A ball struck by a cue in billiards with English goes straight at first....

[sophiecentaur]

Completely irrelevant. If that force is significant, then it causes significant linear acceleration of the ball in that direction.Because it is, according to Newtons 2nd Law.

Isn't it necessary to consider the angular momentum in this situation, though? The forces involved are difficult to quantify - as with most collision problems - because they will vary throughout the ball - cue collision so it's best to think in terms of momentum and angular momentum changes and Impulses. That's ok, I think, as long as it can be assumed that angles don't change much during the actual collision. Flexing of the cue adds a complication but any flexing will only produce a significant change in the 'lateral forces' on the cue.
What I have found out is that a tangential impulse on a (free) body will produce the same change in angular momentum as linear momentum. So a sphere with all it mass at the centre and an MI of zero will not move forward at all if it's struck tangentially (simplest case). A dumbbell (MI = mr², struck 'tangentially' will move off at half the speed as if both masses were together and struck with the same impulse. A solid sphere is half way between, with an MI of 2mr²/5 so the translational momentum due to a tangential impulse will be 2/5 of a single mass of m. That implies (to me) that the effect of the tangential component of the cue's impulse will only be 2/5 of the radial component. So the diagram that's been used, showing F_t and F_r, neglects this factor of 2/5. As far as I can see, this will mean that, for a collision in free space, the ball will always move slightly (or a lot) to the left of the cue direction. It needs the steering action of the balloon the cloth to make it actually go straight.
I found this link very useful.

 

[A.T.]

The forces involved are difficult to quantify.

True, but the linear acceleration is parallel to their sum. There is no "weighting" of components depending on their direction, or whatever you are trying to suggest here.

 

[sophiecentaur]

There is no "weighting" of components depending on their direction,

I think you would need to justify that, in the light of my arguments using a dumbell and a sphere with nearly zero MI. The resulting motion of those two is very dependent on the angle of the impulse. If those arguments are valid then so is the argument for a uniform sphere.
The principle of resolving vectors is not in question, is it?

 

[A.T.]

I think you would need to justify that

See Newtons 2nd Law.

 

[sophiecentaur]

See Newtons 2nd Law.

What I'm saying doesn't contravene Newton's Laws - if you apply it correctly. Neither does it contravene Energy Conservation.

 

[A.T.]

What I'm saying doesn't contravene Newton's Laws

You were asking me to justify why linear acceleration is parallel to the sum of applied forces, which is a direct consequence of Newtons 2nd Law.

 

[sophiecentaur]

You were asking me to justify why linear acceleration is parallel to the sum of applied forces, which is a direct consequence of Newtons 2nd Law.

But this isn't just a case of 'applied forces' it is 'applied impulses' and the times involved will depend upon the MI of the object. I would agree that a string pulling the ball along for a while would produce a result other than what I am describing. Unlike with the string, once the cue has left the ball, it no longer has any effect on it. That involves another concept.
Imagine dropping a barbell, with one end held on a string. You would have a form of pendulum and the tension in the string would vary in time. That would be different from hitting one end of a barbell in space.

 

[A.T.]

But this isn't just a case of 'applied forces' it is 'applied impulses' ...

You were specifically talking about components of a force:

It's worth noting that the Force F_ft may be significantly large, as the diagram shows but it is not acting on the cm of the ball and its effect on final direction is tiny, compared with the component that acts directly through the cm.

... and the times involved will depend upon the MI of the object.

The application time is the same for all components of the same applied force.

 

[sophiecentaur]

You were specifically talking about components of a force:
The application time is the same for all components of the same applied force.

I am still having a problem with this.
Are you saying that an impulse applied radially for a brief period will have precisely the same initial effect as an impulse applied tangentially? How does that argument work for a dumbel? The time of the action is the same but the forces are different (being due to the reaction against the cue).
In addition to this basic idea, if there is any slippage between tip and ball, the tangential force will be even less less, which could also be relevant.
I can't be sure, from you remarks, whether you are just insisting on the application of Newton's laws (which is, of course right) or whether you are really appreciating the forces (impulses) involved. The dumbel model is simpler but easier to see.

 

[A.T.]

Are you saying that an impulse applied radially for a brief period will have precisely the same initial effect as an impulse applied tangentially?

The same impulse results in the same linear momentum change:
https://en.wikipedia.org/wiki/Impul...ion_in_the_case_of_an_object_of_constant_mass

 

[sophiecentaur]

The same impulse results in the same linear momentum change:
https://en.wikipedia.org/wiki/Impul...ion_in_the_case_of_an_object_of_constant_mass

Oh, I get that BUT the forces are different, aren't they? So the Impulses will be different. Show me that the forces, radial and tangential are the same for a dumbel and I will believe I have it all wrong.

 

[A.T.]

Are you saying that an impulse applied radially for a brief period will have precisely the same initial effect as an impulse applied tangentially?

So the Impulses will be different.

Then your question makes no sense, because you aren't comparing applied radially vs. applied tangentially, just different impulses.

 

[sophiecentaur]

Then your question makes no sense, because you aren't comparing applied radially vs. applied tangentially, just different impulses.

Why do you have a problem with this? The resolved radial and tangential forces can be found from the applied force vector. They are different and they are applied for the same time. Is it not reasonable that their net effect can also be calculated? Additionally, if the tangential force is beyond limiting friction, the tangential effect will be even less.
Where is the hole in that argument? We are. after all, discussing the net effect on the ball and where it will go.

 

[A.T.]

Why do you have a problem with this? The resolved radial and tangential forces can be found from the applied force vector. They are different and they are applied for the same time.

There is no problem with the above, and It's not what I criticized.

We are. after all, discussing the net effect on the ball and where it will go.

The net acceleration is parallel to the net force.

 

[sophiecentaur]

'Net produces net' is something I would have said was obvious. Whether or not I made that clear, is hardly part of the answer to the OP.
What is your opinion about the actual situation and whether this net force can be in line with the cue?
If a player is aware of a sideways force on the cue (and they tell us that the cue actually bends) then that suggests to me that the net force on the ball is to the left. Some other force is needed for the ball to end up going straight.
Rather than just pointing out that part of the above is "wrong", it could help if you could add something positive.

 

[A.T.]

'Net produces net' is something I would have said was obvious.

Well, I hope we mean the same thing by "net force". I mean a simple vector sum of the force components, without any extra weighting based on whether they act radially or tangentially.

What is your opinion about the actual situation and whether this net force can be in line with the cue?

The friction coefficient required to produce the result in the videos by impact alone is way below 1, so not completely unrealistic. I don't know whether it's feasible for that particular material pairing with the chalk in between.

 

[sophiecentaur]

You don't like the term "weighting" but you must agree that the two forces will not be the same. They are both reactive forces. The radial force is due to the mass and the tangential force is due to the MI divided by the radius. They are different and that is what the weighting is about. It isn't necessary to calculate the actual value as the direction is all that counts.
I have scribbled some calculations which suggest that the value of tangential reactive force is 2/5 of the radial reactive force for a uniform sphere. I have to go out and chop some wood in the sun but I will try to present the sums in a readable form. Would you disagree with my earlier idea that a cue hitting a rigid sphere with all its mass concentrated at the centre would (could only) experience a zero tangential reaction force?

 

[A.T.]

you must agree that the two forces will not be the same.

Sure, in the first video the friction would have to be about 1/4-1/3 of the normal force, in order to explain the result by impact alone. These aren't completely unrealistic friction coefficients.

 

[sophiecentaur]

Sure, in the first video the friction would have to be about 1/4-1/3 of the normal force, in order to explain the result by impact alone. These aren't completely unrealistic friction coefficients.

You seem to be completely ignoring the relevance of Moment of Inertia. Can you disagree that it is crucial and that the ratio of forces is calculable? The effect of friction will be very non-linear in view of the cue tip material and the chalk so you can't do more than say it could reduce the tangential effect. The MI is always with us yet you still don't seem to be acknowledging it. Do you, in fact, get my point about a sphere with all its mass at the centre?

 

[A.T.]

You seem to be completely ignoring the relevance of Moment of Inertia.

To verify this, you could measure the angular velocity, and check if it is consistent with the tangential impulse required for the observed linear velocity.

 

[sophiecentaur]

To verify this, you could measure the angular velocity, and check if it is consistent with the tangential impulse required for the observed linear velocity.

How would one know the tangential impulse? It would not be possible to measure or even apply just a tangential Impulse under these conditions (without any radial force)
In any case, would there be any point in adding experimental error to what the Maths of such well established theory can tell us? I get the feeling that you really don't feel able to go along with what I'm saying. What objection can you have? Could I put it another way that would be more convincing? Nothing in the above couple of posts violated Newton.

 

[A.T.]

How would one know the tangential impulse?

From a video where spin is around vertical axis you can get the linear and angular velocities. Then you can check if their relationship is consistent with a single impulse applied at the queue contact point.

 

[sophiecentaur]

From a video where spin is around vertical axis you can get the linear and angular velocities. Then you can check if their relationship is consistent with a single impulse applied at the queue contact point.

No doubt but I was asking you a question about the theory. That is much more reliable. I thought you could help me, in the light of your earlier remarks.

 

[A.T.]

From a video where spin is around vertical axis you can get the linear and angular velocities. Then you can check if their relationship is consistent with a single impulse applied at the queue contact point.

A single short impulse on an uniform resting sphere should result in the following relationship:

w/v = 5/2 * r/R²

where:

w : angular speed
v : speed
r : lever arm of the force around the center
R : radius of the sphere

In the below video a quick & dirty analysis yielded an error <10%. Thus the observed kinematics can be explained without significant forces other than the cue impact.

 

[sophiecentaur]

A single short impulse on an uniform resting sphere should result in the following relationship:

w/v = 5/2 * r/R²

where:

w : angular speed
v : speed
r : lever arm of the force around the center
R : radius of the sphere

In the below video a quick & dirty analysis yielded an error <10%. Thus the observed kinematics can be explained without significant forces other than the cue impact.

That's interesting and the 5/2 factor is familiar. But could you translate that into direction of motion? Remember, the force on the ball is not necessarily in the direction of the cue (or the cue wouldn't bend*.
What was your quick and dirty analysis? Looking at the linear and rotational motion between frames? That is not the complete kinematics of the situation.

* In fact, the net motion will be in the direction of the net force, which is patently not in the line of the cue. That's putting it in a nutshell.

 

[A.T.]

Looking at the linear and rotational motion between frames?

Yes

net motion will be in the direction of the net force,

Yes.

 

[sophiecentaur]

YesYes.

Glad to read those positive responses.
And do you have an answer about the direction?

 

[A.T.]

And do you have an answer about the direction?

The second "Yes" is about the direction.

 

[sophiecentaur]

Only if you know the direction of the force and that goes without saying. Can you help with calculating the actual direction of the force? You seem to imply that you could but I should do it myself. Can you do it?

 

[A.T.]

Can you help with calculating the actual direction of the force?

As you said yourself, it's the direction of motion.