What is the motion of the cylinders after the collision?

[Game_Of_Physics]

1. A moving rough cylinder of radius a, and mass m collides with an identical cylinder, on a smooth horizontal surface. Its centre of mass moves with linear velocity v0, and its angular velocity is ω0. What is the motion of the cylinders after the collision?

I have be told that the answer to this problem does not involve using conservation of angular momentum because there is an external torque during the collision, and the way to do it is using conservation of rotational energy. I do not understand why angular momentum is not conserved if you consider that it is a closed system.

Thanks in advance for help

 

[scottdave]

Since they both have rough surface, it would seem that they stick together at one point. Can you draw out a diagram of what that system would look like? Can you see how the external forces (horizontal surface and gravity) work to create a torque on the new system?

 

[Game_Of_Physics]

Okay so the new system is now made to rotate because of the torque, which therefore conserves angular momentum?

 

[scottdave]

Imagine these two cylinders floating out in space. One (call it A) is moving toward B (which is stationary and not rotating) and also A is rotating. When they hit, and if the surfaces are able to interact, then A and B could spin together for an amount of time. The same angular momentum, but now a new moment of inertia is the two cylinders together. If they come apart, then both could be spinning, but the sum of angular momentums of both will equal the initial. But in actuality, there is the surface, which will push a force if the angular momentum is 'wanting' to spin down. And if something is wanting to spin up, then gravity will work against that. Will they stick? Will the first cylinder try to 'roll over' the second? Will the first one just stop, and the second one starts rolling at the same speed as the first? I am not sure of the answer, but these are things to look at.
What I am saying, is that you might be able to use angular momentum, if you know how the external torques are contributing to the change in momentum.

 

[vela]

1. A moving rough cylinder of radius a, and mass m collides with an identical cylinder, on a smooth horizontal surface. Its centre of mass moves with linear velocity v0, and its angular velocity is ω0. What is the motion of the cylinders after the collision?

What is "it"? The first cylinder or the second cylinder?

I have be told that the answer to this problem does not involve using conservation of angular momentum because there is an external torque during the collision, and the way to do it is using conservation of rotational energy. I do not understand why angular momentum is not conserved if you consider that it is a closed system.

I'm inclined to agree with you. What's this supposed external torque acting on the system?

Since they both have rough surface, it would seem that they stick together at one point.

I don't think you can assume they'll stick together. The two surfaces will slide against each other and exert kinetic friction forces.

 

[haruspex]

the way to do it is using conservation of rotational energy.

It seems unlikely energy would be conserved. In linear impulses there are three circumstances:
- elastic, so energy is conserved
- partly elastic, in which case you need to know the coefficient of restitution
- completely inelastic, in which case the maximum energy is lost consistent with conservation of momentum and constraints such as impenetrable surfaces.
The same applies here, but the elastic cases get nasty because you get a rotational bounce as well as a linear one. Unless the question specifies otherwise I would assume completely inelastic.

What's this supposed external torque acting on the system?

It depends what axis one uses. The first cylinder will exert a frictional impulse down on the stationary one, resulting in a vertically upward impulse from the ground on it (in addition to the ongoing normal force supporting the weight). So ok if the point of contact of the second cylinder with the ground is the axis.
Correspondingly, there will be a vertically upward frictional impulse on the first cylinder, which is guaranteed to make it airborne.

The two surfaces will slide against each other

Not necessarily. We are told the surface is rough, but no actual coefficient, so we should take as rough enough to prevent slipping, as long as that is possible with a finite coefficient. And it is. The horizontal impulse between them is an unbounded force, permitting a correspondingly unbounded frictional impulse.

@Game_Of_Physics, let the impulse between them on collision be J, at some angle θ to the horizontal. Can you think how to find θ? What equations can you write relating the horizontal and vertical components of J to the subsequent motions?

 

[Game_Of_Physics]

I am sorry I forgot to mention that neither of the cylinders leave the surface. What I find puzzling about this is why an external torque is created. I would have thought the only way to achieve this would be for the 2nd cylinder to be on a rough surface, (or more precisely, for the cylinders to be on surfaces with unequal coefficients of friction), or for one of the two cylinders to be inelastic? Could someone explain to me if this is incorrect, and if so, why?

 

[Game_Of_Physics]

It seems unlikely energy would be conserved. In linear impulses there are three circumstances:
- elastic, so energy is conserved
- partly elastic, in which case you need to know the coefficient of restitution
- completely inelastic, in which case the maximum energy is lost consistent with conservation of momentum and constraints such as impenetrable surfaces.
The same applies here, but the elastic cases get nasty because you get a rotational bounce as well as a linear one. Unless the question specifies otherwise I would assume completely inelastic.

The question does not specify otherwise, but how can you use energy conservation to answer the question if it is an inelastic collision? Surely it must be elastic?

 

[zwierz]

I will follow this hypothesis:

it would seem that they stick together at one point.

The notations

By superscripts $+,-$ we denote quantities after the collision and before the collision respectively.
So that
$\boldsymbol v^{\pm}_1=v^{\pm}_1\boldsymbol e_x$ is the velocity of the center $S_1$ before and after the collision;
$\boldsymbol v^{\pm}_2=v^{\pm}_2\boldsymbol e_x$ is the velocity of the center $S_2$ before and after the collision;
$\boldsymbol \omega_1^\pm=\omega_1^\pm\boldsymbol e_z$ is the angular velocities of the first cylinder before and after the collision;
$\boldsymbol \omega_2^\pm=\omega_2^\pm\boldsymbol e_z$ -- the same for the second cylinder;
$\boldsymbol R_{1,2}=R_{1,2}\boldsymbol e_y$ are the reactions of the floor for the first and the second cylinder respectively;
$\boldsymbol T=T_x\boldsymbol e_x+T_y\boldsymbol e_y$ is the reaction that the second cylinder exerts the first one.

The equations

$$J(\boldsymbol \omega_2^+-\boldsymbol \omega_2^-)=-\boldsymbol{S_2A}\times \boldsymbol T,\quad J(\boldsymbol \omega_1^+-\boldsymbol \omega_1^-)=\boldsymbol{S_1A}\times \boldsymbol T,$$
$$m(\boldsymbol v^+_1-\boldsymbol v^-_1)=\boldsymbol R_1+\boldsymbol T,\quad m(\boldsymbol v^+_2-\boldsymbol v^-_2)=\boldsymbol R_2-\boldsymbol T.$$
Here we have 6 scalar equations for 8 unknowns: $\omega_{1,2}^+,T_x,T_y,R_{1,2}, v^+_{1,2}$.
To close this system add kinematic equations: $\omega_1^+=-\omega_2^+,\quad v_1^+=v_2^+$

 

[Game_Of_Physics]

Right okay. So you have used conservation of angular momentum and conservation of linear momentum, without having to use conservation of angular/linear energy? I have been told the answer involves using conservation of energy because the creation of the torque shown in your diagram invalidates the use of conservation of angular momentum...

 

[zwierz]

I used the equations of the impact theory and the hypothesis by scottdave
If you wish to have the conservation of energy then you should provide another hypothesis about the collision and replace two equations in the last line of #9 with equations of your hypothesis.
For example you can replace those equations with equation of conservation of energy and $\omega_1^+=-\omega^+_2$

 

[vela]

Correspondingly, there will be a vertically upward frictional impulse on the first cylinder, which is guaranteed to make it airborne.

How do you know it's going to go airborne?

Not necessarily. We are told the surface is rough, but no actual coefficient, so we should take as rough enough to prevent slipping, as long as that is possible with a finite coefficient.

I don't agree that "rough" means "enough friction to prevent slipping." I would take "rough" to mean "there's friction between the two cylinders." But it sounds like your interpretation is correct from what the OP has subsequently said.

 

[haruspex]

View attachment 194828

I will follow this hypothesis:The notations

By superscripts $+,-$ we denote quantities after the collision and before the collision respectively.
So that
$\boldsymbol v^{\pm}_1=v^{\pm}_1\boldsymbol e_x$ is the velocity of the center $S_1$ before and after the collision;
$\boldsymbol v^{\pm}_2=v^{\pm}_2\boldsymbol e_x$ is the velocity of the center $S_2$ before and after the collision;
$\boldsymbol \omega_1^\pm=\omega_1^\pm\boldsymbol e_z$ is the angular velocities of the first cylinder before and after the collision;
$\boldsymbol \omega_2^\pm=\omega_2^\pm\boldsymbol e_z$ -- the same for the second cylinder;
$\boldsymbol R_{1,2}=R_{1,2}\boldsymbol e_y$ are the reactions of the floor for the first and the second cylinder respectively;
$\boldsymbol T=T_x\boldsymbol e_x+T_y\boldsymbol e_y$ is the reaction that the second cylinder exerts the first one.

The equations

$$J(\boldsymbol \omega_2^+-\boldsymbol \omega_2^-)=-\boldsymbol{S_2A}\times \boldsymbol T,\quad J(\boldsymbol \omega_1^+-\boldsymbol \omega_1^-)=\boldsymbol{S_1A}\times \boldsymbol T,$$
$$m(\boldsymbol v^+_1-\boldsymbol v^-_1)=\boldsymbol R_1+\boldsymbol T,\quad m(\boldsymbol v^+_2-\boldsymbol v^-_2)=\boldsymbol R_2-\boldsymbol T.$$
Here we have 6 scalar equations for 8 unknowns: $\omega_{1,2}^+,T_x,T_y,R_{1,2}, v^+_{1,2}$.
To close this system add kinematic equations: $\omega_1^+=-\omega_2^+,\quad v_1^+=v_2^+$

I mostly agree with that. In particular, it was not immediately obvious to me that the two cylinders should (on the basis of complete inelasticity) have the same horizontal velocity immediately after impact, just as they would with no rotations or vertical motion, but I concluded that is the case.
However, in your definitions you do not seem to have allowed for vertical motion of the first cylinder, even though you have T_y.

 

[haruspex]

conservation of rotational energy

There was a more fundamental point I should have made. There is no law of conservation of rotational energy, just as there is no law of conservation of linear energy in a specific coordinate direction. There is just conservation of energy.
Edit: This statement was wrong; it is not conserved: But, curiously, in this question, if by rotational energy you mean that embodied in the spin of each cylinder on its own axis, it is conserved.

neither of the cylinders leave the surface

Unfortunately that is not physically possible without some other external impulse, so adding that "information" creates an internal contradiction. When a question statement contains an internal contradiction different answers may be validily obtained by different methods. You must overrule that statement.

why an external torque is created.

As I posted, torque depends on the choice of axis.
Because of the downward (frictional) impulse of the first cylinder on the second, there is a corresponding impulsive normal force from the ground on the second cylinder.
If you take the centre of the first cylinder, at moment of impact, as the axis then that normal impulse has a torque about your axis.

how can you use energy conservation to answer the question

You can't.

How do you know it's going to go airborne?

Because the upward frictional impulse of the second cylinder on the first has no opposing vertical impulse. A change in vertical velocity is inevitable.

I don't agree that "rough" means "enough friction to prevent slipping."

In general, no, but since we are not given a coefficient of friction I see no alternative.
By the way, I believe when scottdave writes about the surfaces "sticking" he means the same thing.

 

[zwierz]

However, in your definitions you do not seem to have allowed for vertical motion of the first cylinder, even though you have Ty.

yes I considered biliteral constraint:

I forgot to mention that neither of the cylinders leave the surface.

 

[haruspex]

forgot to mention that neither of the cylinders leave the surface.

As I just posted, that is not possible and must be rejected. Your own equations prove it will.

 

[zwierz]

I do not see problems with biliteral constraints

 

[haruspex]

the creation of the torque shown in your diagram invalidates the use of conservation of angular momentum...

If you mean the impulse T shown in the diagram, that is internal to the system consisting of the two cylinders. The only external impulse is from the ground on the second cylinder. This can be eliminated from angular momentum equations by using an axis in the vertical line through it, such as the centre of the second cylinder.

 

[haruspex]

I do not see problems with biliteral constraints

Not sure what you mean. Are you saying to add a smooth ceiling to hold both cylinders down?

 

[zwierz]

Are you saying to add a smooth ceiling to hold both cylinders down?

Yes it is exactly the statement I studied. I understood OP in such a way. But surely another statements are also possible. For example as I mentioned above one can consider perfectly elastic collision with conservation of energy. Surely we can study the case when cylinder can fly up. Then one must bring additional hypotheses about interaction between cylinders and the floor

 

[haruspex]

Yes it is exactly the statement I studied. I understood OP in such a way.

Then the question statement should have specified that ceiling. It is not enough to say "it does not leave the surface"; the reason must be given. This is because it will change the equations, producing a different answer, and the way it changes the equations may depend on the reason it does not leave the surface.
I repeat, your equations are not consistent as they stand, since you have a vertical component in T with no consequent momentum change.

one can consider perfectly elastic collision with conservation of energy.

Of course, but it will make things quite complicated. For one thing, the downward frictional impulse on the second cylinder would lead to a subsequent bounce, so both would tend to leave the surface.

 

[vela]

Because the upward frictional impulse of the second cylinder on the first has no opposing vertical impulse. A change in vertical velocity is inevitable.

Unless the upward frictional force is greater than the weight of the cylinder, the cylinder will stay on the surface. The only difference will be that the normal force on it is reduced.

Perhaps you worked out the problem and found that to be the case, but I don't think it's obvious that the cylinder will leave the surface. Consider the case where it's barely spinning when it runs into the other cylinder. My intuition tells me it's not going to leave the surface in that scenario.

 

[zwierz]

It is not enough to say "it does not leave the surface"

perhaps it is not enough for engineers, in fundamental science it is enough to impose biliteral constraint mathematically. Concrete realisation of this constraint is another question. Discussing of such a question is not of interest for me.

I repeat, your equations are not consistent as they stand,

They are consistent. Solve them and make sure that they have a unique solution

 

[haruspex]

Unless the upward frictional force is greater than the weight of the cylinder, the cylinder will stay on the surface.

A collision is generally presumed to take a very short time, involving unknown large forces, but combining to form a moderate inpulse, i.e. ∫F.dt. Modest forces, such as gravity and the corresponding normal force for that, produce no impulse worth mentioning in that time.
Just as a normal force N supports a frictional force up to Nμ_s, a normal impulse J supports a frictional impulse Jμ_s. That easily overwhelms mgΔt.

 

[haruspex]

They are consistent.

You have an equation $m(v_1^+-v_1^-)=R_1+T$. Your R₁ and T each have an e_y component, but the v₁ terms do not. You can make R₁ zero, but not T_y.

Solve them and make sure that they have a unique solution

I did solve what were effectively the same equations, but allowing for a vertical motion of the first cylinder. As I mentioned somewhere, it does turn out that the rotational KE (measured in terms of each cylinder's rotation about its centre) is conserved, though I could not think of any simple reason for that. The first cylinder stops spinning and the second rotates at the same speed the first cylinder had been rotating, but in the opposite sense.

I have not tried to solve with an added force holding the first cylinder down, but I believe a lot will change.

 

[zwierz]

You have an equation m(v+1−v−1)=R1+Tm(v_1^+-v_1^-)=R_1+T. Your R1 and T each have an ey component, but the v1 terms do not. You can make R1 zero, but not Ty.

This vector equation is equivalent two scalar equations
$$m(v_1^+-v_1^-)=T_x,\quad 0=R_1+T_y.$$
So what?

 

[haruspex]

This vector equation is equivalent two scalar equations
$$m(v_1^+-v_1^-)=T_x,\quad 0=R_1+T_y.$$
So what?

R₁≥0, T_y>0. Isn't that rather a problem?

 

[zwierz]

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

 

[Game_Of_Physics]

Thank you all for the helpful replies, and sorry I haven't replied sooner. Just to pick up on a few things, the cylinders do not necessary have to leave the surface. Secondly, I feel I should reiterate that the method needed to solve this is conservation of rotational energy, and not conservation of angular momentum.

I am still unclear why the normal forces on the cylinders change. Are we having to assume that the cylinders are inelastic?

 

[Game_Of_Physics]

The only external impulse is from the ground on the second cylinder. This can be eliminated from angular momentum equations by using an axis in the vertical line through it, such as the centre of the second cylinder.

There is also an impulse from the ground on the first cylinder, causing a change in the angular momentum of a larger system including the surface, and therefore angular momentum isn't conserved.

 

[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.