# Torque on a rod explanation

1. Jan 25, 2012

### aaaa202

I have now finished my lectures on classical mechanics, but I'm a bit ashamed to say that I still don't understand the concept of torque fully.

Let's imagine this. We have a rod being acted on by the equal and opposite forces, but in different distances from the centre of rotation (attached picture). This then means that the rod will start rotating anticlockwise. To understand why the upwards force is more efficient in rotating the object we can apply an energy observation: Since the object is a rigid, rotating object, the power from the force farthest away from the rotation centre will be greater, and thus the rod will rotate anticlockwise.

Now, that observation helps me understand torque. The problem is that it does not really apply to all cases. Consider this: When the object has not started to rotate yet, there will be no power from neither of the two forces. Yet everyone knows that the object will INDEED start to rotate.

Does this mean that torque can't be understood in terms of concepts like momentum and energy? - i.e. concepts that I recognize from linear dynamics.

I ask this because the derivations in my book somehow always assume that the object to which we apply a torque is rotating already - thus they indirectly use the energy observation made earlier. But that cannot explain the case, where the object has not yet started rotating. Would this be an empirical observation? I mean, that we define angular momentum from torque, which applies to an object already rotating, but then we find out that the concept works on an even deeper level - i.e. in the case where the angular velocity is zero..

That was a lot of questions, but I have never got a good answer from anyone to this question - that is my teachers etc. etc.
If in some deeper way this actually can be explained through Lagrangian- and Hamiltonian mechanics, you are welcome to say that - I just don't like the feeling of not understanding a concept fully...

Cheers :)

#### Attached Files:

• ###### Torques on rod.png
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2. Jan 25, 2012

### Studiot

Wow, perhaps your difficulty is that you are trying too hard.

The text book 'assumes' that the object is rotating because it cannot be doing anything else

In particular it is not in equilibrium.

There is no such thing as 'just started to rotate' with, for instance, the lever system you have shown. If it is not rotating it is being held against rotation by another force that you have not shown.

Does this help?

Last edited: Jan 25, 2012
3. Jan 25, 2012

### aaaa202

Certainly you can observe a situation like mine, but perhaps I didn't make it clear enough what I meant.

Suppose you have something supporting the rod and then removed it. It would obviously initiate a rotation, because it is NOT in equilibrium - i.e. the net torque on it is not zero. The torque can explain why it will rotate as soon as the object has started its rotation - but the initian phase can as I see it not be understood in terms of energy concepts.

But perhaps a torque is just a knew concept not really related to linear dynamics. In linear dynamics after all you can't really explain why something is accelerating due to a force just using energy concepts.

I must admit, that I look forward to see if lagrangian and hamiltonian dynamics give a whole new way of understanding clasiccal mechanics.

4. Jan 25, 2012

### Studiot

Langrangian and Hamiltonian views of mechanics don't involve comparing two systems with different characteristics any more than the Newton view does.
Another way to put this would be to say that we don't change the system part way through the path, as removing a restraint or force would do.

5. Jan 26, 2012

### aaaa202

Then I'm just saying that there must be some empirical, in the way that torque behaves in nature, which can't be shown with the equations I have seen.

Either that or you can't use energy concepts to understand torque completely.

I'm thinking the last one, is probably the right one.

6. Jan 26, 2012

### Studiot

I am sorry I am not explaining myself very well.

Firstly you have two different, independent mechanical systems.

1) The system at rest with restraint to balance forces and moments. Since it is at rest there are no energy considerations, although as an arrangement of forces and reactions energy theorems such as that of Maxwell, Betti et al may be deployed.

2) A second, different and independant, system after the restraint has been removed.

Let us take a linear motion example first.

Consider a projectile mounted in a catapault, or an arrow held in a taught bow by an archer.
We can define the forces in the bow or catapault string and therefore the necessary restraint.
The model is adequate to fully describe the situation

This is situation (1) above.

Now remove the restraint. the catapault or bow fires and the projectile accelerates forwards to its terminal velocity.
The projectile does not instantly achieve its terminal velocity.
It goes through a startup process as you envisage.
To model this startup process you need details of the elastic properties of the catapault or bow to find the elastic strain energy and model its transfer to the projectile.
You also need details of the inertial resistance of the projectile.

This is situation (2) above.

Now consider your rod in its pivot.

In situation (1) the rod is restrained and no energy is involved.

The model is adequate to fully describe the situation.

In situation (2) the restraint is suddenly removed and the rod commences rotation.
It does not reach its terminal angular velocity instantly.
To model this startup process you need details of the rotational inertia of the rod and the stiffness or otherwise of the driving forces F, just as above.

In short you model is missing some things.

Does this help?

7. Jan 26, 2012

### aaaa202

hmm no, I think we are discussing two different things. Mainly I think that is because the situation i set up doesn't really deliver the point I want.

So I think what's wrong with my understand, is perhaps that I don't understand why the center of mass will make a downwards motion.
Consider the attached picture, which is a modified model of the rod, where now, the downwards torque is bigger. You could see this as the weight of the rod, and then the other force would be a person pushing upwards on the rod.

Due to the sum of torques being different from zero it will start to rotate downwards.

But at the same time, the rod would obey the linear equation of forces for its center of mass. That is that zero net force means zero acceleration. But clearly if it rotates the center of mass will fall downwards. So what explains this? Is it wrong to see the geometric center of the rod as the center of mass in this case?

Similarly I don't understand the opposite case. That is, if you have an equilibrium because the resultant torque is zero. This could be achieved by having a force in one distance from the rotational centre, and then another with half the magnitude but in twice the distance. Yet certainly the center of mass would not be in equilibrium, and would have to fall downwards or upwards depending on the convention.

#### Attached Files:

• ###### Torques on rod.png
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8. Jan 26, 2012

### Staff: Mentor

The net force is not zero. Don't neglect the force from the pivot.

9. Jan 26, 2012

### aaaa202

So you are suggesting that the pivot pushes down on the rod? How on earth can it do that? Usually a force from a pivot comes through a reaction, but there's not really anything pushing up against the pivot, which could explain that it pushed down.

10. Jan 26, 2012

### Staff: Mentor

Yes, the pivot pushes down on the rod. (You have an upward force on the rod, which leads to that reaction force at the pivot.)

If you don't think the pivot is pushing down, then remove it and see what happens.

11. Jan 26, 2012

### Studiot

Once again your diagram is incomplete.

Have you tried the analysis with Doc Al's comment incorporated?

Attached is an equilibrium analysis.
You cannot apply forces as you have shown without restraint or motion because the system is not in equilibrium.
For equilibrium without other restraint the applied upward force (I have called it F) has to be greater than the applied downward force (I have called it W) by the value of the vertical hinge force.
In this case you have the moment equilibrium maintained as the product of the larger force and the smaller lever distance equal to the product of the larger lever distance and the smaller force.

It really is easier to see as a diagram than in words.

This is fundamental to the setup as a machine with mechanical advantage or if you like the fact that to maintain equilibrium you have to push up with a greater force than the weight of the plank if you push up at a point between the plank centre and the pivot.

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• ###### plank1.jpg
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12. Jan 26, 2012

### aaaa202

hmm. Upwards force on the rod? :(
Surely the sum of forces on the rod is zero. What is not zero is the torques. So would you then understand the non-zero net torque as something that makes the rod push up against pivot making the whole system rotate downwards?

13. Jan 26, 2012

### Staff: Mentor

Only if you ignore the force from the pivot.
You can think of it that way. The only way that the center of mass can accelerate downward is if there is net downward force on the rod.

14. Jan 26, 2012

### Studiot

The equilibrium torques (I prefer to use the term moments) sum to zero as I have shown.

15. Jan 26, 2012

### aaaa202

Yes okay I see now, that I should have incorporated the force from the pivot point. Yet I don't really understand what causes the pivot to push down on the rod.

That force must come from rod pushing up against the pivot. But again that must mean that the rod is moving upwards.. Clearly it's not. Though the sum of torques is nonzero. So that'd be the only thing explaining that the rod pushes up against the pivot.

But then we're finally back to the original discussion of how I don't understand torque when energy observations can't be applied.

16. Jan 26, 2012

### aaaa202

I think I'm going in circles. I'm afraid my whole approach to mechanics is probably not right.
I see it as this:

First we take the rod without the force from the pivot and try to understand what makes the force from the pivot appear. In that case we will have that the net torques makes the object push against the upper part of the pivot making it push down such that center of mass translates downwards.

But maybe it's wrong to do it like that. You have to understand the forces as something that appear instantly.

But I'm still left with the fact that torque can't be understood through energy in this case.

17. Jan 26, 2012

### Studiot

Yes, so try to follow the chain of reasoning through with us.

First take my amended version of your diagram.

Consider a plank, hinged at one end (P) with weight W acting downwards at L2, halfway along the plank.

Apply a variable upward force F at some point between W and the hinge.

Let the hinge forces be Y horizontal and X vertical

Now consider what happens as we increase F from zero.

1) F=0

2) 0<F<W

3) F=W

4 F>W

5 F>>W

18. Jan 26, 2012

### Studiot

One thing puzzles me.

You started this thread mentioning Hamilton and Lagrange. These are not usually taught until quite a high level of mechanics.

So what level are you studying at?

There is something known as 'the equation of energy', which is a precursor to Hamilton's methods which you could employ.

However it is important to get the basics of mechanics right first.

If I posted a derivation of this equation from the principles of angular and linear momentum would that help. It uses simple calculus.

19. Jan 26, 2012

### rcgldr

Assume the rod is rigid. When the forces are applied to the rod, the effect of those forces travel through the rod as the speed of sound in the rod. For most physics problems, the speed of sound within the rod is treated as if it is infinite, so that the force from the pivot occurs simultaneously with the other forces on the rod.

There is no issue here. Assume the initial state of the rod is that it is not moving, so it's inital energy state is zero. The resulting torque force on the rod increases it's (angular) kinetic energy over time (assming the rod has mass and angular inertia). If there are no drag forces, then rate of angular velocity continues to increase as long as the forces are applied. If the forces are constant, then power increases with the angular velocity of the rod.

The pivot force is initially upwards and always in the tangental direction of the end of the rod. The rod will be accelerating counter-clockwise in the diagram you included in your first post.

20. Jan 26, 2012

### aaaa202

Yes that might be very helpful.

I'm currently at the first year of university. I think assignments in mechanics tend to be very easy, but the underlying concepts can still be quite hard.

I've tried to follow your way, but I'm afraid to say that I still doesn't really help me in understanding.

How I understand things now:
If you have the rod attached to the pivot with the forces applied you can see it like this:
Initially the rod will try to rotate around its center of mass, since there is a net torque around this point. The rod will however push against the upper side of the pivot, making the system rotate downwards as you would expect.

21. Jan 26, 2012

### Studiot

Yup that takes care of case (1) where F = 0.

It is important to realise that the hinge forces, X and Y will be what they need to be to make things work. Either may be zero.
Also we test whether the system is in equilibrium or motion by testing for zero resultants vertically, horizontally and zero net moment.

If all these three conditions are met the system is in equilibrium.
If not it is in motion with the resultants determining the motion.

It is important to get this sequence of logic this way round.

So case (1)

F=0
X=W at all times. Edit: X and W are in opposite directions as shown so really X = -W with signs
Y=0
Both X and Y pass through hinge and exert no moment about it.

But W exerts a moment about the hinge.

So the system is in motion, not equilibrium.

So the plank swings down until it is hanging vertically below P.
As the plank swings down the perpendicular distance from P to W decreases so the moment decreases until the distance is zero and the moment is zero and the plank is hanging.

This is a more complete analysis of case (1)

We should finish the others before moving on.

Last edited: Jan 26, 2012
22. Jan 26, 2012

### aaaa202

hmm :)
I thought that took care of the case on my picture, where F equals the weight.... I can see that it of course also would explain the case where F=0, but why not the original case with F=W?

Btw, I really appreciate all the help..

23. Jan 26, 2012

### Studiot

Let's slowly build up through each case as each one is instructive. We will get there.

Case (2) is a little more complicated so here is another diagram and I will be more careful with the signs.

Did you note my edit - shows how easy it is to rush and miss something?

Let the plank make an angle θ with the vertical at any time.
And let F be normal to the plank at all times.

Case (2) 0<F<W

There are now horizontal forces acting, but since there is no horizontal translation:

Fcos(θ) = Y

Since there is no vertical translation of the system

Fsin(θ) = W ± X

But F < W and sin(θ) < 1 → Fsin(θ) < W so X must be opposite in sign to W ie upwards.

Since X and Y pass through P they exert no moment about P.

Summing Moments about P we have

L1F - L2Wsin(θ)

Which is not zero for all θ.
So the system in not in equilibrium it is in motion.
When the plank is horizontal θ = 90 so L2Wsin(θ) > L1F
So the plank commences to swing down

However as it swings down the moment developed by F remains constant but that developed by W diminishes as θ decreases.

Eventually a condition is reached, before the plank is vertical, where

L1F = L2Wsin(θ)

and rotation stops here - equilibrium is achieved.

In the next post we will consider the energy implications of cases (1) and (2).

#### Attached Files:

• ###### plank2.jpg
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Last edited: Jan 26, 2012
24. Jan 26, 2012

### aaaa202

hmm yes, I think I understand all this. Now I'm not sure if it builds up to a point, but I'm only interested in the instant that the rod starts rotating, only because of the moments around it. For me something can't possibly rotate just because of moments. But of course I'm wrong because if you push at the end of a wrench in outer space it will start to rotate - a puzzle to me, who can only understand torque through energy.
Now I know I'm repeating myself, but I can't really see if this is the end will make me understand, why torque is more than just energy considerations..

25. Jan 26, 2012

### Studiot

What subject are you studying?