Climbing a Pulley: Conservation of Momentum & Energy?

[physics_liker]

On a standard pulley, on one end of the rope hangs a mass equal to a person's weight. That person is climbing up on the other end of the rope at constant speed. My guess is that the man and the mass will rise up at the same speed.

And does the conservation of momentum apply in this case? How does the conservation of momentum work if a case is in a similar situation? For example, a man walking on a plank of wood of his equal weight on frictionless ice. Relative to the ground, the man is moving at half of his walking speed.

Let's say you are swinging on a swing, and suddenly fly off at the lowest point of the swing path tangentially and leave the swing seat. Now in this case, the conservation of momentum applies, but not the conservation of energy. In this "reverse inelastic collision", (as I like to think of it as) where does the energy go?

 

[Doc Al]

On a standard pulley, on one end of the rope hangs a mass equal to a person's weight. That person is climbing up on the other end of the rope at constant speed. My guess is that the man and the mass will rise up at the same speed.

Why guess? Apply Newton's laws.

And does the conservation of momentum apply in this case? How does the conservation of momentum work if a case is in a similar situation? For example, a man walking on a plank of wood of his equal weight on frictionless ice. Relative to the ground, the man is moving at half of his walking speed.

In the case of the man and plank, there are no external (horizontal) forces on that system, thus momentum is conserved. Is that the case for the man climbing the rope? (First define the system you would like to consider.)

Let's say you are swinging on a swing, and suddenly fly off at the lowest point of the swing path tangentially and leave the swing seat. Now in this case, the conservation of momentum applies, but not the conservation of energy. In this "reverse inelastic collision", (as I like to think of it as) where does the energy go?

What do you mean "conservation of energy" doesn't apply? (Or "conservation of momentum" does, for that matter?)

 

[pallidin]

Let's say you are swinging on a swing, and suddenly fly off at the lowest point of the swing path tangentially and leave the swing seat. Now in this case, the conservation of momentum applies, but not the conservation of energy. In this "reverse inelastic collision", (as I like to think of it as) where does the energy go?

I wonder if you are thinking that the swing seat should propel backwards as you slide-off forwards? If that's the case, why should it? The scenario is not a collision event.

 

[physics_liker]

Why guess? Apply Newton's laws.

I didn't literally guess. I just wasn't sure of the answer, so I used the word "guess". I think the situation is similar to the one walking on ice. For every inch of rope you pull down its original position, your hand reaches 2 inches up the rope (assuming that he stretches two full arm lengths). But since one inch is descended down, the man only advances one inch upward, as does the mass on the other end of the rope.

In the case of the man and plank, there are no external (horizontal) forces on that system, thus momentum is conserved. Is that the case for the man climbing the rope? (First define the system you would like to consider.)
I'd like to consider both cases, actually. I don't quite understand the concept of conservation of momentum. The only case I have used the concept on is for collisions. I don't know how it applies to the pulley and the plank. I'd like someone to give me an answer for this question.

What do you mean "conservation of energy" doesn't apply? (Or "conservation of momentum" does, for that matter?)

I meant that if the conservation of energy applies. The kinetic energy of the person and the swing seat at the bottom point would be equal to the sum of the kinetic energy of the seat and the person after he flies off the swing seat. The latter energy does not add up to the initial energy. So I ask, where does the energy go?

 

[physics_liker]

I wonder if you are thinking that the swing seat should propel backwards as you slide-off forwards? If that's the case, why should it? The scenario is not a collision event.

I do not think the swing seat would be propelled backwards. The person on the swing has acquired his speed due to the drop of his own potential energy. But I have another question: how do we know what speed will the person leave off the swing? And I think that if man leaves off the swing with the same speed as that when he is on the swing. Then there is no change in momentum for him, and therefore no reason for the swing to be propelled backwards.

 

[pallidin]

Bear in mind that other, significant force vectors are involved here. As you are arcing downwards prior to "sliding-off", a force is being applied on the swing-set frame itself. You may not notice it because the swing-set is anchored in the ground.

A good example might be if the swing set was not anchored to the ground, but instead on wheels. Assuming the ground is flat and friction is minimal, what happens?
The swing set moves back and forth as you swing back and forth(in opposite directions), maintaining a constant center of mass.
Now, when you eject yourself, what happens to the motion of the entire swing-set on wheels?

 

[Doc Al]

I meant that if the conservation of energy applies. The kinetic energy of the person and the swing seat at the bottom point would be equal to the sum of the kinetic energy of the seat and the person after he flies off the swing seat. The latter energy does not add up to the initial energy. So I ask, where does the energy go?

I still am not seeing why you think conservation of mechanical energy does not apply.

 

[physics_liker]

Why guess? Apply Newton's laws.

I didn't literally guess. I just wasn't sure of the answer, so I used the word "guess". I think the situation is similar to the one walking on ice. For every inch of rope you pull down its original position, your hand reaches 2 inches up the rope (assuming that he stretches two full arm lengths). But since one inch is descended down, the man only advances one inch upward, as does the mass on the other end of the rope.

In the case of the man and plank, there are no external (horizontal) forces on that system, thus momentum is conserved. Is that the case for the man climbing the rope? (First define the system you would like to consider.)
I didn't literally guess. I just wasn't sure of the answer, so I used the word "guess". I think the situation is similar to the one walking on ice. For every inch of rope you pull down its original position, your hand reaches 2 inches up the rope (assuming that he stretches two full arm lengths). But since one inch is descended down, the man only advances one inch upward, as does the mass on the other end of the rope.

What do you mean "conservation of energy" doesn't apply? (Or "conservation of momentum" does, for that matter?)

You can try an example for yourself. Let's the swing is 2 meters above its bottom point. At the bottom point, all potential energy is converted into kinetic energy of the person and the swing. Now I am saying that KE of the swing-person system does not equal to the combined KE's of the two separate systems, namely the swing and the person after he flies off. I had seen a problem like this, so it's not made up. The original question asked for the speed of the swing after the person jumps off. To find that, conservation of momentum can be applied to solve the problem, but not the conservation of energy. (The two speeds from these two different methods differ.)

Now I have another question in addition to the previous ones, how would you find out the speed of the swing and the person afer they jump off, given only the initial potential energy and the weight of the swing and the person. (This time the speed of the person is not given.) Is this problem possible to solve? Since the conservation of energy does not seem to apply in this case. With the conservation of momentum, I can only obtain one equation with two variable speeds. (I am probably not making a clear explanation, nor asking a clear question. But if anyone can understand, please answer my questions. Thank you.

 

[Doc Al]

You can try an example for yourself. Let's the swing is 2 meters above its bottom point. At the bottom point, all potential energy is converted into kinetic energy of the person and the swing. Now I am saying that KE of the swing-person system does not equal to the combined KE's of the two separate systems, namely the swing and the person after he flies off. I had seen a problem like this, so it's not made up. The original question asked for the speed of the swing after the person jumps off. To find that, conservation of momentum can be applied to solve the problem, but not the conservation of energy. (The two speeds from these two different methods differ.)

If the person just let's go and flies off, then mechanical energy is conserved. (The person leaves with the same speed as the swing.) But if the person pushes off from the swing as he flies off, then he's adding energy.