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Why is/is-not energy conserved in these scenarios?

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Homework Statement


PROBLEM A: A small cube of mass m slides down a circular path of radius R cut into a large block of mass M. M rests on a table, and both blocks move without friction. The blocks are initially at rest, and m starts from the top of the path. Find the velocity v of the cube as it leaves the block.

PROBLEM B: A small block slides from rest from the top of a frictionless sphere of radius R. How far below the top x does it lose contact with the sphere? The sphere does not move.

Homework Equations


Problem A: Conservation of Energy, Conservation of Momentum
Problem B: Conservation of Energy, FBD on m.

The Attempt at a Solution


I've solved both problems with what is the correct answer. Problem A: v = [(2gR)/(1+m/M)]1/2; Problem B: R/3.

However, I'm looking back at how both of these problems are correctly solved and am second-guessing the thinking behind both of them.

PROBLEM A: Examining the M-m system, the net external force-x equals zero, so momentum-x is conserved in the system. I understand this. From the energy perspective though, we consider the m-M-earth system. I define h=0 to be the vertical height at which m is leaving M with a horizontal velocity at. So, our initial energy (ONLY the grav. potential energy of m) must equal our final energy (the kinetic energy of m AND M). As M is a rigid body, we may examine any one point on it and see that its change in grav. potential energy is zero, so we are not concerned with that.

But, what if I were for some reason analyze just the m-earth system with the same h=0 height? Now, we have the equation that the initial grav. potential of m is equal to the final kinetic energy of ONLY m, clearly disagreeing with our equation from before. But what would make this second, "wrong" equation wrong and our first equation right? In the m-earth system, no friction from M acts on m (we are told its frictionless) AND the only force on m by M would be a normal force that is always perpendicular to m's displacement. So wouldn't energy be conserved in our "incorrect" way of looking at things? What's going on here?

PROBLEM B: The problem in this case is solved correctly by ONLY analyzing the m-earth system. (I'll say that m is the mass of the sliding block.) The mass of the sphere isn't even assigned a variable (although neither was the sliding block) and we are told that the sphere does not move. Analyzing the m-earth system (with h=0 being the vertical position of the "flight" point), we get that our initial energy (grav. potential of m) must equal our final energy (kinetic energy of m at "flight" point or at the angle with respect to the vertical at which it takes flight). Taking this with a FBD equation for the centripetal acceleration of m solves the problem.

So, similar to my "incorrect" way of thinking about problem A, we are only considering the m-earth system. Because there's no friction AND because the normal force acting on m is perpendicular to m's displacement, energy should be conserved in this system, making our equations useful. ...But if it's conserved in the m-earth system in this case, why is it presumably not conserved in my "incorrect" way of thinking of problem A?

Additionally, what if we were to consider problem B from an m-M(sphere)-earth perspective. Is energy conserved in this system? My assumption is that energy would be conserved BECAUSE we are told that the sphere does not move. As the diagram shows the sphere resting on some sort of stand attached to the table, I assume that this stand/table must be exerting a force onto the sphere to keep it from moving, BUT (this just hit me as I was about the argue that energy should not be conserved in this system) that his force is balancing what every force little m is exerting downwards, at an angle, on the M(sphere), meaning that the net force on the sphere is zero, meaning that no work is done on the sphere and that energy is conserved in the entire m-M(sphere)-earth system. Then, from this perspective, you would write down your energy equations only to see that kinetic-initial, kinetic-final, potential-initial, and potential-final of the sphere all equal zero, returning us back to the same equation in the "correct" way of solving the problem.

So, I think I've convinced myself that the m-Earth system analysis works in Problem B (it came to me as I was about to type why I think it should not work), but why does it not work for Problem A?

Additionally, as a side question, to have gravitational potential energy due to the earth, it makes sense for us to include the earth in our system, right? (Otherwise energy would clearly not be conserved.) So my notation of m-earth system and such is correct, right? Additionally, we would then have to take into account the kinetic and potential energy of the earth, but this would be calculated from our reference frame, right? (meaning that we would find the velocity of the earth and potential energy of it to always be zero?)

Thank you so much.
 

Answers and Replies

  • #2
PeroK
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PROBLEM A: Examining the M-m system, the net external force-x equals zero, so momentum-x is conserved in the system. I understand this. From the energy perspective though, we consider the m-M-earth system. I define h=0 to be the vertical height at which m is leaving M with a horizontal velocity at. So, our initial energy (ONLY the grav. potential energy of m) must equal our final energy (the kinetic energy of m AND M). As M is a rigid body, we may examine any one point on it and see that its change in grav. potential energy is zero, so we are not concerned with that.
One problem with the argument is that you are essentially looking at the problem in the reference frame of the large block M. If that does not move (e.g. is anchored to the Earth) then that represents an inertial reference frame, in which Newton's laws hold. If, however, the large block moves (hence accelerates) then that no longer represents an inertial reference frame. In that accelerating reference frame, Newton's laws do not hold (unless you add the necessary "fictitious" or "inertial" force acting on the small mass ##m##).
 
  • #3
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One problem with the argument is that you are essentially looking at the problem in the reference frame of the large block M. If that does not move (e.g. is anchored to the Earth) then that represents an inertial reference frame, in which Newton's laws hold. If, however, the large block moves (hence accelerates) then that no longer represents an inertial reference frame. In that accelerating reference frame, Newton's laws do not hold (unless you add the necessary "fictitious" or "inertial" force acting on the small mass ##m##).
Okay. So when we take into account inertial reference frames (which we always should), then my interpretation would be the following:

Problem A: Taking energy into account using an m-M-earth system ("correct" way) IS correct because of the fact that M is accelerating. We can see that it is accelerating as little m is exerting a downward-angled force onto it, which will be unbalanced onto M in the x-direction. Analyzing problem A from the "incorrect" m-earth system IS incorrect. because we'd be essentially analyzing the energy from the perspective of M, which is a non-inertial reference frame. Mathematically, it is clear that we would be analyzing it from M's perspective because we get math that speaks nothing of M's kinetic and potential energy. This makes sense because M cannot have kinetic energy (no velocity relative to itself) or potential energy (no change in position relative to itself) relative to itself. Is this way of thinking right?

Problem B: Taking energy into account suing an m-earth system is correct. This would be the same as taking energy from the perspective of myself if I were observing the system and my velocity relative to the ground was zero. This is the same as taking energy from the perspective of M which is correct because M also has a velocity of zero relative to the ground. M has a velocity of zero because the stand/table is exerting a force to balance the downwards, inwards normal force on m ONTO big M. Is this way of thinking right?

I think that I understood what you said, but that you linked it to an example of me describing a non-inertial reference frame in Problem A but it being analyzed correctly through using an m-M-earth system INSTEAD of an m-earth system. If I am interpreting what you said incorrectly, please tell me.
 
  • #4
PeroK
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I think that I understood what you said, but that you linked it to an example of me describing a non-inertial reference frame in Problem A but it being analyzed correctly through using an m-M-earth system INSTEAD of an m-earth system. If I am interpreting what you said incorrectly, please tell me.
In general, if you have two blocks that can both move, then analysing the forces is tricky (*). That's one reason that the energy approach is much better.

If you had a scenario where block ##M## was being accelerated by an external force at a given rate, then you could solve the problem in the reference frame of block ##M## by applying a constant fictitious force to block ##m##. But, in this example, the acceleration of ##M## is constantly changing.

(*) This is definitely something to remember, as it is easy to fall into the trap of looking at forces and angles wrongly when the large block is accelerating. I'd be happy with the energy-based solution!
 
  • #5
jbriggs444
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M cannot have kinetic energy (no velocity relative to itself)
In the general case, block M could rotate. But in this case we have the table acting as a dump for angular momentum about any horizontal axis and we are expected to assume that the forces involved are centered so that there is no torque about the center of mass in the vertical axis.
 
  • #6
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In general, if you have two blocks that can both move, then analysing the forces is tricky (*). That's one reason that the energy approach is much better.

If you had a scenario where block ##M## was being accelerated by an external force at a given rate, then you could solve the problem in the reference frame of block ##M## by applying a constant fictitious force to block ##m##. But, in this example, the acceleration of ##M## is constantly changing.

(*) This is definitely something to remember, as it is easy to fall into the trap of looking at forces and angles wrongly when the large block is accelerating. I'd be happy with the energy-based solution!
Alright. Our class plans to cover fictitious forces and non-inertial reference frames more in the future, so I look forward to that.

What about with how it relates to including the earth in our system as we do in essentially all of these problems (towards the bottom of my first post). Is my way of thinking there flawed?
 
  • #7
PeroK
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What about with how it relates to including the earth in our system as we do in essentially all of these problems (towards the bottom of my first post). Is my way of thinking there flawed?
The Earth is considered so large that any distance it moves or KE it gains is negligible.
 
  • #8
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One problem with the argument is that you are essentially looking at the problem in the reference frame of the large block M. If that does not move (e.g. is anchored to the Earth) then that represents an inertial reference frame, in which Newton's laws hold. If, however, the large block moves (hence accelerates) then that no longer represents an inertial reference frame. In that accelerating reference frame, Newton's laws do not hold (unless you add the necessary "fictitious" or "inertial" force acting on the small mass ##m##).
Thinking about this all again, I realize that the reference frame of the large block M is non-inertial. But why can't I make the argument that I'm examining this situation from an inertial reference frame (let's say I'm right next to this all, I'm at rest with respect to the earth, and I'm watching this transpire) and that I'm only examining the little m-earth system? If big M's forces on m are always perpendicular to the displacement of little m and I am viewing this from an inertial reference frame, why wouldn't energy be conserved? Why am I "not allowed" to analyze just the little m - earth system? What am I missing here?
 
  • #9
haruspex
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If big M's forces on m are always perpendicular to the displacement of little m
But in your inertial frame they will not be. Some of m's displacement will be more vertical than that, so have a component into the surface of M, matching the displacement of M. Therefore m does work on M.
 
  • #10
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But in your inertial frame they will not be. Some of m's displacement will be more vertical than that, so have a component into the surface of M, matching the displacement of M. Therefore m does work on M.
Could you state that again in another way? I'm not sure I understand what you're getting at. Some of m's displacement will be more vertical than what? And isn't the displacement of little m always tangent to the surface of M, meaning that it's displacement will always be perpendicular to M's normal force on m? My thinking is then that, because the normal force on little m by big M does no work on little m, that energy would be conserved. Please tell me where my thinking goes wrong.

I can see why m does work on M with how the normal force of little m onto big M is, for most of the time, downwards at an angle, exerting a horizontal normal force onto M (we don't care about this vertical component of the normal force onto big M by little m as we assume that the surface supporting the entire system will simply exert a larger normal force to compensate for this).
 
  • #11
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But in your inertial frame they will not be. Some of m's displacement will be more vertical than that, so have a component into the surface of M, matching the displacement of M. Therefore m does work on M.
Wait a second. Are you arguing that if little m is doing work on big M, then big M must be doing negative work on little m? This would come simply from Newton's 3rd law, right? (This only seems weird to me because my intuition is telling me that the normal force by M on m is always perpendicular to m's displacement. I'll think about this some more and convince myself otherwise soon.) So, if I'm analyzing a system and a component of my system is doing work on some particle/object not included in my system, do I then make the argument that minus work is being done by the outside object on the inside object of my system, and that energy is not conserved? I'll think about this more when I have the time or whenever you're able to reply.
 
  • #12
haruspex
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Some of m's displacement will be more vertical than what?
More vertical in the inertial frame than in the frame of reference of M.
And isn't the displacement of little m always tangent to the surface of M, meaning that it's displacement will always be perpendicular to M's normal force on m?
It stays in contact with the surface, but that surface is moving. Consider m to be at a point on M's surface at time t, and the tangent at that point is at angle θ to the horizontal. Time dt later, M has moved some small distance horizontally. To stay in contact, m must descend (in the lab frame) at an angle steeper than θ. Thus its motion has a component (anti)parallel to the normal force.
 
  • #13
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More vertical in the inertial frame than in the frame of reference of M.

It stays in contact with the surface, but that surface is moving. Consider m to be at a point on M's surface at time t, and the tangent at that point is at angle θ to the horizontal. Time dt later, M has moved some small distance horizontally. To stay in contact, m must descend (in the lab frame) at an angle steeper than θ. Thus its motion has a component (anti)parallel to the normal force.
I can't believe that I wasn't taking that into account (the movement of big M) when I was picturing m's motion in my head. And then this just leads to a Newton's 3rd Law argument where there now is a net* external work (the normal force on big M which isn't perpendicular* to the displacement of little m) that is changing the amount of mechanical energy in the m-earth system, making our equations of assuming that energy is conserved in the m-earth system invalid, right?
 
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  • #14
haruspex
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I can't believe that I wasn't taking that into account (the movement of big M) when I was picturing m's motion in my head. And then this just leads to a Newton's 3rd Law argument where there now is a new external work (the normal force on big M which isn't parallel to the displacement of little m) that is changing the amount of mechanical energy in the m-earth system, making our equations of assuming that energy is conserved in the m-earth system invalid, right?
Yes.
 
  • #15
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Doesn't really address the OP's question(s) but perhaps worth noting the similarity between Problem A and this thread on alpha decay. In the latter, energy from mass conversion can be apportioned by mass ratio - here the same with the gravitational potential energy of cube mass m. Letting ΔE = mgh , the respective kinetic energies of cube and block upon separation will be:

$$KE_{cube}=ΔE\times\frac{M}{M+m}$$ and $$KE_{block}=ΔE\times\frac{m}{M+m}$$

From which we may readily verify the OP's solution to Problem A. The solutions above should be consistent with conservation of energy and momentum (in earth frame if that needs to be specified ?).
 
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