# Billiard ball hit by a cue

1. Dec 21, 2014

### anon90

1. The problem statement, all variables and given/known data

A billiard ball of mass $M$ and radius $R$ is hit by a cue as shown in the figure.
http://i.imgur.com/vJ7qB8W.png
The blow can be thought as an impulse $J$ of given value, and let $μ$ be the coefficient of static friction.

2. Relevant equations

Find the maximum angle for which the ball's initial velocity isn't null.

3. The attempt at a solution

It seems I have a big, serious doubt here.
From the definition of Impulse I know that
$J=\Delta p$ (1)
$JRsin(θ)= I\Deltaω$ , with I being the moment of inertia of the body.
Since the body is at rest before being hit we can just write $p$ and $ω$ to indicate the initial values of the rotational and translational velocities.
This is where I get stuck: I fail to comprehend how to impose the non-null initial velocity, and thus how to get the disequation which will give me the maximum value of the angle.
The problem should be solved once I know which formula to use.
I know the velocity of a given point P is $V= v_r+v_t=v_g+ωr_P$, (rotational and traslational components, $v_g$ stands for the velocity of the c.o.m.) equation (1) can lead me to the value of $v_g$, but what about the other term?
That said the problem is saying that $V$, hence its magnitude, isn't zero, but I'm having trouble putting this into pratice.
I know this is a vague answer but I'm completely lost and I can't find useful examples on my notes.

The problem seems very easy, yet I can't solve it and this is pretty depressing; what am I forgetting about?
Forgive me for the bad pic but it's the best reproduction I can do at the moment, if you don't understand something feel free to say it.

Last edited: Dec 21, 2014
2. Dec 21, 2014

### haruspex

Are you sure this is an accurate statement of the question?
I would think that initial velocity here means the velocity just after being struck (as opposed to any time later). But it still doesn't make sense. Unless theta is a right angle the ball will be sure to move. Even if we change it to 'velocity in the y direction' (i.e. vertical in the picture) that will still be the case.
Don't forget the given information about friction.
You mean $I$.

3. Dec 21, 2014

### anon90

I think they meant the velocity after being struck, for the ball is at rest before it, but I'm having a hard time imagining how this problem works...

As for the hint I know that $|F| \geq μN$ is the rule for static friction.
By using this I get $μNRsin(θ)Δt \geq IΔω$, with $θ_0$ (the one you found with the = ) being the maximum angle I'm looking for, but then I would have to deal with acceleration too, am I wrong?
I suppose you don't consider the impulsive force when finding the value of $N$.
Besides I think I'm forgetting about equation (1), as the one used is for the rotation only... I don't know if what I'm saying makes sense though, I'm pretty confused right now.

Last edited: Dec 21, 2014
4. Dec 21, 2014

### haruspex

No, that's backwards.
With my correction above, yes. Do you see how to rewrite that using J?
Is it? Why would that mean the ball does not move? Please confirm yo have stated the question word for word (or is it a translation?)
Impulses work very like forces in statics. If there's a normal impulse JN then the maximum frictional impulse is $\mu_s J_N$. You can think of the acceleration as unlimited, so the inertia acts like a reactionary force maintaining static equilibrium. To put that more mathematically, if in some small interval Δt the normal force is N then the maximum frictional force in that interval is $\mu_s N$, so the maximum frictional impulse exerted during the interval is $\mu_s N \Delta t$. Integrating, $J_F = \int F.dt \leq \int \mu_s N.dt = \mu_s \int N.dt = \mu_s J_N$

5. Dec 21, 2014

### anon90

If I were to follow your explanation then it would simply be
$μNRsin(θ)Δt \leq IΔω$
$JRsin(θ) \leq μJ_nRsin(θ) \leq IΔω$ since $J \leq μJ_n$
Then $JRsin(θ) \leq \ IΔω \over J R$
... but I'm not convinced with this outcome. How would you phisically read this equation, regardless of the question I posted?
I'd say for greater angles the ball's velocity won't have a rotational component, but I'm not too positive about this.
If that's the case then there's $J=Δp$, the other equation; the ball might as well have a traslational velocity, if I don't consider this one too, right?

It's a translation, but the text itself is clear, at least in my language.
You have to find the maximum angle, that is greater angles lead to a null initial velocity.
I don't get how this is even possible (if you hit a ball, it will most likely move no matter the angle) but the text says so.

6. Dec 21, 2014

### haruspex

No, you need to distinguish between the applied impulse and the effective impulse. The applied impulse is J, having components $J_N = J \cos(\theta)$ and $J_T = J \sin(\theta)$. The effective impulse has components $J_N$ and $J_F = \min\{J_T, \mu J_N\}$.
Why not?
I agree.

7. Dec 21, 2014

### anon90

It's the only conclusion I can come to, but as you can see I'm completely at a loss, partly because I don't get the physics behind this problem, partly because of my lack of knowledge.
If my belief is wrong, then what would happen with greater angles?

8. Dec 21, 2014

### TSny

Following haruspex, it's too bad that the wording of the problem is not very clear. It could be that "null velocity" is referring only to translational velocity of the center of the ball. If so, then you don't need to try to worry about whether or not the ball has spin immediately after it is struck.

9. Dec 21, 2014

### anon90

It's unfortunate the text isn't clear, but as I said the original meaning is what I wrote before (i.e. null velocity refers to both the translational and rotational velocity)
I think the problem is more focused on the rotational part, otherwise the impulse would've been parallel to the x axis.
The translational velocity is ruled by $J=Δp$, the ball will move if the x-component of the impulsive force is stronger than the friction force, am I right? I don't get where the effective/applied impulse come from, though. Never read of them during classes at least.
Could you help me figuring out what the disequation in the previous post mean, regardless of the problem? Now I'm kind of puzzled.

10. Dec 21, 2014

### haruspex

Again, it's a fault with the way the problem is posed.
Consider the static force problem:
A force F is applied at an angle θ to the normal to a surface of an object mass M. The coefficient of friction is μ.
What does this mean? If tan θ > μ then the tangential force felt by the mass is μF cos θ, not F sin θ. So what force is really being applied?
The problem is that you cannot just apply a specified force or impulse (as a vector) to an object. You can hit it with a mass moving at some speed in some direction, but the impulse that results will depend on circumstances.
I don't see how that helps. For θ < 90 degrees, there will always be a resulting translational velocity. If μ > 0 it will include a component at right angles to the line joining the impact point to the ball's centre, and consequently....
... there will always be a torque impulse about the ball's centre.

11. Dec 21, 2014

### TSny

I guess I’m missing something obvious. (If so, it won't surprise me!) I don’t see why the ball would necessarily pick up translational motion if $\theta$ is any angle less than 90o. It seems to me that the ball could end up slipping with a counterclockwise spin with zero translational velocity immediately after the applied impulse. But this would involve kinetic friction instead of static friction, and there is no mention of a kinetic coefficient of friction. So, I guess we are not allowed to assume slipping. Then, zero translational velocity of the CM will also imply no rotation.

Suppose you want to find the conditions for the ball to end up with zero translational velocity and also zero rotation. Consider the point O of the table at the point of contact of the ball with the table. If the impulse produces no translational velocity of the CM and no rotation about the CM, then the angular momentum of the ball about O will certainly be zero after the impulse. So, the net angular impulse about O must be zero. You can then see that this restricts $\theta$ to one specific value. I think it also puts a lower bound on the value of $\mu_s$ which is rather large.

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Last edited: Dec 21, 2014
12. Dec 22, 2014

### haruspex

No, I was - thanks. Because there was no table surface shown in the posted diagram, I was taking it to be a plan view. Your diagram shows it as elevation.
So, presuming the cue is supposed to be striking down at an angle theta, there are two coefficients of friction we need to know in general. You appear to be taking that between cue and ball to be arbitrarily large. At least that allows the impulse to be specified without leading to difficulties, so let's go with that.
In the impulse diagram, we have J cos θ and J sin θ as you show. But I would write the normal from the table as an impulse K, and frictional impulse as L <= μsK. There's no need to show mg, or involve a Δt.
anon90, does this allow you to finish the question?

13. Dec 22, 2014

### anon90

I don't know if extra questions (for the problem don't put kinetic coefficient of friction) are allowed, but...
If I were to consider slipping too then I should study another equation, specifically the torque with respect to some point O and see when it's zero and when not; the only main difference is that, this time, I have to consider the kinetic friction instead of the static one.
$F=αK$ F is the kinetic friction, α its coefficient, K the normal.
$JRsin(θ)=IΔω$
Any hint about the next step? For now it seems I won't get an inequality.

Let me see.
By looking at the diagram I get:
$Jcosθ-f=0$ -> $Jcosθ \leq μK$
$-Jsinθ+K=0$ -> $Jsinθ =K$
In this case I don't have to study the rotational part because as TSnv said null translational velocity implies null rotational's.
So $Jcosθ \leq μJsinθ$ and that's where I get the $θ$ restriction.
mg is neglected because we're considering impulsive forces only, right?
N and K on the other hands can be considered impulsive in that interval.

14. Dec 22, 2014

### TSny

In the last equation above, you've left out the angular impulse due to the friction force.

Don't forget the relation between linear impulse and linear momentum.

I think you will find that there is only one value of $\theta$ that will result in no translational velocity of the ball and this value of $\theta$ will depend on $\mu_k$.

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For the case of static friction where both $V_c = 0$ and $\omega = 0$ after the impulse, then I believe that this can occur for only one value of $\theta$ and one particular value of $\mu_s$.

15. Dec 22, 2014

### haruspex

Not sure about that. My intuition is that only static friction applies in impulses, even when spin results. If I try to justify that, I think of the ball as elastic. The impulse from the collision is transferred to internal motion of the ball before the ball starts to slip. Probably the right value to use is indeterminate, somewhere between the two coefficients, and depends on detailed conditions.
TSny did not say that.

16. Dec 22, 2014

### TSny

I agree. There are just too many assumptions required for this problem!

17. Dec 23, 2014

### anon90

But weren't we talking about the rotational velocity?
I thought the translational part was ruled by $Jcosθ≤μJsinθ$
The linear impulse/momentum relation is $J=Δp$, but why should I use it now that I'm studying the rotational aspect of the problem?

But, since slipping isn't allowed, if the translational velocity is null then the ball isn't moving at all.

Forgive me if I asked you an unclear problem.

18. Dec 23, 2014

### TSny

For kinetic friction there is only one value of the friction force: $f = \mu_k N$ (not $f \leq \mu_k N$).

If the center of the ball is at rest both before and after the impulse, then what is $\Delta p$? Use that information.

19. Dec 23, 2014

### anon90

But I am studying statics here.
https://www.physicsforums.com/threads/billiard-ball-hit-by-a-cue.788645/#post-4953283 That's how I got that inequality by looking at the force diagram, I don't think I should be using the kinetic friction force in that case.

Well, zero, but I wouldn't even use it, for what I get so far.
If I don't want the ball to slip without moving forward I would just study the torque to respect of point O and see when it's zero, but if I used kinetic friction I would obtain only one value at which the angular velocity is zero both before and after the hit. To be honest I would use the static one, but at this point I'm not in the position for giving suggestions.
Either way I don't think I understand how to solve this problem, unfortunately.

20. Dec 23, 2014

### TSny

I'm getting confused because I think there are at least two different scenarios being considered. I thought that here you were looking at the scenario where the CM of the ball remains at rest after the impulse, but the ball is allowed to have some slipping rotation. Slipping means kinetic friction is in operation, not static.

I think there's some very useful information to be obtained from the condition that $\Delta p = 0$.

If the ball does not slip then you would have static friction. And the only way for the ball not to slip and also have $V_{CM} = 0$ is to have no rotation. So, both the net linear impulse and the net angular impulse must equal zero.

Anyway, it would be nice to know exactly what the problem means by "null velocity".