How can a marble on a circular track return to its point of origin using only its own momentum?

Question

A marble rolling on a curved track appears to violate conservation of momentum. Please help me understand why this is illusion and/or what mechanism is acting such that momentum is cancelled and recreated in a new direction.

Consider a marble on a flat track on Earth's surface. You push the marble to create momentum. All forward motion is from that initial push. The track is curved such that the marble returns to its starting point. An observer standing next to the track would say the marble's entire journey was apparently funded by its initial forward momentum, but this seems a violation of the vector velocity part of conservation.

Contrast that with a space ship traveling away from Earth, coasting on the momentum of an initial thruster blast. In an attempt to return to Earth, you apply side rocket thrust (like the track did to the marble). But while you can spin the ship and/or alter its trajectory to one side, you are still traveling away from Earth at the same steady speed. Only by using your rockets to directly cancel momentum away from Earth (which takes exactly as much energy as it did to create the momentum), and then adding more energy to push the ship back towards earth, can you get home.

So while the space ship can never return using only its own existing momentum, a marble on Earth can?

Possible explanations:

1. The marble track is somehow translating the marble's forward momentum into some other form of energy, then giving it back to the marble in the form of reduced momentum with a new vector.

2. Or I'm completely misunderstanding how momentum should be described in the case of the marble and thus creating the illusion by my mischaracterization.

3. I have an intuition that gravity is a significant part of the issue, so perhaps the marble's momentum (and all planetary surface objects) should be considered as a case of angular momentum?

4. Or maybe there's no illusion, I'm confusing various motions and energies being displayed by the marble, with its much smaller forward momentum, and in fact the cancellation and recreation of momentum can be fully explained by the appropriation of energy from these other factors that are inherent to and traveling with the ball?

Answer 1

It is quite simple. It is that you are entirely overlooking the fact that the marble is interacting with the track and with whatever is supporting the track and ultimately with the Earth. The combined momentum of the the marble, the track and the Earth is conserved throughout. If you repeated the experiment with the marble on a light track which was on a surface with very low friction, you would find that the marble and track both would move, with the track continuously recoiling from the instantaneous point of contact with the marble, so that the combined momentum would be conserved.

Answer 2

The momentum of a particle running along a track is definitely not conserved. Momentum is only conserved when there are no external forces applied to a system, and the track here applies forces to the marble.

What is conserved for a frictionless track is mechanical energy- the sum of kinetic and gravitational potential energy.

But while you can spin the ship and/or alter its trajectory to one side, you are still traveling away from Earth at the same steady speed. Only by using your rockets to directly cancel momentum away from Earth (which takes exactly as much energy as it did to create the momentum), and then adding more energy to push the ship back towards earth, can you get home.

This is incorrect. You can apply thrust to the side continuously and turn around that way, in a similar way to how the track works. Applying thrust perpendicular to your velocity will give you a circular trajectory, and eventually you will be facing earth again. The momentum of the spacecraft is not conserved, but the total momentum of the spacecraft and its exhausted fuel is.

Answer 3

Assumptions

I take the static friction is enough that the motion of the marble does not cause the ramp to oscillate on the floor.

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Let's try to imagine the situation, attach an momentum arrow to the marble as it moves. Let us try deduce facts about this arrow as the marble moves.

The time rate change of magnitude of the momentum vector is given as:

$$ \frac{d}{dt} |\vec{p}| = \frac{\frac{d \vec{p} }{dt} \cdot \vec{p}}{|\vec{p}|}$$

But, $\frac{ d \vec{p} }{dt} = \vec{F}$, what are the forces acting on the marble? The normal and gravity, we know the marble is in equilibrium for the direction normal to the surface it is rotating and hence the only effective force is the tangential force caused by gravity, let this be given as $\vec{F_{g,T} }$ , then:

$$ \frac{d}{dt} |\vec{p} | = |\vec{F}_{g,T}| $$

Clearly, the length of the momentum arrow attached to this particle changes with time as the quantity given in the above expression. And hence, momentum is not conserved in length neither direction.

Since, the block doesn't oscillate due to static friction (as assumed), the weight of marble block system pushes down the earth. If we were to include, the earth in the system, then the momentum would be conserved as you would expect.

Answer 4

This question assumes that the momentum of the ball is conserved, and hence any amount of push will result in the ball returning to its original point. This is not the case, since the ball's momentum is conserved only when there are no external forces operating on it.

Whether the ball returns to its original point is dependent on how much force the ball is pushed with in the first place. Too small of a force, and the ball may stop half way.

Your intuition with the spaceship not being able to return to earth using tangential thrust forces only, is correct. At points along its journey, it will need to fire thrusters in opposite directions to return it back to earth.

How crafts return to earth in real missions, is accomplished by using the gravitational force of a massive object e.g., the moon, which will "slingshot", or accelerate the craft back on a trajectory toward earth.

This is exactly how the command modules returned to earth in the Apollo missions to the moon.

Answer 5

I will just write my thoughts, as I cannot say I am completely right.

In the first case, as soon as the marble collides with the curved surface the momentum vector of the marble which is normal to the surface becomes zero due to normal force and due to elastic of the marble, the marble rebounds with the same momenta anti-parallel to the normal vector the marble had before. Here I have treated the tangential part of marble momentum as invariant. Now it continues at every point on the curved surface as marble travels through the circumference.

But you should be thinking that after the first collision the net momentum vector of the ball has changed and yes it has but total momentum is still conserved because at the point of contact the ball gives momentum to the track in outward direction and if the track is itself placed in a smooth surface it will move in outward direction. This effect will also continue throughout the circumference resulting in net momentum of the track to become zero and the marble will have the same initial momentum vector without violating conservation of momentum.

Also I am not able to understand what you are saying in case 2 but you can take gravitational potential energy into consideration along with initial kinetic energy and see if it works out.

Answer 6

It might help to think of the x-component and y-component of the momentum as separately conserved quantities, with the x-component and y-component of the force as their rates of change.

Consider the diagram below, start at the bottom, and just look at the horizontal component of the momentum and force. At the start, the moment is entirely in the positive x direction, there is no x-component of the force, as the track is pushing it upwards towards the centre. As the ball moves round the first quarter of the track, an x-force appears (red arrows), directed backwards against the motion. This slows the ball down, bringing it to a halt (x-component only) at the rightmost point. The contact of the ball with the track applies an equal magnitude force in the opposite direction on the track, changing its momentum oppositely, so total momentum is conserved. At the rightmost point the backwards x-force still applies so the ball now accelerates backwards (x-component only) passing around the second quarter. The ball is now moving in the negative x direction, the change in momentum has come entirely from the force applied by the track. Now as the ball enters the third quarter an x-force arises again now in the positive x direction, opposite to its motion, slowing it down. Eventually it is brought to a halt (x-component motion) at the leftmost point, and then accelerated rightwards back towards the start.

If you think of the x and y components separately, you can see each component of momentum get cancelled and reversed by the corresponding component of the force from the track opposing its motion. I think the problem comes from looking at the momentum vector as a magnitude and direction, seeing the magnitude stay constant, and think that somehow that means the momentum is in some way "the same lump of momentum"; that it is just changing direction, like the ball going round the circle is always the same lump of matter and just changing direction. That doesn't work - a constant sideways force involves a flow of momentum, but no change in its magnitude.

It's interesting to consider a rigid body spinning. The internal forces holding it together constitute an ever-cycling flow of momentum inside the body. Again, there is an intuitive tendency to think that because it is rigid, the parts are 'fixed' and not moving with respect to one another. It feels odd to note that places on the other side of the planet are constantly moving at speeds of up to two thousand miles an hour with respect to us. Again, considering just one component of our velocity, we can see that the downward force of gravity accelerates each of us up to two thousand miles an hour and then back down to zero again each day, every day. And in six months when we are on the other side of the sun we will be travelling at sixty kilometers per second relative to our current frame of reference, accelerated to that speed by the appropriate component of the sun's gravity, which then goes into reverse and slows us down over the rest of the year. Each separate component of momentum is cyclically sloshing backwards and forwards between us and the sun. It's only the combination of x and y momenta making up the vector magnitude that stays constant.

Answer 7

You appear to be confusing momentum and energy.

Both are conserved. But momentum is a vector quantity, and energy is not.

You can only change either by interacting with something else.

A space ship traditionally changes its momentum by firing rockets (splitting itself into payload and propellant), or by interacting with other objects (light from a sun, or gravity from a heavy body).

A space ship in orbit around another body which does an impulse burn of thrust will actually return back to its starting location, unless it generates enough thrust to escape the gravity well of the thing it is orbiting. In a more complex situation it may not (as it interacts with more bodies).

For the marble, it is being turned through a different mechanism. It is interacting with the track, which is in turn anchored to a planet whose mass is much much larger than the marble. If the track is rigid, and marble is smooth, and the corners are rounded, the marble will change direction without changing velocity. Momentum will be exchanged between the marble and planet (the planet, being very large, will be changed by a degree nobody could detect), and the marble will change direction.

Under the assumptions of a rigid connection and zero friction, very little to no energy will be lost by the marble.

A space ship with frictionless wheels on it that rode a ramp that was much, much more massive than the space ship would also be able to do a 180 degree turn without losing much energy.

You may have heard of the terms "elastic collision" and "inelastic collision". An elastic collision is one where next to no energy is lost in the deformation of the colliding bodies or sound of impact.

You can model a marble track as an elastic collision.

When you solve for the result of an elastic collision between a small object and a heavy object, you'll find that almost no momentum and energy is transmitted from the small object to the heavy object.

If you solve the equations of elastic collision ($m$ is mass, $u$ is initial velocity, $v$ is end velocity), with $m_2 \gg m_1$ and $u_2 = 0$ (massive object starts out stationary),

$$v_1 = u_1 \frac{m_1-m_2}{m_1+m_2} \approx u_1$$ $$v_2 = u_1 \frac{2m_1}{m_1-m_2} \approx 0$$

the large object doesn't budge noticeably, and the small object reflects.

Energy is conserved, momentum is reversed on the small object, and the large object ends up moving at a ridiculously slow speed.

Because the energy of an object is proportional to the square of the velocity, that ridiculously slow speed results in an even ridiculously smaller change in kinetic energy, leaving almost all of it for the light object to maintain its velocity.

In the case of a track firmly connected to the Earth, the 6 gram marble is $10^27$ times lighter than the planet. This means in the "ideal" situation, as little as 1 part in $10^27$ of its momentum is transferred to the planet.

In practice, the transfer of momentum from the ball to the planet won't be perfect, but you can get close enough that rolling friction and wind resistance is going to dominate over that problem.