Mechanical Paradox: Weightless Rod w/ Wheel & Weight

[Ketman]

You have a weightless rod with a wheel attached to one end (weightless, frictionless) and a weight attached to the other. The weight is mounted on a short axle, and can rotate freely. With the wheel on the floor, the rod is held at 45 degrees to the horizontal and then released. Since the rod and wheel are weightless, the weight falls purely vertically while the wheel rolls across the floor, and we can work out the linear energy developed when the weight hits the floor from its height.

The action is repeated, except this time the wheel is blocked. The weight falls in a curve, but because the starting and finishing heights are the same as in the first trial, the KE developed is the same. And being mounted on an axle, the weight doesn't rotate either, so the energy developed is purely linear, as in the previous trial, and must be the same. The paradox for me is that in the second case the weight has undergone a horizontal displacement, but has apparently used no energy to achieve it. What's the explanation for that?

 

[Filip Larsen]

As long as you are assuming a conservative force with no friction you can move a mass from A to B along any path with the same net different in mechanical energy. If, for instance, A and B are at same height you could "using" zero net energy slide the mass along the horizontal floor, or you could raise the mass (frictionlessly) on a pole using some energy that you would gain back when lowering mass back down. In your example, any place on the floor would be equally "valid" (have same energy potential), even very far away as you could just let the mass slide without friction down a very shallow incline.

You need to introduce friction somewhere (i.e. a non-conservative force field) for the path from A to B to have influence on difference in mechanical energy.

 

[Doc Al]

The paradox for me is that in the second case the weight has undergone a horizontal displacement, but has apparently used no energy to achieve it. What's the explanation for that?

A constraint force can change the direction of the velocity without doing any work, as long as it remains perpendicular to the velocity. The only force on the weight with a component parallel to its velocity is gravity. (Of course, looking at the weight plus rod as a whole, you know that the force applied to the blocked wheel does no work since its displacement is zero.)

A similar situation would be a block sliding down a frictionless slide compared to one in free fall.

 

[Ketman]

Thanks for the replies. The thought of "no cost" horizontal translation of an object under gravity occurred to me as a result of looking into trebuchets. A trebuchet is a missile-throwing device, with a heavy counterweight on the short end of a beam that pivots about an axle fixed to a supporting frame. A sling is attached to the long end, and is "tuned" by adjusting its length so that it releases the missile just as the counterweight is directly under the axle, and (hopefully) brings it to a halt by taking all the kinetic energy developed (less what is lost through friction). It's been noted that trebs throw further if the frame is on wheels, rather than being fixed to the ground. With the arm at some angle above the horizontal, the falling weight first pushes the carriage back, then pulls it forward. So the fall is more vertical. Presumably this is a better bargain because the fixed variety loses a lot of energy via dissipation through the less-than-rigid structure, and through the earth. What did occur to me, though, is that if both varieties did use their energy with the same efficiency, the fixed-frame machine would throw further, because of what I called in my previous post "horizontal displacement". The missile will take off a few feet further forward than it would on the wheeled machine, and will land that much further away. It looks like a free lunch.

Edit: it's the other way round, isn't it? The wheeled machine will end up further forward because it's pulled forward more than it's pushed back.

 

[pallidin]

You need to introduce friction somewhere (i.e. a non-conservative force field) for the path from A to B to have influence on difference in mechanical energy.

Indeed. I think it should be considered that blocking the movement of the wheel IS friction.
In this case 100%.
Friction should not only be considered a scenario which resists movement, but also that which does not allow movement.

 

[Ketman]

Indeed. I think it should be considered that blocking the movement of the wheel IS friction.
In this case 100%.
Friction should not only be considered a scenario which resists movement, but also that which does not allow movement.

...the difference being that the latter scenario uses up no energy. The two cases I explored produce an equivalent result because of that. In the first case, resistance to movement is 0%. In the second, it's 100%. The result in both cases is that energy is 100% conserved.

 

[pallidin]

Right. I was just clarifying for my own mind...

 

[pallidin]

BTW, just curious, are the reason you are asking these questions in regards to optimization of your post#4 with respect to trebuchets?

 

[Ketman]

BTW, just curious, are the reason you are asking these questions in regards to optimization of your post#4 with respect to trebuchets?

Not really. I'm satisfied that the right way is to put wheels on the trebuchet, though that's more a practical constructor's solution than a physics solution. If you could guarantee complete rigidity in your frame, the fixed-frame machine might be just as good. But the time and materials involved would be out of proportion to the improvement, at least for full-size machines throwing 8-10lb missiles. I've made experiments with knee-high models tossing golf balls, and I've found the fixed-frame can compete with the wheeled frame, but only if it's both very heavy and very rigid. At full size, the lighter wheeled machine is a more practical solution to the problem of conserving energy. In other words (to refer back to my post #6), the 0% resistance to movement is a cheaper and more achievable goal than the 100% resistance.

The paradox of achieving horizontal displacement with no expenditure of energy is just something I observed in passing, and just wondered if it was considered a paradox in theoretical physics.

 

[pallidin]

The paradox of achieving horizontal displacement with no expenditure of energy is just something I observed in passing, and just wondered if it was considered a paradox in theoretical physics.

I see. Well, horizontal displacement is certainly evidenced in, say, manipulating a gyroscope in certain ways.
However, to say there is NO EXPENDITURE OF ENERGY is false.
Remember, that, even though the velocity vector is the same in "A to B" horizontal displacement, YOU MUST ALSO DISPLACE THE MASS IN ORDER FOR THIS TO OCCUR.
This requires energy. Sometimes very little, but energy is required because mass is being moved to a new position!

 

[pallidin]

Also, the notion that a horizontal "stop block" creates a situation where horizontal displacement is achieved without energy is also false.
There IS energy in that "stop block", else it could not stop anything!
In your scenario, that energy is the mass of the Earth resisting rotation from your experiment, as the "stop block" is attached to the earth.

 

[Doc Al]

Also, the notion that a horizontal "stop block" creates a situation where horizontal displacement is achieved without energy is also false.
There IS energy in that "stop block", else it could not stop anything!
In your scenario, that energy is the mass of the Earth resisting rotation from your experiment, as the "stop block" is attached to the earth.

The block does not do work or consume energy; the total mechanical energy is conserved even though the mass moves horizontally.

 

[pallidin]

The block does not do work or consume energy; the total mechanical energy is conserved even though the mass moves horizontally.

The mass of the earth, connected to the block, is opposing the horizontal movement against the block.
Agreed?
Opposition is not without force. That would be impossible.

 

[Doc Al]

The mass of the earth, connected to the block, is opposing the horizontal movement against the block.
Agreed?
Opposition is not without force. That would be impossible.

What has that to do with energy? Application of force does not equate to energy.

 

[pallidin]

What has that to do with energy? Application of force does not equate to energy.

Are you suggesting that it is possible to apply force without energy?

 

[Doc Al]

Are you suggesting that it is possible to apply force without energy?

Applying a force does not necessarily require an energy source, if that's what you're saying. Put a book on a table, it exerts a force yet no energy is consumed.

 

[pallidin]

What, then, is the fundamental nature of that application of force, if it is not energy?
Please define that "non-energy" aspect.
Just trying to understand.

 

[Ketman]

However, to say there is NO EXPENDITURE OF ENERGY is false.
Remember, that, even though the velocity vector is the same in "A to B" horizontal displacement, YOU MUST ALSO DISPLACE THE MASS IN ORDER FOR THIS TO OCCUR.
This requires energy. Sometimes very little, but energy is required because mass is being moved to a new position!

I thought this was already settled. There is no expenditure of energy. If that were not so, there would be energy loss in the case I theorized where the weight came down in a curve because the rod was constrained at the other end. It would arrive with less kinetic energy, and therefore less velocity, than in the case where it falls freely. But it doesn't. It arrives with 100% of its original PE converted to KE. It gains a horizontal displacement in the process, but there is no energy loss as a result of that, because no energy is used. What the block is exerting is a purely static force. It does no work itself, and no work is done against it.

 

[pallidin]

Yeah, it appears that I am confused.
Wouldn't be the first time! My bad.
I'll just shut-up and listen...