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The effect of spring force when perpendicular to displacement?

  1. Nov 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi everyone, I am generalizing the following problem because I disagree with how the book solves the problem.



    A square mass 'm' is attached to two springs with identical constants 'k', in a configuration such that the springs mount to the left/right sides of the mass and extend outward and upward towards the ceiling in a 'V' shape. It starts from rest and moves upward a distance 'h', catapulted upward by the springs and opposed by gravity. What is the final velocity of the mass when the mass reaches the ceiling (and now the springs are horizontally aligned with the mass).

    2. Relevant equations

    Conservation of Energy
    PEsprings,initial = KEfinal + PEgravity,final + PEsprings,final

    Initial KE, Initial PE gravity are both zero.
    There is a 'final' potential energy from the springs because the springs haven't returned to their initial free length.

    3. The attempt at a solution

    My problem with the book solution is that, at the final state configuration, the mass is moving upward and the springs are perfectly horizontal to the mass. Why should there be any potential energy from the spring in the system if the force of the spring is doing no work at that instant? The force is perpendicular to the displacement, so I'm thinking the solution should have some kind of cos(theta) term.

    In the book solution, they simply add everything together. I believe that whoever wrote the solution to this problem is misunderstanding the statement that 'spring forces are conservative' and didn't pay attention to the fact that the force is not doing any work at the final state.
     

    Attached Files:

    Last edited: Nov 7, 2013
  2. jcsd
  3. Nov 7, 2013 #2
    Not doing any work and not having potential energy are unrelated. If you hold a spring extended, no work is being done, yet the potential energy is clearly there.
     
  4. Nov 7, 2013 #3
    OK. Let's do it with forces. If the mass is y meters from the ceiling, what is the length of each spring? What is the tension in each spring? What is the vertical component of tension in each spring?

    Chet
     
  5. Nov 8, 2013 #4

    CWatters

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    The problem statement contains the words "Assume that the springs become loose and floppy once they're at their rest length". This is a trick because they never reach their rest length....

    When it hits the ceiling each spring will still be 2m long. The unstretched length of the spring is 1M so the springs still contain some PE. In other words not all the energy stored in the springs at the start is available to do work on the block.
     
  6. Nov 8, 2013 #5
    Thanks for the replies everyone! I must admit that I'm not a student, I am an adjunct Professor at a community college, teaching Engineering Dynamics. I understand how the book says to solve the problem: here's my issue:

    We arrive at the Conservation of Energy by integrating the conservative forces with respect to the change in displacement. In this case, however, there are moments in time (the final state) where the force of the spring can do NO work on the system. My question is: If we were rolling a ball horizontally on a table, we wouldn't include gravity because gravity is doing no work on the system. I feel as though the spring terms should have some kind of 'Sin(theta)" which allows the work done by the spring to be zero when perpendicular.

    One more point which I think proves my assertion: If you calculate the Newtonian equations of motion for this problem and then calculate the Lagrangian equations of motion for this problem -- they are only identical when you take this sin(theta) term into account!
     
  7. Nov 8, 2013 #6
    That is true, but other times work gets done, which contributes to the change in potential energy.

    If the table is completely flat, then the initial potential energy and the final potential energy are equal, so we might as well drop both terms from the conservation law.

    But if you have, say, an initial flat region and a final flat region, at different heights, then, even though no work is done at some initial and final periods of time, it is done somewhere in the middle, so you end up with different initial and final potential energies.

    Work is always done over some spatial extent. If at some point velocity and force are orthogonal, you can at best say that instantaneous work is zero, but work over any finite extent may still be non-zero. So I am not even sure why what happens at one one particular point matters so much to you.

    If, using the same coordinate, you get different results in Newtonian and Lagrangian formalisms, then there is clearly an error somewhere in your derivation. Without seeing the latter, though, it is hard to say where it is.
     
  8. Nov 8, 2013 #7

    CWatters

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    If the ball in your example was sliding down a frictionless slope would you expect the work done by gravity to be mgh or mghSinθ ?
     
  9. Nov 8, 2013 #8
    OK everyone, I'm attaching the Newtonian and Langrangian equations of motion... Aside from a discrepancy with one negative sign (whoops) they are identical under these conditions:

    -In Newtonian, the spring force varies with the Sin(theta) (written in terms of y and the width of the system)
    -In Langrangian, the potential energy of the springs are functions of y, unstretched spring length Lo, and width of the system.

    If you calculate the Potential when y = h (the ceiling in my textbook), there IS a potential energy unless the width = Lo. This is consistent with what you guys are saying (and the book solution).

    So I guess I was wrong/right... The potential energy of the springs IS a function of the angle that the spring is at, but even so, unless the spring returns to unstretched length, there is a NONZERO amount of energy in the system regardless of whether or not the spring can do work on the system.

    Still a bit strange to me because of how we derive potential energy from work, but oh well...
     

    Attached Files:

  10. Nov 8, 2013 #9
    You were doing pretty well for a while there on the Newtonian analysis, although you have a sign error on the 2k term. I'm going to go back to your third from last equation with the sign corrected:[tex]\frac{d^2y}{dt^2}=-g+\frac{2k}{m}\left((h-y)-L_0 \frac{(h-y)}{\sqrt{x^2+(h-y)^2}}\right)[/tex]
    Now, suppose I multiply both sides of the equation by dy/dt:
    [tex]\frac{dy}{dt}\frac{d^2y}{dt^2}=\frac{1}{2}\frac{d}{dt}\left(\frac{dy}{dt}\right)^2=-g\frac{dy}{dt}+\frac{dy}{dt}\frac{2k}{m}\left((h-y)-L_0 \frac{(h-y)}{\sqrt{x^2+(h-y)^2}}\right)[/tex]
    This equation can be rewritten as:
    [tex]\frac{d}{dt}\left(\frac{mv^2}{2}\right)=-mg\frac{dy}{dt}-k\frac{d(h-y)^2}{dt}+2kL_0 \frac{d}{dt}\sqrt{x^2+(h-y)^2}[/tex]
    This equation can be rewritten as:
    [tex]\frac{d}{dt}\left(\frac{mv^2}{2}\right)=-mg\frac{dy}{dt}-k\frac{d(\sqrt{x^2+(h-y)^2}-L_0)^2}{dt}[/tex]
    This equation can be rewritten as:
    [tex]\frac{d}{dt}\left(\frac{mv^2}{2}+mgy+k(L-L_0)^2\right)=0[/tex]

    Does this answer the OP's question?

    Chet
     
    Last edited: Nov 8, 2013
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