What is the motion of the cylinders after the collision?

[haruspex]

the cylinders do not necessary have to leave the surface.

The first cylinder will definitely leave the surface unless restrained from doing so, as zwierz suggests.

the method needed to solve this is conservation of rotational energy,

There is no such conservation law. Who is telling you this?

Are we having to assume that the cylinders are inelastic?

In the absence of a given coefficient of restitution, we must assume either perfectly elastic or perfectly inelastic. Perfectly elastic will make the problem very complicated. The second cylinder will also need to be constrained to avoid bouncing up.
If either cylinder is restrained to stay on the surface then work cannot be conserved.

There is also an impulse from the ground on the first cylinder,

I do not see how.

 

[haruspex]

We have already discussed that. I consider a model with bilateral constraint. In such a case the direction of reaction is not prescribed

Your diagram clearly shows R₁ as a reaction from the ground. If you change it to be a force from some ceiling then your equations work.

 

[zwierz]

Your diagram clearly shows R1 as a reaction from the ground. If you change it to be a force from some ceiling then your equations work.

ooooooh! it is just a picture. It does not matter what holds the cylinder from flying up whether it is a ceiling from above or some specially designed rail on the ground.

 

[zwierz]

Consider the case of flying up.

Let $v_1^-,\omega_1^->0,\quad \omega_2^-=0,\quad v_2^-=0$ Assume that after the collision the first cylinder remains on the floor and the vector $\boldsymbol v_2^+$ can have both components to be non zero. In this case I expect $R_2=0$ .
Assume also that contact points of the cylinders have the same velocities right after the collision (perfectly inelastic collision and rough cylinders):
$$\boldsymbol v_1^++\boldsymbol\omega_1^+\times\boldsymbol{S_1A}=\boldsymbol v_2^++\boldsymbol\omega_2^+\times\boldsymbol{S_2A}.$$
In addition to this equation we have
$$J(\boldsymbol \omega_2^+-\boldsymbol \omega_2^-)=-\boldsymbol{S_2A}\times \boldsymbol T,\quad J(\boldsymbol \omega_1^+-\boldsymbol \omega_1^-)=\boldsymbol{S_1A}\times \boldsymbol T,$$
$$m(\boldsymbol v^+_1-\boldsymbol v^-_1)=\boldsymbol R_1+\boldsymbol T,\quad m(\boldsymbol v^+_2-\boldsymbol v^-_2)=-\boldsymbol T.$$Here are 8 scalar equations.
And unknowns are as follows
$\boldsymbol T,\boldsymbol v_2^+$ -- 4 scalar unknowns;
$\omega_{1,2}^+,v_1^+,R_1$ -- 4 scalar unknowns

For perfectly elastic collision and for the middle cases when restitution coefficient$\in(0,1)$ it is better to use in the Lagrangian formalism.

 

[Game_Of_Physics]

For the perfectly rough cylinders we have tangential forces acting on the cylinders. Assuming that they remain in contact with the plane at all times, this changed vertical force on each cylinder requires the balancing normal reaction force from the plane to change. Before the collision it was balancing the weight of each cylinder, but after the left-hand cylinder needs less support and the right hand more. Thus an external couple acts on the system, and so we may not use conservation of angular momentum.

Assuming the cylinders are perfectly rough, then no work is done against friction, so the rotational kinetic energy of the original cylinder: ½Iω₀² = ½Iω₁² + ½Iω₂²

 

[haruspex]

Assuming that they remain in contact with the plane at all times,...Thus an external couple acts on the system, and so we may not use conservation of angular momentum.

That is still the wrong way to say it. Rather, "assuming they remain in contact with the plane because there is a physical constraint on the first cylinder, which the question author omitted to mention, to stop it doing otherwise then..."

Assuming the cylinders are perfectly rough, then no work is done against friction, so the rotational kinetic energy of the original cylinder: ½Iω02 = ½Iω12 +

There are two things wrong with that logic.

First, as I keep pointing out, there is no law of conservation of rotational work, there is only a law of conservation of total work. Consider a ball sent rolling up a slope. At first it has linear energy and rotational KE about its centre. (Note that at any instant the point of contact with the ground is the centre of rotation, so considering it as linear plus rotational is a mere matter of choice; you could equally consider it all as rotational.) At the top of the slope it has almost come to a halt, so now it is all GPE. Then it topples over and falls to its original height; now nearly all the energy is linear KE. There has been no loss of work anywhere, yet the energy has been traded from linear to rotational.

Secondly, kinetic friction is not the only reason for losing energy. Inelastic impacts lose energy, and they can be rotational in form. Consider two rough discs on the same axle, spinning at different speeds. One is suddenly pushed against the other so that they lock together and now spin at the same rate. Applying conservation of angular momentum, we find there has been loss of rotational energy, yet there was no slipping.
Here the two cylinders are brought into contact tangentially rather than coaxially, but in other respects the situation is very similar.

In the present case, taking both cylinders as being constrained against vertical motion, there are four vertical impulses, all equal in magnitude. On the first cylinder the impulses are down at its centre and up at the impact point; on the second cylinder they are down at the impact point and up at its centre. It follows that the change in angular momentum is the same for each. This is independent of whether there is any energy loss.
If we suppose the two cylinders subsequently have rolling contact then we also know that their spins become equal and opposite. Consequently, if the original rate of spin of the first cylinder is ω then it finishes with rate ω/2, while the second cylinder gains spin -ω/2. It follows that half the rotational KE has been lost.

To conserve work, the two cylinders would need to bounce apart both linearly and rotationally. The first cylinder would come to a complete stop while the second would move off with the original linear and rotational speeds of the first cylinder, but with the spin reversed.

Edit: I should have clarified that in the para above I merely assume work is conserved. Even assuming perfect elasticity, how that is physically possible is not immediately clear. Any vertical displacement of the mass centres during the collision must be avoided, since that would lead to endless vertical oscillations not reclaimed as holistic motion. So not only must the impacts beperfectly elastic, they must also be perfectly rigid; a tricky combination.

 

[Game_Of_Physics]

In the present case, taking both cylinders as being constrained against vertical motion, there are four vertical impulses, all equal in magnitude.

Why are they equal in magnitude? You do not need no have vertical motion for them to have different reaction forces from the ground?

 

[haruspex]

Why are they equal in magnitude? You do not need no have vertical motion for them to have different reaction forces from the ground?

The vertical impulses they exert on each other must be equal and opposite (action and reaction). If there is no resulting vertical motion for either then for each the vertical impulse from the other cylinder must be equal and opposite to the impulse from ground/ceiling. So all four impulses have the same magnitude.

 

[Game_Of_Physics]

The vertical impulses they exert on each other must be equal and opposite (action and reaction). If there is no resulting vertical motion for either then for each the vertical impulse from the other cylinder must be equal and opposite to the impulse from ground/ceiling. So all four impulses have the same magnitude.

Okay yeah, I think I follow. But that isn't conserving angular momentum is it? If the first cylinder starts with ω and ends with ω/2, whilst the other starts with -ω/2 then the system starts with angular momentum ω and ends with angular momentum 0, and so it isn't conserved.

 

[haruspex]

But that isn't conserving angular momentum is it?

That's right (and I did not claim it was). (I think you meant to write that the second starts with 0 and ends with -ω/2.)

If the first cylinder had not been constrained by a downward impulse to stay grounded, you could have applied conservation of angular momentum about the point of contact between the second cylinder and the ground. There would have been no external impulse with a moment about that axis. But now that we know there is also a downward impulse on the first cylinder, there is no axis which can avoid moments from all external impulses.

And I stress, angular momentum is, in general, only meaningful with respect to a stated axis. Only if there is zero linear momentum is the axis irrelevant. For purposes of conservation of angular momentum, you must be careful about the choice of axis.

 

[Game_Of_Physics]

(I think you meant to write that the second starts with 0 and ends with -ω/2.)

Yes that it was I meant to say.

If the first cylinder had not been constrained by a downward impulse to stay grounded, you could have applied conservation of angular momentum about the point of contact between the second cylinder and the ground.

Okay, I see your point. Actually the question itself does not mention any such constraint, only the mark scheme I have be given by my tutor. What if the cylinders are on earth? Wouldn't that mean the cylinders do not necessarily leave the surface, because the upward force on the first cylinder upon impact may not exceed the gravitational attraction force?

 

[zwierz]

Actually the question itself does not mention any such constraint, only the mark scheme I have be given by my tutor.

It is very strange that your tutor gave you such a problem and had not explained what equations of impact are and what impact is. There is some theory of impact. This theory is not complicated but without knowledge of this theory you could hardly solve a problem of such a type.

 

[zwierz]

Wouldn't that mean the cylinders do not necessarily leave the surface, because the upward force on the first cylinder upon impact may not exceed the gravitational attraction force?

that is impossible

 

[Game_Of_Physics]

that is impossible

But that is what your diagram suggests? You have drawn R1 greater than R2.

 

[haruspex]

the upward force on the first cylinder upon impact may not exceed the gravitational attraction force?

See post #24.

 

[Game_Of_Physics]

Oh...but surely it could just as easily not be a great enough force do achieve this. I mean what if the ball has only a very small angular speed?

 

[haruspex]

Oh...but surely it could just as easily not be a great enough force do achieve this. I mean what if the ball has only a very small angular speed?

You may be confusing force with impulse (momentum). In impact questions like this, the assumption is that the duration of the impact is infinitesimal. That makes the force during the impact unbounded

 

[Game_Of_Physics]

You may be confusing force with impulse (momentum). In impact questions like this, the assumption is that the duration of the impact is infinitesimal. That makes the force during the impact unbounded

For the collision time to be infinitesimal it has to be elastic though doesn't it?

 

[haruspex]

For the collision time to be infinitesimal it has to be elastic though doesn't it?

No, why?

 

[Game_Of_Physics]

No, why?

If one ball collides with another stationary ball completely in-elastically, then the deceleration of the first ball and the acceleration of the second would be slower due to the increased contact time. For example compare dropping a tennis ball and its collision with the ground with that of a golf ball. The collision time of the tennis ball is longer because of its inelasticity.

 

[haruspex]

If one ball collides with another stationary ball completely in-elastically, then the deceleration of the first ball and the acceleration of the second would be slower due to the increased contact time. For example compare dropping a tennis ball and its collision with the ground with that of a golf ball. The collision time of the tennis ball is longer because of its inelasticity.

No, it doesn't work like that.

During any collision, there is a compression phase then an expansion phase. Take a spring as an easy model. In most questions we assume that there is a spring constant k such that the force at displacement x is kx. But real springs are not like that: they have hysteresis. In effect, the ratio between displacement and force depends whether it is moving towards or away from the relaxed state. That is, if you compress a spring some distance, and its resistant force has reached F, as soon as you start to let it expand the force will immediately drop to something less. This means you cannot get the same energy back as you put in, i.e. that it is not perfectly elastic. None f this has anything to do with the time taken.

If anobject impinges on the end of a spring,compressing it, the duration of the compression phase depends on the momentum of the object and the spring constant. The higher the constant, the less time taken to reach maximum compression, and the less time for the overall process.

In summary, the duration of each phase depends on the magnitude of the corresponding spring constant, while the elasticity depends on the ratio of the two constants.

 

[Game_Of_Physics]

No, it doesn't work like that.

During any collision, there is a compression phase then an expansion phase. Take a spring as an easy model. In most questions we assume that there is a spring constant k such that the force at displacement x is kx. But real springs are not like that: they have hysteresis. In effect, the ratio between displacement and force depends whether it is moving towards or away from the relaxed state. That is, if you compress a spring some distance, and its resistant force has reached F, as soon as you start to let it expand the force will immediately drop to something less. This means you cannot get the same energy back as you put in, i.e. that it is not perfectly elastic. None f this has anything to do with the time taken.
If anobject impinges on the end of a spring,compressing it, the duration of the compression phase depends on the momentum of the object and the spring constant. The higher the constant, the less time taken to reach maximum compression, and the less time for the overall process.

In summary, the duration of each phase depends on the magnitude of the corresponding spring constant, while the elasticity depends on the ratio of the two constants.

Ah...yes that's a good point! But then given that is the case, you could in theory have an impulse between an object colliding with another object, where the collision time is long due to a "small spring constant". In this case the force against time graph would be a line with a lower gradient, but as long as the area under it is the same as an equivalent graph that has a steeper gradient, then the impulses would be the same, but for the lower spring constant the collision time (the top limit of the integral) would be greater. My point is that the force that the first object feels because of the impulse from the second object is not necessarily as large as you make out in post 24, and therefore not necessarily great enough to overcome weight (in the case of the colliding cylinders).

 

[zwierz]

Spoiler: about impact problems

Nevertheless not all problems about collision are so trivial. Consider a wedge of mass $M$ that rests on a smooth horizontal floor. The slope of the wedge is also smooth. The wedge can not lose contact with the floor. It is an ideal constraint. Somebody drops a point mass $m$ on the wedge such that this point mass falls onto wedge with vertical velocity $u$. It then bounces from the wedge with velocity $v$ and the wedge begins to move with velocity $w$.
The velocity $u$ is known, the impact is perfectly elastic. Find $v,w$.

 

[haruspex]

Spoiler: about impact problems

Nevertheless not all problems about collision are so trivial. Consider a wedge of mass $M$ that rests on a smooth horizontal floor. The slope of the wedge is also smooth. The wedge can not lose contact with the floor. It is an ideal constraint. Somebody drops a point mass $m$ on the wedge such that this point mass falls onto wedge with vertical velocity $u$. It then bounces from the wedge with velocity $v$ and the wedge begins to move with velocity $w$.
The velocity $u$ is known, the impact is perfectly elastic. Find $v,w$.View attachment 196007

Spoiler: Impact on wedge

$w=\frac{mu\sin(2\theta)}{M+m\sin^2(\theta)}$

 

[haruspex]

the impulse from the second object is not necessarily as large as you make out in post 24

But then it would no longer qualify as an "impact problem", and you would need rather more data to be able to solve it.

Edit: and you mean the force, not the impulse.

 

[zwierz]

Yes. And I forgot to introduce angle :(