Torque or force?

[vlad777]

If we have a rod that has no mass (and hawing no mass is the reason for this question).
(this can be approximated with wery low mass rod in space).
Now we apply force to only one end of this rod and this force is not
collinear (edit: and is perpendicular) with the rod and there are no other forces (like friction or inertion).

Is this rod suppose to rotate (around the end with applied force)
or translate(in direction of force)?


In other words, is center of rotation at zero or at infinity?

 

[duffy]

If we have a rod that has no mass

This immediately falls into a category of questions that give nonsensical results because you set nonsensical constraints. Torricelli's trumpet is another one. A forum on philosophy is more in order for these types of questions.

 

[vlad777]

Isn't there a limit to an equation?

Infinity is a philosophical concept accepted in exact science and math.

If you apply a force to something with no mass, then acceleration is infinite;
but does it translate or rotate?

 

[jpanhalt]

If you apply a force to something with no mass, then acceleration is infinite;
but does it translate or rotate?

No it isn't. Division by zero is undefined. Are you confusing weight with mass, i.e., something in micro-gravity with mass , but negligible weight?

John

 

[vlad777]

Weight is mass times gravitational acceleration. I meant mass, but I am questioning myself now.

Anyway, what equations describe motion of the body when applied force is not collinear with center of mass?
What should I search for (on google or wiki) ?

 

[duffy]

Anyway, what equations describe motion of the body when applied force is not collinear with center of mass?

Your force is a vector, so unless it is tangent or normal to the center of mass, there will be components for both rotation and translation. These will be found by trig functions relative to the angle of the force vector in relation to the position of the center of mass, and would be computed separately for components tangent and normal.

Classical mechanics, inertia and moment of inertia are the things you might want to search on.

 

[jpanhalt]

I unsuccessfully tried to find one of the early videos from space flight. In that video, the astronaut touches the end of a pencil floating by. It rotates about its center of mass. An off-axis force to an object with mass, regardless of how small that mass may be, will cause rotation about the center of mass as well as translational acceleration. No mass, no answer.

John

 

[Ratchit]

Vlad777,

Yes, Duffy has it right. It is like asking what happens when the unstoppable force pushes against the immovable wall. It is more of a philosophical question than a physics question.

Ratch

 

[vlad777]

OK delay that zero mass stuff (I thought that it could be used as dm but no matter now).

Have you ever threw a stick in the air and then hit it with another stick while still in air?
Farther from the center you hit it it gets more rotation,
and closer to the center more translation.

So this has only one force, but moment is defined as two forces in opposite direction on some distance.
How do I construct the moment?

Edit: Is second force reaction to first, and that's why they are the same?

 

[vlad777]

Another thing, is following true?

Plane trough center of mass always divides this mass in two equal parts.

 

[ljcox]

Torque requires 2 forces, ie. a couple.

If the rod has 0 mass, then there is only one force acting on it.

Therefore it should translate, not rotate.

If it has mass, then the other force is due to Newton's third law & will act at the centre of mass, thus it will rotate. (like the pencil mentioned above)

 

[vlad777]

Thanks ljcox.
You answered two of my questions.

 

[ljcox]

Thanks ljcox.
You answered two of my questions.

You're welcome.

But as someone else said, a massless rod is not sensible because it is not possible to determine the acceleration.

ie. as the mass of the rod approaches 0 the acceleration approaches infinity (assuming Classical Physics). But it is undefined when M = 0.

Your question re the centre of mass can be answered as follows:-

Say you have 2 masses (M1 & M2) attached to the ends of your massless rod such that their centres are distance l apart.

The centre of mass is where the system would balance. ie. where the moments are equal.

So say the centre of mass is x from M1.

Then the system will balance when x * M1 = (l - x) * M2.

Thus x = l * M2/(M1 + M2).

So the answer to your question is:-
Yes if M1 = M2

No otherwise.

 

[vlad777]

ljcox, do you know how 3d moments are "added" ?

 

[ljcox]

ljcox, do you know how 3d moments are "added" ?

Torque is a vector quantity, so you add the vectors.

 

[Ratchit]

vlad777,

do you know how 3d moments are "added" ? .

All moments are "3d".

Ratch

 

[ericgibbs]

Vlad777,

Yes, Duffy has it right. It is like asking what happens when the unstoppable force pushes against the immovable wall. It is more of a philosophical question than a physics question.

Ratch

hi Ratch,
This answer to this puzzle is to be found in the effect of cooling liquid helium placed in an impervious container and is cooled close to absolute zero.

At close to absolute the liquid helium runs out of the bottom of the container.

https://www.youtube.com/watch?v=9FudzqfpLLs

The answer in my opinion is that when an unstoppable force pushes against an immovable wall, the unstoppable force passes thru the immovable wall.

In this way no 'rules' are broken.

E.

 

[MrAl]

Hi,

If you have a rod with no mass it doesnt matter where you hit it with a point force, it's going to move in that direction and it wont rotate at all because mass is what gives it inertia and since there is no inertia then there is no point in the rod that has inertia of rest and that is what makes it rotate. But unlike a rod with mass, it will not continue to rotate and/or translate once the force is removed because again there is no inertia. For a rod WITH mass it would rotate or translate or both rotate and translate depending on exactly where it was struck, and it would continue this action even once the force is removed because some parts of the rod will have inertia of motion.

You can figure this out for a rod with mass by dividing the rod up into small sections of length dL and enumerate the sections, and each section becomes a point mass at the signed distance d from the center and each section represents the mass of the rod at the center of each dL. In a uniform rod the masses will all be equal, but their distance from the center will be different. Start with 3 sections. When you hit it on the end (section 1) with a perpendicular force that end is going to move a fraction of an inch ds in time dt according to the laws of motion. This is going to cause a rotation dr in section 2 which is going to make it move and rotate because of the fact that section 3 acts as a fulcrum, and that is going to cause a rotation only in section 3 because section 3 in the time dt is still considered fixed and there is no translational force. Since we hit it on the end there is no secondary lever action so the rod rotates and does not translate. Note this would be very hard to actually do in real life. If we hit it in the middle section (section 2) this would be different because we have inertia of rest for sections 1 and 3, and so we have secondary lever action. This causes the rod to translate and not rotate.
If we use 5 sections and hit it in section 2, there will be partial secondary lever action so both ends of the rod will move, but the end nearest the point force will move farther in time dt than the end farther away from the force, so the rod will translate and rotate.

Dividing the rod up into 1000 equally sized pieces leads to a refinement of the process, and in the limit as the length of each piece dL goes to zero you'll get the formula.

Just to recap though, a rod with no mass will not behave this way because there is no real inertia for a rod with no mass and the whole calculation is based on some real finite inertia, so it is like applying the force to a vacuum which doesnt do much Things like this are often what theory itself is made up of so there are usually predetermined rules to follow, but we dont make up the rules for fictitious objects without having some observed measurements to guide us and a more important reason for doing so.

For the rod with mass, you use classical mechanics and the laws of motion. Hitting it perfectly perpendicular and at the *very* end causes a rotation only because there is no inertia to make the other end rotate, and hitting it in the middle causes it to translate only because there is no inertia to make either end rotate, and hitting it between the center and end causes it to translate and rotate because the shorter end (from the point force) acts to partially rotate the other end which makes it move.
It is interesting that hitting it even a tiny tiny fraction in from the end causes at least some small translation as well as rotation.
A simpler way to look at it is that when hit at the very end we have only one lever action to consider, but when hit somewhere in from the end we have two lever actions to consider.

 

[vlad777]

Torque is a vector quantity, so you add the vectors.

vlad777,
All moments are "3d".
Ratch

I figured this out, thanks.
(moment is cross product of arm and force, and as such goes into third dimension,
and also is a vector that can be added)

MrAl ,

infinitesimal lengths, masses and all, was what I was thinking about when asking
this question; for only purpose of integrating,
so I'll give your explanation a go.
Many thanks.

I like these problems when you model somehow and then integrate;
I am bad at it, but I think it's really cool.

EDIT: Maybe it would be better if I said infinitesimal volume not mass.