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Conservation of momentum
by
chazemz
on 05 Apr, 2020 08:07
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I would like to run a thought experiment past you.
If you could refer to the two diagrams attached, you will see a box that is at rest in deep space. In diagram A, the cannon is fired and the cannon ball flies toward the vanes that are attached to the inside of the box. In diagram B, the cannon ball has passed through the vanes causing them to rotate in opposite directions and is heading towards the opposite wall to the cannon. The momentum of the cannon ball after interacting (colliding) with the vanes must be less than when it was fired from the cannon. In this collision some of the linear momentum of the cannon ball has been converted into the angular momentum of the rotating vanes. Therefore, it is difficult to conclude that F2 is equal to F1. This imbalance of force would result in the box continuing to move in the direction of F1 after the cannon ball collides with the opposite wall.
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#1
by
Mark7777777
on 05 Apr, 2020 09:02
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#2
by
b.lorenz
on 05 Apr, 2020 09:30
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It does not work out. Namely, momentum can not be transformed into angular momentum. In the cannonball-vane collision, it is still required that (working in the rest frame of the box), that the vector sum of the forces on the vanes is zero, otherwise they are torn from their axes. So if the ball excerts a force F on the vane (downward), the bearings must excert a force with magnitude F upward on the vanes, so that F_sum = m*a stays zero. Therefore the momentum the ball loses will go "right trough" the vanes and will decelerate the box.
Now of course the vanes do start to spin, and the kinetic energy of the ball is partially transformed into rotational energy of the vanes. But that is an entirely different matter.
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#3
by
chazemz
on 05 Apr, 2020 09:35
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you are supposed to view it on its side. It did not allow me to view the attachment before posting.
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#4
by
chazemz
on 05 Apr, 2020 09:55
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When the cannon ball collides with the vanes they will turn which (through the bearings ) will apply a torque to the box A torque is a force. The Force and Torque will balance so magnitude is not a problem. Direction however will be more difficult to balance.
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#5
by
b.lorenz
on 05 Apr, 2020 10:11
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When the cannon ball collides with the vanes they will turn which (through the bearings ) will apply a torque to the box A torque is a force. The Force and Torque will balance so magnitude is not a problem. Direction however will be more difficult to balance.
Torque is not a force. It is rather a "side effect" of forces. They are in different equations of motion:
https://en.wikipedia.org/wiki/Rigid_body_dynamics#Force-torque_equations But it is still true that acceleration is caused by simply the vector sum of all forces acting on the body.
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#6
by
chazemz
on 05 Apr, 2020 10:30
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Thankyou for pointing out what I should have added in the original post that when the cannon ball hits the opposite wall to the cannon and comes to rest, the vanes will continue to spin. The energy slowly dissipating as heat by means of the friction of the bearings.
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#7
by
giulio
on 05 Apr, 2020 11:20
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Hi, I'm new to the forum, I find interesting the conversion of movement into heat used by the linear eddy current brake. I'm curious about what happens to F1 by braking the cannonball with that system.
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#8
by
chazemz
on 05 Apr, 2020 12:51
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I am not an expert on eddy currents. They are used in fair ground rides and I believe people far more qualified to comment than I have posted all the necessary calculations. If you have an idea do not be shy and create a post. Some people may be overly critical but most people will try and help. See it as when you first learned to walk, you will fall down, but as long as you get up and try again, you will be stronger
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#9
by
tyrred
on 06 Apr, 2020 10:27
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So you fire a cannon ball in deep space. Inside a box that catches the cannon ball. With vanes that are impacted. And the cannon ball does... what, exactly? Chaotically destroy the apparatus?
This is a solved problem.
As the thread title states... Conservation of momentum.
Is this a serious proposal for a refutation of conservation of momentum?
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#10
by
chazemz
on 07 Apr, 2020 07:58
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I have attached two further diagrams which may help to explain the process better.
In figure 1 you have an open-ended box and in the box are two cannon systems. In each system the cannons are attached to the belts, the belts are connected to the wheels and the wheels are connected to the box. Two of the cannons are fired and the cannon balls are free to leave the box. The cannons will recoil which will also push the slave cannons as shown. The cannons can only apply a torque to the box and this torque is equal and opposite. No force acts upon the box, so the box must remain at rest. As with the vanes, the cannons will continue to rotate, the energy slowly dissipating due to the friction of the bearings in the wheels.
In figure 2 the opening has now been closed and the cannons as before are now fired. The same recoil process will occur as in figure 1, so as the cannon balls fly toward the interior wall the box remains at rest. The cannon balls will now collide with the wall and impart their momentum onto the stationary box. The box must now move in the same direction as did the cannon balls. The cannons will continue to rotate, slowing due to friction as in figure 1.
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#11
by
b.lorenz
on 07 Apr, 2020 08:32
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As I have already said, the assumption that axles and bearings will only transmit torque is flawed. They can transmit torque, if they are not ideal (there is friction slowing the rotatation), and this torque will in fact balance out if the two parts are symmetrical, but they WILL TRANSMIT FORCES too. This is not some obscure fringe of science: You can very easily demonstrate this: Take a small cart, fix a bar on it so that it can rotate around a vertical axis, and then kick one side of the bar. The bar will start to rotate, but the cart will also roll away.
I did this in LEGO. It works. I proposition this topic to be closed.
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#12
by
93143
on 07 Apr, 2020 08:40
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The cannons can only apply a torque to the box and this torque is equal and opposite. No force acts upon the box, so the box must remain at rest.
Not true.
Firing one of the cannons results in it acquiring a finite momentum in the opposite direction to that of the cannonball. The other cannon on the same belt moves in the opposite direction at the same speed, so the total system has zero net linear momentum relative to the wheel axles. Where did the reaction momentum go?
Remember, linear and angular momentum are not interchangeable. They are separately conserved quantities.
What has happened here is that the firing cannon has put the belt in tension around the left-hand wheel, pulling the other cannon leftward and exerting a rightward force on the axle of the wheel. This imparts a momentum to the box. It can be shown that the total momentum of the box, cannons, wheels and belts is equal and opposite to that of the cannonball(s) fired.
b.lorenz and tyrred are correct. This is a solved problem. It doesn't matter if you're the direct descendant of both Norman Dean and Rube Goldberg; try as you might, you will never violate conservation of momentum using Newtonian mechanics. The mathematical description is fundamentally, inherently conservative. If you get nonconservation, you know you screwed up somewhere.
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#13
by
Mark7777777
on 07 Apr, 2020 09:17
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I have attached two further diagrams which may help to explain the process better.
In figure 1 you have an open-ended box and in the box are two cannon systems. In each system the cannons are attached to the belts, the belts are connected to the wheels and the wheels are connected to the box. Two of the cannons are fired and the cannon balls are free to leave the box. The cannons will recoil which will also push the slave cannons as shown. The cannons can only apply a torque to the box and this torque is equal and opposite. No force acts upon the box, so the box must remain at rest. As with the vanes, the cannons will continue to rotate, the energy slowly dissipating due to the friction of the bearings in the wheels.
In figure 2 the opening has now been closed and the cannons as before are now fired. The same recoil process will occur as in figure 1, so as the cannon balls fly toward the interior wall the box remains at rest. The cannon balls will now collide with the wall and impart their momentum onto the stationary box. The box must now move in the same direction as did the cannon balls. The cannons will continue to rotate, slowing due to friction as in figure 1.
I wonder if you could upload PNGs rather than or in addition to PDFs going forward?
That way we don't have to download them to see what they are, but can rather view them immediately, embedded in the post.
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#14
by
chazemz
on 07 Apr, 2020 10:59
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Hi I will try and upload diagrams in the future as requested.
With reference to the two previous questions If I may try to explain as follows.
The whole system is given a value of 10 before firing. When the cannons are fired Third Law dictates that the value must be split, so 5 goes to the cannon balls and five goes to the cannons. To recoil, the cannons must rotate the wheels and since the bearings have friction energy must be lost as heat. This value must be subtracted from the 5. Let us say the cannons will apply a force and torque to the box at the same time on a 60/40 basis. The value of the force on the box will be 3. But let us be overly generous and give a ratio of 90/10. What do you think will happen to the box if you have a value of 5 on one side and a value of 4.5 on the other? I am sorry you cannot subtract from a value and retain the original value, no matter how much you wish to do so, things do not work that way.
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#15
by
chazemz
on 07 Apr, 2020 11:22
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