# Mathematical expression of impulsive forces during a collision

I'm facing with problem regarding collision between bodies. Specifically, I need to understand the impulsive forces that arises during a collision. Sometime ago, I posted two questions (this and this) , but I had no luck. It was my fault, maybe I was not so precise. I will try to be more precise.

Consider three object with equal mass $$m$$. These object can be assumed dimensionless. Two of them (red and green, hereafter called R and G) are tied by a rigid massless rod, which can rotate around a fulcrum located in the origin of $$xy$$ plane. The length of the rod is $$2L$$. The distance between the fulcrum and R is L, and the distance between R and G is L, too. The plane is a table, so the gravity is not working. The rod is rotating with angular velocity $$\omega > 0$$. On the table, there is another mass (blue, hereafter called B), which is steady in the point $$(0, 2L)$$. No friction is present.

At time $$t=0$$, a collision between G and B occurs. For the sake of simplicity, I assume that collision is elastic, and after the collision, B moves along $$y$$ with velocity $$v$$, while the rod remains steady.

During the collision, we observe impulsive forces acting on the three objects and on the fulcrum.

The collision is elastic, and energy is conserved. This implies that:

$$\frac{1}{2}mL^2 \omega^2 + \frac{1}{2}m(2L)^2 \omega^2 = \frac{1}{2}mv^2,$$

which yields:

$$v = L\omega \sqrt{5}.$$

My main concern is to find an expression for such forces. Let's call them $$\vec{F}_R$$, $$\vec{F}_G$$, $$\vec{F}_B$$ and $$\vec{F}_F$$.

For $$\vec{F}_B$$, the calculation is easy. Indeed, since the momentum of G changes instantaneously along $$y$$-axis by $$mv$$, then:

$$\vec{F}_B = mv \delta(t)\vec{{\bf y}}.$$

Using similar thoughts, I can evaluate the forces on R and G:

$$\vec{F}_R = -m\omega L \delta(t)\vec{{\bf y}},$$ $$\vec{F}_G = -m\omega2L \delta(t)\vec{{\bf y}}.$$

Questions

1. Are the expressions of $$\vec{F}_R$$ and of $$\vec{F}_G$$ correct? Do they have only vertical component or do they have also horizontal component?
2. Suppose that $$\vec{F}_G$$ is only vertical as I wrote. What can I say about the Newton's third law? Does it apply on this case? Apparently, the answer is no. Indeed: $$mv = m\omega L \sqrt(5) \neq m\omega 2L.$$
• Why do you assume that the rod is stationary after the collision?
– user197851
Commented Feb 22, 2019 at 20:53
• @LonelyProf It is just for the sake of simplicity. There is no particular reason. Commented Feb 22, 2019 at 21:02
• You can't just make such an assumption "for simplicity". B and G have equal and opposite impulses, say $\pm P$. You know $P=mv$. Write down the new angular velocity of the rigid body comprised by R and G, in terms of $P$ and the old angular velocity. Write down the conservation of energy equation. You should be able to obtain an equation for $P$ in terms of the old angular velocity. Two solutions: $P=0$ (no collision happens), and the solution you are interested in. Over to you.
– user197851
Commented Feb 22, 2019 at 21:28
• @LonelyProf My assumption has been made to reduce the complexity. If it is a problem, then assume that the rod has angular velocity $\omega$ before the collision, and after it has angular velocity $\omega_1$, while the blue mass moves with velocity $v$ after the collision. Anyway, my main concern is about the impulsive forces. Commented Feb 22, 2019 at 23:12
• I am simply pointing out that your initial assumption was incorrect and leads to a violation of Newton's third law. Solving this problem is just a question of writing down the equations in a way that respects the laws of motion. I've outlined one way of doing this. That's as far as I can go, otherwise we are in check my work territory which would be off topic for this site.
– user197851
Commented Feb 22, 2019 at 23:32

I'll offer an answer here to the conceptual parts of the question (as I see them), and try to avoid the details that some people might regard as falling into the "check my work" category.

The assumption that the rod is stationary after the collision is incorrect, and leads to a violation of Newton's third law.

I'll assume that the masses are small enough that the rod is horizontal, and the line between the centres of B and G is vertical at the time of collision. Then the only impulsive force acts in the vertical ($$y$$) direction. If we call this force $$P\delta(t)$$ acting on B, then by Newton's third law, a force $$-P \delta(t)$$ acts on G. We may integrate both these forces over the infinitesimal time of the collision. The post-collisional velocity of B therefore obeys $$mv=P$$. It is simplest to treat the RG system as a rigid body rotating about the origin. The moment of inertia can be calculated from the masses and distances. The angular momentum will change by an amount equal to the integrated impulsive torque, $$-2LP$$. This gives a relation between the angular velocities before and after the collision. Finally, energy conservation can be used to give an equation from which $$P$$ may be calculated, in terms of the initial angular velocity. The final angular velocity will be nonzero.

Throughout all of this, there will be forces along the rod, between the origin and R, and between R and G, but they are not impulsive, and contribute nothing during the infinitesimal time of the collision. At the time of collision, they act in the $$x$$ direction. They can be worked out from the angular velocity before and after, from the usual laws of circular motion: they will change discontinuously, like a step function, when the collision happens. Their function is solely to maintain the distances of R and G from the origin, providing the necessary centripetal acceleration.

Although you are interested in the impulsive force acting on the mass R at the centre of the rod, I don't believe that it is helpful to solve the problem by considering it explicitly. It is easier to think of the rigid massless rod, with added mass points R and G, as a rotating rigid body. Of course, once you know the instantaneous change in velocity of R, you can infer what the impulsive force in the $$y$$ direction, transmitted to it by the rigidity of the rod at the moment of collision, was.

You are also interested in the forces acting at the pivot or fulcrum F. There will always be the force equal and opposite to the force acting along the rod between the pivot and R, which constrains the pivot to be stationary, and keeps R and G in circular motion. At the moment of the collision this will be directed along $$x$$. It has no impulsive component. But you'll be wondering whether there is an impulsive reaction force acting in the $$y$$ direction, at the pivot point. The answer is yes. You can see how to work this out on the Wikipedia page dealing with Center of percussion. In your case, the centre of mass of the rod (midway between R and G), the location of the centre of percussion, and the point of impact, are all at different positions. If you removed your R mass, all these positions would coincide, and there would be no impulsive reaction at the pivot. But because of the mass at R, a reaction force at the point F is needed to keep it stationary.

Hopefully this gives enough information for you to solve the problem, and understand better what is going on.

• Thanks a lot for your excellent answer. There is only one point that I'm not getting properly. You say that the variation of angular moment of the whole rod (i.e., rod + R + G) is $-2PL$. Suppose that $\vec{F}_R = Q\delta(t)\vec{\bf y}$, for some Q. Then, should the variation of angular moment be equal to $-2PL + QL?$ Commented Feb 23, 2019 at 23:22
• Yes exactly. I'm treating the rod (with its masses) as a single rigid body rotating about the origin. If an external impulse of strength $+Q$ is applied in the $y$ direction at point $R$, which is a (perpendicular) distance $L$ from the pivot, then this is equivalent to an impulsive torque of strength $QL$ about the pivot. The total strength of the impulsive torque ($QL-2PL$ if they both act simultaneously) will be equal to the change in angular momentum of the rigid body. If you have worked out the moment of inertia of the body about the origin, this leads to the change in angular velocity.
– user197851
Commented Feb 24, 2019 at 13:23
• Ok. Therefore, $QL-2PL$ is the variation of angular moment of the rod, and $2PL$ is the variation for the mass B. Therefore, the variation for the whole system is $QL$. Hence, the angular momentum is not conserved, right? Commented Feb 24, 2019 at 13:35
• Let me re-emphasize that my previous comment applies to an external impulsive force on R. This is not what you were trying to introduce in your question, $\vec{F}_R$. That would be an internal force transmitted through the rigidity of the rod. In your original question, there is no external impulsive force acting on R. As I said in my answer, I prefer not to solve the problem that way, as I believe that it would be confusing. Others may be able to break things down in this way, but I'm not going to try.
– user197851
Commented Feb 24, 2019 at 13:37
• So, in case things are still not clear, angular momentum (of the system, including mass B) about the pivot is conserved. There are no external torques about the pivot acting on the system as a whole.
– user197851
Commented Feb 24, 2019 at 13:39