Maximum Angle for Non-Null Initial Velocity of Billiard Ball Hit by Cue

[anon90]

Homework Statement

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.

Homework Equations

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

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.
Thank you for your help.

 

[haruspex]

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

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.

$JR\sin(θ)= I\Deltaω$

Don't forget the given information about friction.

with ω being the moment of inertia of the body.

You mean $I$.

 

[anon90]

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.

Thank you for your reply.
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.

 

[haruspex]

As for the hint I know that $|F| \geq μN$ is the rule for static friction.

No, that's backwards.

By using this I get $μNRsin(θ)Δt \geq IΔω$,

With my correction above, yes. Do you see how to rewrite that using J?

with $θ_0$ (the one you found with the = ) being the maximum angle I'm looking for,

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?)

but then I would have to deal with acceleration too, am I wrong?

Impulses work very like forces in statics. If there's a normal impulse J_N 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, $J_F = \int F.dt \leq \int \mu_s N.dt = \mu_s \int N.dt = \mu_s J_N$

 

[anon90]

No, that's backwards.

Gosh, my bad.

Do you see how to rewrite that using J?

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?

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?)

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.

 

[haruspex]

$JRsin(\theta) \leq \mu J_n Rsin(\theta) \leq I\Delta ω$ since $J \leq μJ_n$

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 $J_F = \min\{J_T, \mu J_N\}$.

for greater angles the ball's velocity won't have a rotational component

Why not?

I don't get how this is even possible (if you hit a ball, it will most likely move no matter the angle)

I agree.

 

[anon90]

Why not?.

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?

 

[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.

 

[anon90]

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.

Thank you for your reply.
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.

 

[haruspex]

I don't get where the effective/applied impulse come from, though. Never read of them during classes at least.

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.

It could be that "null velocity" is referring only to translational velocity of the center of the ball.

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...

what would happen with greater angles?

... there will always be a torque impulse about the ball's centre.

 

[TSny]

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...
.

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 90^o. 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.

 

[haruspex]

I guess I’m missing something obvious.

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 <= μ_sK. There's no need to show mg, or involve a Δt.
anon90, does this allow you to finish the question?

 

[anon90]

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.

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.

anon90, does this allow you to finish the question?

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.

 

[TSny]

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.
$JRsin(θ)=IΔω$

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

Any hint about the next step? For now it seems I won't get an inequality.

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 $\mu_k$.

------------------------------------------------------------------------------------------------------------------------------------------------------------------------
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 $\mu_s$.

 

[haruspex]

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

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 said null translational velocity implies null rotational'

TSny did not say that.

 

[TSny]

Probably the right value to use is indeterminate, somewhere between the two coefficients, and depends on detailed conditions..

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

 

[anon90]

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 $\mu_k$.

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?

TSny did not say that.

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.

 

[TSny]

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?

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.

 

[anon90]

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

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.

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

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.

 

[TSny]

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.

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.

Well, zero [for $\Delta p$], but I wouldn't even use it, for what I get so far.

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

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.

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".

 

[anon90]

II 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.

But if I'm looking for solutions for which the CM remains at rest and the ball doesn't slip too then I have to use the static friction, don't I?

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

I think it means when the ball stays still, i.e. no slipping and no CM translation. If either one of them exists then the velocity isn't null.

 

[TSny]

But if I'm looking for solutions for which the CM remains at rest and the ball doesn't slip too then I have to use the static friction, don't I?
I think it means when the ball stays still, i.e. no slipping and no CM translation. If either one of them exists then the velocity isn't null.

OK, I was confused about what scenario you were considering. Yes, no slipping and no CM translation will require two conditions to be satisfied - one to prevent translation of the CM and one to prevent rotation about the CM. When I set up these two conditions I find that there is only one possible value of $\theta$ and only one possible value for $\mu_s$ that will result in a null velocity. [Edit: I should have said there is a certain minimum value for $\mu_s$ in order to get a null velocity.]

 

[anon90]

OK, I was confused about what scenario you were considering. Yes, no slipping and no CM translation will require two conditions to be satisfied - one to prevent translation of the CM and one to prevent rotation about the CM. When I set up these two conditions I find that there is only one possible value of $\theta$ and only one possible value of $\mu_s$ that will result in a null velocity

I'm glad everything is finally clear.
Now, from what I read so far, this problem should be no different from your average one where you just solve the forces diagram; the only main difference is that we care for impulsive forces only. I thought I would've needed to use something else so I wasn't thinking of this kind of solution.

As for the CM translation I get $Jcosθ≤μJsinθ$ as you can see in my previous post.
When I study the rotation (torque to the respect of CM; I could study it to the respect of a random point of the ball though, right?) I should get $JRsin(θ)-fR=IΔω$, in this case Δω=0.
When the two conditions are satisfied the ball isn't moving nor slipping.
Now, if I use the static friction in the second conditions I should get two inequalities, so more than one possible value; on the contrary, if I use kinetic friction then only one value will satisfy both conditions.
μ is known, the only variable is θ.

Any mistakes?

 

[TSny]

Suppose you let K represent the impulse due to the static friction force. It need not be equal to the maximum possible value.

Write the conditions for $V_{cm} = 0$ and $\omega = 0$ in terms of K. Can you conclude anything about the angle $\theta$?

I made a mistake earlier when I said that I thought there could be only one possible value for $\mu_s$. Instead, I think there is a certain minimum value for $\mu_s$ in order to have null velocity.

 

[haruspex]

I think it means when the ball stays still, i.e. no slipping and no CM translation.

I don't believe it can mean that. As TSny says, that would mean there is only one angle that works, and it does not depend on μ; yet the question asks for a minimum angle, and since it specifies μ it seems likely that this angle should be a function of μ.

 

[anon90]

Suppose you let K represent the impulse due to the static friction force. It need not be equal to the maximum possible value.

Let's say K is the impulsive normal, then $f \leq μK$
$Jcosθ−f=0$ -> $Jcosθ≤μK$
$−Jsinθ+K=0$ -> $Jsinθ=K$ (1)
$Jcosθ≤μJsinθ$
This should be the condition for the CM not moving, I don't think you told me whether it's correct or not.

No rotation: same as before $JRsin(θ)−fR=IΔω=0$
Since $f \leq μK$ I get $Jsin(θ) \leq μK$
I can use (1) to write K in terms of things I know.

 

[TSny]

Let's say K is the impulsive normal, then $f \leq μK$

OK, $K$ is normal impulse and $f$ is frictional impulse. (Sorry I had changed the notation on you.)

I agree with what you wrote for the linear condition:

$Jcosθ−f=0$

And I agree with what you wrote for the rotational condition:

$JRsin(θ)−fR=IΔω=0$

What can you deduce about $\theta$ from these two equations?

 

[anon90]

What can you deduce about $\theta$ from these two equations?

From just the two of them?
$JRsin(θ)-Jcos(θ)R=0$ -> $J(sin(θ)-cos(θ))=0$
So $θ=\pi/4 +n\pi$

But, if I consider $−Jsinθ+K=0$ (forces on y-axis) together with the fact that the frictional impulse is linked to the impulsive normal through that inequality, I get a different outcome; what's wrong with my previous reasoning?

 

[TSny]

From just the two of them?
$JRsin(θ)-Jcos(θ)R=0$ -> $J(sin(θ)-cos(θ))=0$
So $θ=\pi/4 +n\pi$

So, only the value $\theta = \pi/4$ will work.

But, if I consider $−Jsinθ+K=0$ (forces on y-axis) together with the fact that the frictional impulse is linked to the impulsive normal through that inequality, I get a different outcome; what's wrong with my previous reasoning?

$\theta$ is now fixed. Use the inequality $f \leq \mu_s K = \mu_s J \sin \theta$ in either of your impulse equations to deduce a restriction on $\mu_s$.

 

[anon90]

$\theta$ is now fixed. Use the inequality $f \leq \mu_s K = \mu_s J \sin \theta$ in either of your impulse equations to deduce a restriction on $\mu_s$.

The thing is, I already know μ's value, I don't see why I should be looking for its restriction.

 

[TSny]

You have four conditions that must be satisfied if "null velocity" means no translational velocity and no rotational velocity:
(1) Net horizontal linear impulse = 0
(2) Net rotationall impulse = 0
(3) Net vertical impulse = 0
(4) $f \leq \mu K$

The first two already restrict $\theta$ to one value.
Then, imposing the last two conditions restricts the allowed values of $\mu$.

The fact that the problem statement makes no mention of a restriction on $\mu$ is just another oddity of the wording of the problem.

 

[anon90]

You have four conditions that must be satisfied if "null velocity" means no translational velocity and no rotational velocity:
(1) Net horizontal linear impulse = 0
(2) Net rotationall impulse = 0
(3) Net vertical impulse = 0
(4) $f \leq \mu K$

The first two already restrict $\theta$ to one value.
Then, imposing the last two conditions restricts the allowed values of $\mu$.

The fact that the problem statement makes no mention of a restriction on $\mu$ is just another oddity of the wording of the problem.

So, by your logic, I have $μ \geq sqrt(2) /(2K)$
But I know what (1) -> $Jsinθ=K$, hence $μ \geq 1 /J$ (2)
All the values of μ satisfying (2) are good, and there's only one possible value for θ.

 

[TSny]

So, by your logic, I have $μ \geq sqrt(2) /(2K)$

How did you get this? Note that $\mu$ should be dimensionless.

 

[anon90]

How did you get this? Note that $\mu$ should be dimensionless.

Right.
$f≤μsK=μsJsinθ$

Together with $θ= 45°$ and $Jcosθ−f=0$, I get $μ \geq 1$

 

[TSny]

OK, that's what I got, too.