Force and point of application

[luis20]

Hi everyone :). I have one question about classical mechanics, which is illustrated in the attachment.

The question is: for rigid bodies, how can the same external force F produce more overall movement in situation A) than in situation B)? This seems to contradict the conservation of energy!

The same external force F is applied in situations A) and B). Since the point of application in A) is away from the center of mass C, we can move that force to C and add a torque T (it's the same).

But if I had applied this force F in the center C in the first place (situation B) I wouldn't have this extra torque!

Conclusion:
The sum of acceleration/velocity/distance traveled of all points of the body is higher in A) than in B). So we give more cinetic energy if we apply this force F away from the center of mass C?

Thanks in advance and I'm sorry if my english isn't the best ^^

 

[haruspex]

In A, the applied force has progressed a greater distance, so has put more energy into the system.

 

[luis20]

Yes, but that force was applied in the same amount of time. So that force should give the same amount of velocity to the body's points in both situations. But total velocity is bigger in A), why?
Edit: About the "greater distance", that might be wrong. The distance is bigger sure, but some of that distance has a differente direction from the force (force has constant direction), so that doesn't give any energy to the system.

 

[haruspex]

that force was applied in the same amount of time. So that force should give the same amount of velocity to the body's points in both situations.

Energy is force times distance (in the same direction). Force times time is momentum. So in A the force has imparted the same linear momentum, but more energy (since some of the energy has gone into rotation).

The distance is bigger sure, but some of that distance has a different direction from the force

It isn't entirely clear from your picture what r2 measures. I took it as distance in the direction of the force.

 

[ecneicS]

In A), you can't move the force to C, because then what force would supply the torque? In other words, torque costs force.

But what I don't know is how much of the force is given to create rotational motion and how much is given to create linear motion. It certainly will depend on time.

EDIT: Take what I said with a grain of salt, I'm confusing myself.

 

[haruspex]

EDIT: Take what I said with a grain of salt, I'm confusing myself.

I agree with that part

 

[Studiot]

Diagram A is impossible as drawn.

That is because it is missing the pivot reactions necessary to create rotation.

 

[luis20]

Diagram A is impossible as drawn.

That is because it is missing the pivot reactions necessary to create rotation.

Pivot reactions? What do you mean? :O

 

[luis20]

In A), you can't move the force to C, because then what force would supply the torque? In other words, torque costs force.

But what I don't know is how much of the force is given to create rotational motion and how much is given to create linear motion. It certainly will depend on time.

EDIT: Take what I said with a grain of salt, I'm confusing myself.

That would make sense to me. But in classical mechanics, any force in a point A can be moved to a point B and a torque must be added. That's the problem :\

 

[Studiot]

You have shown the rigid body in A rotating, presumably because of the moment of F about some fixed point.
Label the ends and centre of the body a, b and c. F is applied at c and a is the rotation centre.
This rotation cannot happen unless there is a reaction force at this fixed point (a).

It does not matter where you apply a single force F to a rigid body that is in free space, you will get the same effect. The only effect you will get is a linear acceleration in the direction of the force. You will not get a rotation.

 

[luis20]

You have shown the rigid body in A rotating, presumably because of the moment of F about some fixed point.
Label the ends and centre of the body a, b and c. F is applied at c and a is the rotation centre.
This rotation cannot happen unless there is a reaction force at this fixed point (a).

It does not matter where you apply a single force F to a rigid body that is in free space, you will get the same effect. The only effect you will get is a linear acceleration in the direction of the force. You will not get a rotation.

Are you sure? If I hit a pencil in one of the ends the pencil goes forward and rotates. The only point of the pencil who doesn't rotate and only goes forward is its center of mass!

The same happens even if I do this in the air, with low friction. Where is the reaction force?

Edit: In vacuum there wouldn't be rotation? Also: I didn't fix any point in the illustration, the body moves forward and rotates

 

[Studiot]

The same happens even if I do this in the air, with low friction. Where is the reaction force?

There is always a reaction point on Earth, that is why I mentioned space. You can't spin a space rocket with one 'jet'.

So back to your diagrams A and B, in the absence of any other forces it does not matter where you apply F on your body the result will be the same. Linear motion up the paper.

 

[luis20]

There is always a reaction point on Earth, that is why I mentioned space. You can't spin a space rocket with one 'jet'.

So back to your diagrams A and B, in the absence of any other forces it does not matter where you apply F on your body the result will be the same. Linear motion up the paper.

Wait, wait!

(begin at 14:40)

He says the body actually rotates. This is from MIT :O

 

[Studiot]

To understand how the turning couple arises (on Earth)

Place a book flat on a table with its spine towards you.
Since gravity is a right angles to the table top gravity does not affect this experiment.

No push the book away gently from you by pushing the spine with one finger in the middle of the book.

The book slides away over the table without rotation.

Now repeat moving the point of pushing with the finger a little towards one end of the book.

If you repeat this several times you will come to a point where the book rotates like your pencil as well as sliding forwards.

The closer to one end you push the more rotation and the less translation you will note.

Why?

Well the distance from the push point to one end is equal to the distance from the push point to the other when you push in the middle.

If you imagine the resistance to motion distributed in little arrows all along the spine, the moment of that resistance force to the right of the push force is of the opposite direction to the moment generated by the push force to the left.

When the distance are not equal the moments are not equal. So the book rotates.

 

[Studiot]

I note your video example applies an impulse not a sustained force?

 

[luis20]

I note your video example applies an impulse not a sustained force?

Yes! That's it. What is the difference? If a single impulse rotates the body, a single force (which will have to turn to an impulse to do anything) will also rotate the body.

So maybe a reaction force in the pivot point is not needed. At least is what that professor says. What do you think?

 

[luis20]

The closer to one end you push the more rotation and the less translation you will note.

I think if I apply the same force/impulse the translation should be the same, regardless the distance to one end. Like that professor stated.

This wouldn't work with round objects (because there is the wheel effect, but I think doesn't matter).

 

[Doc Al]

It does not matter where you apply a single force F to a rigid body that is in free space, you will get the same effect. The only effect you will get is a linear acceleration in the direction of the force. You will not get a rotation.

If the single force creates a torque about the center of mass, then you'll get both rotation and translation.

 

[Studiot]

Originally Posted by Studiot
It does not matter where you apply a single force F to a rigid body that is in free space, you will get the same effect. The only effect you will get is a linear acceleration in the direction of the force. You will not get a rotation.

If the single force creates a torque about the center of mass, then you'll get both rotation and translation.

Yes I was wrong about that.

 

[luis20]

Yes I was wrong about that.

Studiot, read my last reply. See if you can understand what I thought was true but now I see it isn't ! :P

Edit: I will put it here:If that professor is right, than I can create more overall movement with less movement. I intuitevely thought this wasn't possible! But it appears to be. For instances, I thought that conservation of linear momentum wouldn't consider the signal of the velocity, but it does!

So imagine the following collision:

Body A: mass=1kg v=9m/s
Body B: mass=2kg v=0

What I call overall movement does not consider if velocity is positive or negative. In the inicial conditions I have 1kg of mass with 9m/s, this is what I call overall movement.

After collision, the real result is:

Body A: v=-3m/s
B: v=6m/s

Linear momentum is -3*1 + 6*2= 9, like before. But wait! I would intuitevely say that this couldn't be true because I should work with modules | | because I don't care if velocity is positive or negative.

What I see in this result is:
1kg with 3m/s and 2 kg with 6m/s. This for me is on average 3kg of 5m/s or 1kg of 15m/s: this is more than the initial 9m/s in 1 kg! (I see mass as a number of something, like atoms). This intuitive idea is wrong.
I thought that if body A has lost 6m/s (9-3 and not 9-(-3)), it could only give 3m/s to the body B, because this body doubles the mass. But this is not what happens.
So I can conclude that In a closed system I can create more of this visual "overall movement", I don't know if this gives potential to anything since energy remains the same.

I don't know if I'm making myself clear, all this can be very silly :9

 

[Studiot]

Yes! That's it. What is the difference? If a single impulse rotates the body, a single force (which will have to turn to an impulse to do anything) will also rotate the body.

Not sure what you mean by the bit in brackets, but we are agreed on the rest now.

However the difference between an impulse and a steady force (such as gravity) is that you wanted to discuss energy.

Now energy = Force x distance when the force is a steady force such as gravity.

By steady I mean that it is always acting so if a body of mass m moves downward a distance d gravity is acting the whole time and the energy involved is mgd.

However if I extend a spring a distance d by applying a force F the energy involved is only 1/2Fd since the force 'builds up from zero'

For an impulsive force the situation is even lower because not only does the force start at zero but it ends at zero. So the energy is ∫Fδd and you need an expression of F in terms of d which is difficult. That is why we use momentum.

Another point is that F usually represents a steady force such as gravity or a push rod that is constrained to act along the line shown.
Such a the force will tend to brush the object out of its way unless applied at the centre of mass. But it does not remain tangential to the rotation.
An impulse of course does not have this problem.

 

[Darwin123]

Hi everyone :). I have one question about classical mechanics, which is illustrated in the attachment.

The question is: for rigid bodies, how can the same external force F produce more overall movement in situation A) than in situation B)? This seems to contradict the conservation of energy!

The same external force F is applied in situations A) and B). Since the point of application in A) is away from the center of mass C, we can move that force to C and add a torque T (it's the same).

But if I had applied this force F in the center C in the first place (situation B) I wouldn't have this extra torque!

Conclusion:
The sum of acceleration/velocity/distance traveled of all points of the body is higher in A) than in B). So we give more cinetic energy if we apply this force F away from the center of mass C?

Thanks in advance and I'm sorry if my english isn't the best ^^

I am not sure why you think there is more kinetic energy with a force away from the center of mass. I conjecture that you did not include the kinetic energy associated with rotation.
The total kinetic energy of a rigid body can be decomposed into the kinetic energy associated with the translation of the center of mass (CM), and the kinetic energy associated with rotation of the rigid body about the CM.
If a force adds a specific amount of kinetic energy to the rigid body, then the energy has to be divided between the rotation about the CM and the translation of the CM. If the force were directly applied at the CM, the change in rotational kinetic about the CM won't change.

 

[luis20]

I think now I have a different idea of this.

Thanks everybody for all the replies