# Horizontal impulse on a ball at rest on a plane (with friction)

• FranzDiCoccio

#### FranzDiCoccio

Thread moved from the technical forums to the schoolwork forums
Summary:: I'd like to check my understanding of standard problems where a billiard ball resting on a plane is hit horizontally at some height above its center

So the situation is that a ball of mass ##m## and radius ##r## is at rest on a horizontal surface. There is friction between the ball and the surface. The ball receives a horizontal impulse at some height ##h## above the surface (##r<h<2r##) and starts moving.

One possible question is: "find ##h## such that the ball rolls without sliding". This is discussed e.g. in thread 1 and thread 2.

After looking at these threads and some more sources I'd consider the angular momentum of the ball wrt an axis that lies on the surface and is perpendicular to the impulse (i.e. to the ball velocity).
This choice gets rid of the contribution of the friction to the rotational impulse on the ball. Friction in principle still affects the overall impulse. However, as suggested in thread 2, one assumes that the impulse is sudden, i.e. it lasts for a very short time. Therefore the impulse of the friction is negligible when compared to the impulse of the cue stick.
Hence, taking into account that the initial momentum is ##\vec p_0=0##, we have ##\vec J = \vec p_1=m \vec v_1##.
Since the impulse is horizontal and at an heigth ##h## above the table, and since the initial angular momentum is ##\vec L_0=0##, we have that
##J h = L_1##.
After the impulse, the ball moves with velocity ##v_1## and rotates about its center with angular velocity ##\omega_1##. Its angular momentum wrt the chosen axis is ##L_1=I \omega_1 + m v r_1##.
If we want the ball to roll without sliding, it should be ##\omega_1 r = v_1##, so that ##L=(1+\kappa) m v_1 r## where ##\kappa =2/5## for a homogeneous ball. Therefore we get ##(1+\kappa) m v_1 r=m v_1 h##, which gives ##h=(1+\kappa) r## (or ##\kappa r## above the center).

In thread 1 the result depends both on the force from the cue stick and the friction. However, I guess that we can say that the former must be extremely large, since it produces an effect on such a short time scale. On the other hand, as we mention, the friction is much smaller.

As far as I understand, the OP of thread 2 obtained the solution like in thread 1, but ignored the friction from the start (and he was not very sure why). I found two discussions on the web (here and here) which also seem to ignore the friction from the table (but perhaps the authors make a comment about it being negligible... I do not understand the language).

A variant of the problem is "what happens if the ball is hit at a different height above its center?". In this case the ball will initially slide on the table, and end up rolling without sliding at a different velocity.

If I'm not missing something the strategy should be similar. The force responsible for the change in velocity is the kinetic friction between the ball and the table. Since the corrisponding torque about the chosen axis vanishes, after the ball starts moving ## \Delta \vec L=0## (##L_2=L_1##).
But on the one hand ##L_1-L_0=L_1= J h = m v_1 h## and on the other hand ##L_2 = m v_2 r (1+\kappa)##. Therefore ##v_2 = v_1 \dfrac{h}{r (1+\kappa)}##.
This makes sense: if the ball is hit above the "critical height" it will end up rolling faster. If the cue stick hits below that height, the ball will end up rolling slower.

It should be possible to hit the ball so that it stops, or even ends up going backwards. I'm not seeing this, but it's very late here and I'm a bit tired. Perhaps this is not possible with an horizontal impulse.

Does all this makes sense?

[...]

Since the impulse is horizontal and at an heigth ##h## above the table, and since the initial angular momentum is ##\vec L_0=0##, we have that
##J h = L_1##.

Be careful here. $h$ is the distance of the cue-stick (i.e., impulse) above the table. But if you want to find the resultant angular momentum, you'd be more interested in the height difference between the cue-stick (i.e., impulse) and the center of the ball. So, I think there should be an $r$ thrown in there somewhere.

[...]

In thread 1 the result depends both on the force from the cue stick and the friction. However, I guess that we can say that the former must be extremely large, since it produces an effect on such a short time scale. On the other hand, as we mention, the friction is much smaller.

As far as I understand, the OP of thread 2 obtained the solution like in thread 1, but ignored the friction from the start (and he was not very sure why). I found two discussions on the web (here and here) which also seem to ignore the friction from the table (but perhaps the authors make a comment about it being negligible... I do not understand the language).

'Sounds reasonable to me. Keep in mind that thread 1 wasn't dealing with impulses, thus not dealing with arbitrarily large forces. As a result, in thread 1 there is a range of heights where the ball will roll without sliding. As long as the frictional force is below the threshold of static friction, the ball rolls.

In the case of impulses, where the force can be considered arbitrarily large (approaching infinity), the range of heights narrows to a single point.

If it helps, solve the problem where the ball is on a frictionless plane. There exists a particular height where the ball will roll without sliding.

A variant of the problem is "what happens if the ball is hit at a different height above its center?". In this case the ball will initially slide on the table, and end up rolling without sliding at a different velocity.

[...]

It should be possible to hit the ball so that it stops, or even ends up going backwards. I'm not seeing this, but it's very late here and I'm a bit tired. Perhaps this is not possible with an horizontal impulse.

Does all this makes sense?

Sure, it makes sense. Assume the ball initially slides due to the impulse, but then later, eventually rolls due to the impact/consideration of dynamic friction (not to be confused with the static friction used in thread 1).

This choice gets rid of the contribution of the friction to the rotational impulse on the ball.
Yes, provided the impulse is purely horizontal. But when two objects collide, ignoring friction between them, there is only a normal force, i.e. normal to the tangent plane at point of contact. So if the ball is struck above half way there will be a downward component to the impulse, producing a corresponding normal impulse from the table and hence a frictional impulse from it.
In actual billiards, there is the complication that chalk on the cue tip ensures there is also a frictional impulse between cue tip and ball.
It should be possible to hit the ball so that it stops, or even ends up going backwards.
Ignoring friction between cue tip and ball, I don't think it is possible to finish with the cue ball rolling backwards unless it strikes another ball first.
But with chalk on the cue it might be done.
I think there should be an r thrown in there somewhere.
No, moments are being taken about an axis at table level.

• FranzDiCoccio

Be careful here. $h$ is the distance of the cue-stick (i.e., impulse) above the table. But if you want to find the resultant angular momentum, you'd be more interested in the height difference between the cue-stick (i.e., impulse) and the center of the ball. So, I think there should be an $r$ thrown in there somewhere.

not really... I am considering an axis of rotation on the table. The correct lever arm for the force is ##h##, right?

'Sounds reasonable to me. Keep in mind that thread 1 wasn't dealing with impulses, thus not dealing with arbitrarily large forces. As a result, in thread 1 there is a range of heights where the ball will roll without sliding. As long as the frictional force is below the threshold of static friction, the ball rolls.

In the case of impulses, where the force can be considered arbitrarily large (approaching infinity), the range of heights narrows to a single point.

If it helps, solve the problem where the ball is on a frictionless plane. There exists a particular height where the ball will roll without sliding.

The approach of thread 1 confuses me a little. I guess it's hard for me to imagine a horizontal force acting on the ball for a finite amount of time with the same magnitude, while the ball is rotating and translating. Here I am probably worrying about the friction between the ball and the tip of the cue stick, which could be assumed to be negligible.

Yes, provided the impulse is purely horizontal. But when two objects collide, ignoring friction between them, there is only a normal force, i.e. normal to the tangent plane at point of contact. So if the ball is struck above half way there will be a downward component to the impulse, producing a corresponding normal impulse from the table and hence a frictional impulse from it.
In actual billiards, there is the complication that chalk on the cue tip ensures there is also a frictional impulse between cue tip and ball.

Ok, yes... I get it. The part of the friction that is proportional to the ball weight can be safely ignored. However, the large horizontal impulsive force generates a large reaction force from the table, which in turn generates a large frictional impulse. This still does not contribute to the torque, owing to the choice of the axis, but it reduces the horizontal impulse.

So I guess that most of the discussions about this problem I have found (e.g. threads 1 and 2) do apply to a frictionless case. If this is true, hitting the ball at a "wrong" height would cause the ball to slide on the table "forever", because there is no mechanism causing the ball to roll without sliding in the end.

EDIT: maybe it could work on a very long table with a very small friction. Since the friction is small, the impulsive friction can be ignored. Since the table is very long, the ball will end up rolling without sliding at some point.

Ignoring friction between cue tip and ball, I don't think it is possible to finish with the cue ball rolling backwards unless it strikes another ball first.
But with chalk on the cue it might be done.

Yes, I see this now. The cue stick should have the ball roll backwards, and this is possible only with friction between its tip and the ball (for which you need chalk).

PS My analisys could work for an object that is made of two discs joined by a thin slab. The slab could be initially vertical, and the impulsive force would hit it horizontally without causing a force on the table. Of course the moment of inertia would be different from that of a ball, and the "critical" height for rolling without sliding would be different. But one can always write ##I=\kappa m r^2## where ##m## is the total mass of the object, ##r## is the radius of the discs and ##\kappa## is some suitable number. Last edited:
the large horizontal impulsive force generates a large reaction force from the table
Not exactly. I am saying that the cue striking horizontally above the centre of the ball produces both a horizontal and a vertical impulse. The vertical impulse produces an impulsive normal reaction from the table.

Not exactly. I am saying that the cue striking horizontally above the centre of the ball produces both a horizontal and a vertical impulse. The vertical impulse produces an impulsive normal reaction from the table.
ok yes, of course. I was focusing only the impulsive friction related to the the impulsive normal reaction from the table.
Is there any other effect I did not consider? Are you suggesting that the ball tends to jump upwards because there is no force balancing the impulsive reaction from the table?

not really... I am considering an axis of rotation on the table. The correct lever arm for the force is ##h##, right?

I stand corrected. Although it's not the approach I would have taken myself, particularly when solving for the situation with finite forces and static friction, etc., taking moments around a point on the table is valid.

Not exactly. I am saying that the cue striking horizontally above the centre of the ball produces both a horizontal and a vertical impulse. The vertical impulse produces an impulsive normal reaction from the table.

If this were an actual homework problem, I would advise treating any vertical impulses as negligible. Even in real-world situations, they're almost negligible if not fully negligible (this assumes the cue-stick's tip is chalked up properly).

If it helps, consider what happens if left or right "English" is applied to the shot (instead of top or bottom English, as we are discussing specifically for this problem). Suppose the cue-stick (and thus the impulse) is parallel to the x-axis, but offset just enough such that it strikes the cue-ball 30 deg off center (left or right), even though the direction of the shot is parallel to the x-axis. Assuming the cue-stick's tip is chalked up correctly, the cue-ball won't move 30 deg off the x-axis, rather it will move almost straight along the x-axis; it's just that it will spin a lot. But the ball's initial, translational motion will be almost identical to the direction of the impulse.

Similarly, if the same reasoning is applied to top or bottom English as discussed in this thread. If the impulse is in the horizontal direction, the vertical impulses that might occur can be considered negligible for most purposes.

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Are you suggesting that the ball tends to jump upwards because there is no force balancing the impulsive reaction from the table?
Whether the ball tends to jump upwards will depend on the elasticity of the vertical interaction between table and ball. It is not an inherent result of Newton's third law. A typical material will deform when a force is applied. It will compress slightly under force of an impacting object. If the material is elastic then, after the impacting object has come to a stop, the material will rebound from its compressed state toward its original undeformed state. As it does so, it will exert a force on the impacting object, causing it to "bounce".

If you hypothesize a perfectly rigid table and a perfectly rigid ball, you have a bit of a conundrum. You will have an infinite force applied over an infinitesimal distance in an infinitesimal time. The amount of energy that is dissipated is indeterminate. So when one has a collision, even with an ideally rigid surface, one should still specify elasticity as a collision parameter.

In pool table physics problems, one normally assumes 100% inelasticity for vertical interactions of ball with table.

Whether the ball tends to jump upwards will depend on the elasticity of the vertical interaction between table and ball. It is not an inherent result of Newton's third law.

Yes, even while asking that question I was not very convinced that the ball would jump upwards. I was just trying to understand what to make of the reaction of the table.

If this were an actual homework problem, I would advise treating any vertical impulses as negligible. Even in real-world situations, they're almost negligible if not fully negligible (this assumes the cue-stick's tip is chalked up properly).
[...]
Similarly, if the same reasoning is applied to top or bottom English as discussed in this thread. If the impulse is in the horizontal direction, the vertical impulses that might occur can be considered negligible for most purposes.

Ok so if I get it right: the chalked tip of the stick exerts an additional vertical force which makes the reaction from the ball horizontal or almost horizontal (the additional force would be tangential, but anyway this means it has a vertical component and a horizontal one. The former ends up canceling or greatly reducing the vertical impulse from the table. The latter becomes part of the overall horizontal response to the stick).

So the text of the exercise should contain a sentence like "Assume that the tip of the stick is properly chalked, so that the overall reaction from the surface of the ball is horizontal". This should get rid of the effects of the friction from the table when the ball is hit. The friction could be there, though, and this would probably allow for problems where the ball is hit so that it initially slides a bit because its translational and tangential velocity do not match.

Although it's not the approach I would have taken myself, particularly when solving for the situation with finite forces and static friction, etc., taking moments around a point on the table is valid.

Ok... I'm still not clear on how to think about the problem with finite forces. I can use the second law of dynamics for both translations and rotations, and require that the tangential and translation acceleration match. But this seems even more "instantaneous" to me. If the force is finite it should act for a finite amount of time. Does the ball move in that time? And how?

how to think about the problem with finite forces
The safe way to handle idealisations is to start with the general case (imperfectly elastic, some deformability, noninfinitesimal time...) then take limits. But that can get very messy. If the cue tip is in contact with the ball for a while, chalked, it may exert retarding friction.

The safe way to handle idealisations is to start with the general case (imperfectly elastic, some deformability, noninfinitesimal time...) then take limits. But that can get very messy. If the cue tip is in contact with the ball for a while, chalked, it may exert retarding friction.
yes, I can imagine that... And if it is not chalked one should consider the reaction of the plane.

I guess I'm trying to find a set of reasonable assumptions that make the problems like the one in the OP not too messy. These problems are like
1. Find the optimal height such that the ball rolls without sliding
2. The ball is hit at a "non optimal" height, starts off sliding and eventually rolls without sliding because of the friction with the table. Find the final velocity
It seems to me that an impulsive force only requires the further assumption of a chalked cue stick, so that one does not have to bother with the normal reaction of the plane. Maybe a bit "handwaving" but reasonable, especially if the exercise focuses on the conservation of angular momentum (of an object that rotates and translates).

The solution to problem 1 is obtained in an admittedly easier way through the second law, for a frictionless plane and (possibly) a non impulsive force.
But if the plane is frictionless problem 2 does not seem to make sense.

If it helps, solve the problem where the ball is on a frictionless plane. There exists a particular height where the ball will roll without sliding.

Do you mean something like this?

If there is no friction between the table (and if we assume a chalked cue, to avoid the normal force from the table), the only relevant force is the one applied horizontally by the stick.
One can set ##\tau = F(h-r) = I \alpha## and ##F = m a##. One can then require ##a = \alpha r## which should ensure that the ball accelerates without sliding. The result is the same as in the OP, ##h=(\kappa+1) r##.

I initially thought of this, but I am having some trouble imagining what really happens for finite force and time.
How long does the contact last?
If the time is very short the ball will roll extremely slowly.
If the time is non negligible, the ball will roll with respect to the (chalked) stick and I imagine some dragging happening.

That is why I liked better the approach with impulsive forces and torques.

an impulsive force only requires the further assumption of a chalked cue stick, so that one does not have to bother with the normal reaction of the plane
What exactly happens will depend on the effect of the chalk and on the mass, balance, elasticity and angle of the cue. Let's just say we can pretend there is some combination which leads to a purely horizontal impulse.
This video shows that, even with chalk, applying side spin leads to the ball going off in a direction a little off from the cue's momentum, suggesting it might be hard to avoid some vertical component from applying topspin.
Find the optimal height such that the ball rolls without sliding
If you add "even with no friction from the table" then it ceases to matter whether the impulse is purely horizontal.