# Why does a rod rotate?

I'm a physics tutor tutoring High School students. A question confused me a lot.

Question is:

Suppose a mass less rod length $l$ has a particle of mass $m$ attached at its end and the rod is hinged at the other end in vertical plane. Another point object of mass $m$ is moving with velocity $v$ and hits the rod at its end and continues in its path with velocity $v/2$. The rod gets enough velocity that it can complete a vertical circular path. What would be the force due to hinge on the rod just after collision? Assume that time taken for collision is very small.

From conservation of momentum, the particle attached to rod gets velocity $V' = v/2$ which is equal to $2\sqrt{gl}$ and therefore $$T-mg=\frac{mv'^2}{l}$$ The tension in the rod is the Normal reaction due to hinge on the other end of the rod. Hence, the force exerted by rod is $$\frac{mv'^2}{l}+mg$$

But other teacher claims that I'm wrong. He says there will be horizontal force due to hinge on the rod. I say, there won't be any horizontal force. He gives an example that suppose there is a rod and you flick it at one end with finger then won't the other end move (or tend to move)? Well it does.

EDIT:

I've understood your answers mathematically.I still don't know how to counter my friend's argument. I'm convinced by it. Because it sounds intuitive. He says, suppose you have a rod (massless or with mass) in space, and you flicked at one of its end, then the other end surely will have velocity.Similarly, in the above problem, the collision will impart velocity to one end, so other end also must move but it is hinged and can't move because of hinge. So, there is a horizontal force due to hinge on the rod.

• @ claws : one has to careful in extending intuition from massive cases to massless ones. If you have a completely massless rod, you can get the rod to anything you want it to do: spin, translate, do a jiggle, do a dance, whatever. It's not going to violate conservation of energy, momentum, 2nd law of thermodynamics. Why? Because the massless rod is simply as good as empty space. So flicking a massless rod is a completely meaningless statement. If you have a massive rod, then flicking it causes it to translate + rotate, with definite quantities based on the impulse imparted. this is meaningful. Apr 21, 2013 at 4:31
• @ claws : in your question, you have a mass attached to a massless rod. now you can give the massless rod any arbitrary rotation about the mass (which is also the center of mass) and it'll be as good as a non-rotating one. so, the 'other end also must move' is once again a meaningless statement. what the problem does is it removes that indeterminancy: it explicitly says that the end w/o the mass is fixed in space. so during the instant of collision, the mass is essentially 'free' and cannot feel any force from the hinge. only just after the collision will the rod start to constrain the mass Apr 21, 2013 at 4:37
• @ claws : to move in a circle. This requires a tension force. Now, if the rod was massive, then yes, your intuition is right, and there will be some tangential force due to the hinge on the rod+mass system in the instant of collision. Apr 21, 2013 at 4:38
• @nervxxx: If the rod is said to be a light rod. Then it should mean that for all calculation purposes neglect mass, but consider it stiff and acting like a rod (in otherwords, its not as good as empty space, but matter with definite dimensions). Now, for light rod, would his argument apply? Apr 21, 2013 at 5:13
• @ claws: stiffness is not the same as mass. I've always been assuming that the rod is stiff and doesn't bend. Look, in the massless case, what you're doing is you're really boxing a region of space and saying that that region is privileged and deserves to be called a 'rod'. but there's nothing distinguishing it from it and the space it surrounds. If the rod was light but not massless, then its motion is well determined. The massless limit is simply a way of adding a constraint on the system (i.e. that the mass moves in a circle) and can't give rise to dynamics as it is not a dynamical object. Apr 21, 2013 at 5:53

Well you are right. The other teacher is right only if he is a considering a massive rod. Furthermore, if he is considering a massive rod, he is right only if he is considering the infinitesimal time interval during collision, not after, and even furthermore only for a very specific combination of parameters - it is actually quite subtle!

Here's why. Because the rod is massless in your scenario, during the collision the two masses can be simply thought of as being free. Your have worked out this scenario in your post. Then starting from just after the collision, what the rod serves to do is to simply constrain the motion to that of a circle. Hence we need the tension force to provide the centripetal force as you have analyzed.

Ok, now let's see why the other teacher might be right. If the rod is massive, then (it plus the mass at the end)'s center of mass is not at the end. Say that it is some distance $x$ away from the end where the mass is attached (or equivalently, $L-x$ away from the hinge). Furthermore let's assume that the rod has a uniform density. Removing this assumption makes the problem much harder, but it is still tractable.

What the collision does is it provides an infinite force $F$, but with finite impulse (think delta function). Let's assume that the hinge also provides an infinite force $F'$with finite impulse, in the opposite direction. We will see if this force $F'$ is $0$.

The impulse delivered is \begin{align} J = \int F - F' dt. \end{align} But we know that if the mass that strikes the rod+mass carries along its merry way with $v/2$ (although the situation is a bit unphysical as it would have to 'teleport' to the other side), then by conservation of momentum the rod+mass setup must acquire a velocity \begin{align} v' = \frac{m}{M(x)+m}\frac{v}{2}, \end{align} where $M(x)$ is the mass of the rod. ($M$ is actually a function of $x$, because once you specify where the center of mass is, you've specified the mass of the rod, and vice versa.)

Because the impulse causes a change in momentum of the mass+rod system $J = (M+m)\Delta v$, we have \begin{align} J = m \frac{v}{2}. \end{align}

Besides a linear impulse, these two forces provide a rotational impulse too, that causes the mass+rod to spin. That is, \begin{align} (L-x)\int F' dt + x \int F dt = I(x) \omega, \end{align} where $I(x)$ is the moment of inertia of the mass+rod system about its common center of mass, given by the parallel axis theorem: $I(x) = \frac{ML^2}{12} + M(L/2 - x)^2 + mx^2$.

Ok, this is kind of messy. But let's push on. Let's figure out what $\omega$ is. Multiplying $J$ by $-x$ and adding it to the last equation, we get \begin{align} & L\int F' dt = I(x)\omega -xm \frac{v}{2} \end{align} Now if $\omega (L-x) = v'$, i.e. that the velocity of the point of the rod+mass at the hinge, which is the sum of the translational velocity plus the rotational velocity is $0$, then $F' = 0$.

Now, all these analyses was done in the infinitesimal time interval during collision. Just after the collision, once everything has settled. Then regardless of whatever situation we have, we have $\omega(L-x) = v$. Then the force of the hinge on the rod will be tangential to the motion because of circular motion requirements.

Ok, long story short: if you're talking about just after the collision, then there will be no horizontal force. if you're talking about during the collision, if the rod is massless, there will be no horizontal force. if you're talking about during the collision, if the rod is massive, then in general there will be a horizontal force.

• M is actually a function of x. Why? I didn't understand your reason? Apr 21, 2013 at 3:42
• once you specify where the center of mass is, for a uniform rod + a point mass $m$, then $M$ is fixed. It is given by: $mx = M(L/2 - x)$, so $M = \frac{mx}{(L/2 - x)}$. Apr 21, 2013 at 4:54
• @nervxxx How could you conserve linear momentum of the system {(rod + mass) + mass}, until you've proven that the F' is zero, because if F' is not zero, there is a net force in the horizontal direction and you cannot conserve momentum in presence of a (non zero) external Force, i.e. F' ? Oct 9, 2018 at 4:44

Your teacher friend would be right if the rod wasn't massless.

Lets assume the rod does have some finite mass. In this case the centre of mass (COM) of the rod/attached weight system (which I'll call a pendulum from now on) would be somewhere between the attached weight and the hinge.

If the COM is not exactly at the point of collision, in addition to the linear acceleration of the mass, there would be a torque applied to the pendulum. So the pendulum will try to rotate about its COM (as well as about the hinge). This rotation is prevented by the hinge which must apply an equal and opposite torque, and therefore a horizontal force (which less than the collisional force as its further from the COM than the collision point was).

Now your rod is massless so the force is applied straight through the COM of the pendulum, hence no torque and no horizontal hinge force. The only forces on the hinge are the centripetal force due to mass now having a rotational velocity, and the weight of the mass which was there before collision too.

$$\boldsymbol{F}_{hinge} = \left( \begin{array}{c} 0 \\ mg + \frac{m}{l}\left(\frac{v}{2} \right)^2 \end{array} \right)$$ No horizontal force, vertical force is weight + centripetal. The above only applies when the pendulum is directly below the hinge.

The other teacher's answer is correct only in a situation with energy losses due to friction.

If energy is to be conserved, there can be no external forces acting on the system parallel to the motion at impact, and hence the only force acting on the rod at impact must be vertical.