before} -\omega_{1}^{\rm before}} \equiv 1 $$

which is not equal to the COR.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.