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Particle confined to a potential under a force

  1. Feb 19, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    1)Consider the potential V(x) = -3x2+5x4. A particle of mass m has initial velocity v at the origin x = 0. Can the particle reach positions that are infinitely far from the origin? If not, find it's maximum distance from the origin.

    2)A particle of mass m, which is on the -ve x axis is moving towards the origin with constant velocity u. When the particle reaches the origin, it starts experiencing a force F = -kx2, where k is +ve constant. How far does the particle get along the +ve x axis?

    3. The attempt at a solution

    1) Since as x → ∞, V → ∞ the particle will always be confined to the potential. So there exists a maximal distance. I am not sure how to find this distance however. I have however, found the equilibrium point of the potential, velocity at the point where the force vanishes etc.. I think this may help here but I am not sure how.

    2) So, $$\frac{1}{2}mv_f^2 - \frac{1}{2}mu^2 = \Delta V = \int \underline{F} \cdot \underline{dx}.$$ If ##v_f## = 0, then we have $$-\frac{1}{2}mu^2 = \int_o^x -kx^2 dx$$ and solving for x gives ##x = \left(\frac{3mu^2}{2k}\right)^{1/3}.##
    OR
    set up a diff eqn: ##m \ddot{x} = -kx^2\,\Rightarrow\, mv \frac{dv}{dx} = -kx^2##, which is separable.
    Integrating both sides: $$\int_u^v v dv = \int_0^x -kx^2 dx \,\Rightarrow\, \frac{1}{2}\left(mv^2 - mu^2 \right) = -\frac{k}{3}x^3$$ which yields the same answer.
    Does this seem fine? I have a couple of questions: Even though the particle is restricted to 1D, it is under the influence of a force and so we can associate a potential with this force. This potential varies with x, and so I cannot say that V=0, right?

    This somewhat confuses me because previously I have done problems where a body has undergone a force in 1D, but I have set the potential to be arbritarily zero.
    Does this make sense? Where have I gone wrong in thinking there is some kind of contradiction?

    Many thanks.
     
  2. jcsd
  3. Feb 19, 2013 #2

    Simon Bridge

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    Lynchpin: conservation of energy.
     
  4. Feb 19, 2013 #3

    CAF123

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    I don't understand what you mean. Is this in response to my first or second question? Could you elaborate? Thanks.
     
  5. Feb 19, 2013 #4

    Simon Bridge

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    The total energy has to be a constant ... you know something about the kinetic energy at some place and the potential energy at quite a few places.

    i.e. part 1.
    at x=0, what is the potential energy? what is the kinetic energy?
    when the particle has gone as far as possible, what is the kinetic energy? Therefore what is the potential energy? This tells you the max value for x.
    You were actually doing this for part 2.

    You are computing the change in potential energy ... your choice of V=0 is arbitrary so you may as well pick the place which makes the math easy.
     
  6. Feb 19, 2013 #5

    CAF123

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    At x =0, potential is zero which means E = T.
    T = 0 at maximum distance so V = E.
    ok, so then $$\int 6x^2 - 20x^3 dx = \frac{1}{2}mv^2\,\Rightarrow\, 3x^2 - 5x^4 - \frac{1}{2}mv^2 = 0$$.
    Now let ##x = \alpha^2## to give a quadratic (quadratic expected since we know no information about particle direction so two solns corresponds to travelling left or right).
    In the end, I get $$x = \sqrt{\frac{3 \pm \sqrt{9 - 10mv^2}}{10}}.$$ Correct? (The dimensions look wrong)

    Do you have any ideas about my problem regarding part 2)? (see bottom post #1)

    Many thanks
     
  7. Feb 19, 2013 #6

    Simon Bridge

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    That's the way.
    This is why you draw energy levels over the potential graphs.
    The place where the energy level crosses the potential is the classical limit of motion.
    In QM, this limit can be exceeded.
    Indeed - commented bottom of post #4.
     
  8. Feb 19, 2013 #7

    CAF123

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    Perhaps an example will illustrate my problem: Consider a rollercoaster going along a flat surface and about to approach a loop de loop. On this flat surface, assign V=0, which means it is travelling in 1D and V=0 along the straight before it enters the loop. Say the rollercoaster is also under the influence of a force, namely the thrust of it's engine or friction from the track ,etc.. Now here is my problem: In the case above, the particle is in 1D and under a force -kx2. Since we can associate a potential, the potential is a function of distance and so changes along the track. So V is not zero along the track.

    However, in my rollercoaster example, the rollercoaster is also under a force but I have set V=0 along the straight. Only when it moves in the loop will its potential change. It's potential at some place in the loop is mgx, x the distance of the rollercoaster from the straight.

    Does this make more sense in illustrating my problem?
    Thanks
     
  9. Feb 19, 2013 #8

    Simon Bridge

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    Your two examples of force are non-conservative.
    That makes it more of a driving or damping force than something that is helpfully described by a potential.

    Have a look at the harmonic oscillator potential with damping.
     
  10. Feb 19, 2013 #9

    CAF123

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    Okay that makes sense. To be sure I understand fully: consider a human being walking along a pavement and friction acting on the person allowing them to move. So since friction is non conservative, I cannot associate a potential function and thus I can (arbritarily) set V=0. However, if this being was somehow under the influence of a conservative force, his/her potential would change as they walked along the pavement? (I presume this has to be a conservative force in direction of motion, since gravity always acts)
    Is this correct?

    Also, for the answer to part 1), I have an answer with a double sqrt for x, but I see that the dimensions do not check? Did I make an error somewhere?
     
    Last edited: Feb 19, 2013
  11. Feb 19, 2013 #10

    haruspex

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    The potential function given you is, on the face of it, dimensionally inconsistent, so you cannot expect anything better of your answer. You can fix this by considering that the two constants, -3 and +5, have dimension, namely, MT-2 and ML-2T-2 respectively. If you redo your calculation using unknowns for those, you should find it all makes sense.
     
  12. Feb 19, 2013 #11

    CAF123

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    Does the above make sense?
     
  13. Feb 19, 2013 #12

    Simon Bridge

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    I had a think about it ... I think I wasn't quite reading you correctly.

    We normally think if potentials as being "out there" - a background for particles to move against. It gets a little conceptually tricky when you use them for macroscopic situations.

    i.e. if a trolley heads into a fan, it may experience a constant force pushing it back - which you can turn into a potential barrier. If you put the fan on the cart pointing forward, the cart will feel a constant force against it's motion too ... which could also be a potential barrier - in fact, the math would be identical.
    But we'd usually think of this as the engine doing work rather than the cart running against a potential.

    This is what you are grappling with?

    Anyway - with any consideration of potential energy, you can choose any position you like to have a zero potential.
    Your example of friction would usually be treated in conservation of energy equations as a work term taking away from kinetic energy without contributing to potential energy. Where you have friction, you no longer have a strictly closed system - otherwise the entire thing could be described as a potential/kinetic relationship.

    Similarly with the motor - it is common to consider the motor doing work without worrying about where it gets the energy from, and so the system is, again, not closed.
     
    Last edited: Feb 19, 2013
  14. Feb 20, 2013 #13

    ehild

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    There are conservative and non-conservative forces. The work done by a non-conservative force depends on the path taken. The work done by a conservative force depends only in the initial and final position. Along any closed path, the work of a conservative force is zero.

    We can assign a potential function V(x,y,z) to a conservative force so the work done between the initial and final position is equal to W=Vi-Vf. Also, the force is equal to the negative gradient of the potential.

    Gravity is conservative, friction is not, neither is the force of the thrust.

    When a stone falls down from height H, the work of gravity is mgH. If a stone is thrown up and rises to height H, the work done by gravity is -mgH. If the stone is thrown up and then it falls down to the ground the net work of gravity is zero. The work done by gravity is the same if the stone is thrown at an angle: The work depends only on the height of rise. You can assign the potential energy to the force of gravity as U(z)=mgz, and W=mg(zi)-mg(zf).

    Let a block move along a horizontal surface with coefficient of friction μ, from A to B along a straight line. The force of friction is μmg and it is opposite to the velocity of the block. When the block moves from xA to xB, in the positive x direction, the force of friction is F=-μmg and the work of friction is
    WAB=-μmg(xB-xA)
    Now you push the block back from B to A. The force of friction changes direction as the block moves in the negative x direction. The work of friction is
    WBA=μmg(xB-xA)=WAB, the same as before.
    The net work done is different from zero. Also, if you move the block along a circle from A to B, the work is -μmg l, the length of the arc.
    The work depends on the path taken and it is not zero along a closed path. The force of friction is not conservative.

    The same with a thrust, it means a force forward along the velocity of the object. Its work is equal to (force of thrust) times (path taken). It is not a conservative force.

    As for your roller-coaster example: It experiences the force of gravity and the normal force from the ground and the force of thrust.

    Gravity has potential, and it depends only on the height above the pavement. The normal force does not do any work. The thrust is not conservative. The potential energy does not change along the horizontal track, but the thrust does some work. Along the loop, the potential energy changes to mgH as gravity does -mgH work, but the thrust does work, too. The net work done is equal to the change of kinetic energy.
    KE(f)-KE(i)=-mgH+W(thrust). The work done by the conservative force is equal to the negative potential difference, U(i) - U(f): KE(f)-KE(i)=U(i) - U(f)+W(thrust). You can move the potential energy term to the other side saying that the change of mechanical energy is equal to the work of thrust. [KE(f)+U(f)]-[KE(i)+U(i)]=W(thrust). In general: The change of mechanical energy E=KE+PE is equal to the work done by the non-conservative force.
    The force F=-kx^2 is also a conservative force, it has potential. If two conservative forces act on an object, it has two potential energy terms. For example, a block hanging on a spring has both elastic and gravitational potential energy.
    In that case, the change of all potentials add up and determine the potential energy of the object.


    ehild
     
    Last edited: Feb 20, 2013
  15. Feb 20, 2013 #14

    CAF123

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    It's probably quite simple, but I am not seeing it yet. Perhaps I am not explaining myself very well. Going back to the rollercoaster example, and consider some energy conservation question about a rollercoaster.

    (See attached)

    MAny thanks.
     

    Attached Files:

  16. Feb 20, 2013 #15

    ehild

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    A conservative force has potential. If a conservative force acts on a body, the body has potential energy. In the gravitational field, it has gravitational potential energy. If it is connected to a spring, it has elastic energy, but it is also kind of potential energy. If there are several conservative forces the potential energy of the body is the sum of all potential energies.

    ehild
     
  17. Feb 20, 2013 #16

    Simon Bridge

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    (Off the attachment: if the particle has a charge, it can experience an electrostatic potential without changing height.)

    I believe ehild and I have explained the situations.
    You wrote that $$V_b-V_a = \int_a^b d.fx$$ ... this means that the force is related to the gradient of the potential. If there is no force, then the gradient must be zero, so the potential does not change.

    Technically that should be "potential energy" though.
     
  18. Feb 20, 2013 #17

    CAF123

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    Thanks for your help Simon and ehild.
     
  19. Feb 20, 2013 #18

    Simon Bridge

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    No worries.
     
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