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Friction, energies and variable distance

  1. Apr 14, 2004 #1
    Suppose you have a rail with two springs on the sides, both with the same constant K. The length of the rail is L, and the length of each spring is D (when not contracted or elongated). On that rail you have an object of mass M (which may be treated as a point particle), and the coefficient of friction between that object and the rail is μ. You take the object, and give it an initial potential energy of KD2/2, by forcing it against one of the springs until it is fully contracted and releasing the object.

    Where will the object be brought to a halt? Assume that once it stops, it does not start moving again (i.e infinite static friction).

    Obviously the beginning of the answer should be:
    [tex]F_{friction}x = \Delta {E_p}_{els}[/tex]
    But where do we go from there? We don't know where the object stops, so we don't know if it still has potential energy. Furthermore, how do we find the final location of the object from x?
     
  2. jcsd
  3. Apr 14, 2004 #2

    arildno

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    First, I assume that the length dimension in the problem is L+2D,
    that is L is the length of that part of the rail which lies strictly between the unstretched springs (?).
    Secondly, as to your infinite static friction:
    Where do you want it to occur?
    If you exclude it from the starting point, it still will make the system stop terribly soon.
    Thirdly, since you say you give the object an initial potential energy of K/2D^(2), this should be the same as saying the velocities of the object and the tip of spring remains equal up to x=D (if starting at x=0); the particle then parts from the spring and is no longer imparted any energy from the stretching of the spring.
    In particular, the particle's potential energy at D is 0.

    Without having more detailed information of how the spring is constructed, I don't see how we otherwise might solve the problem.
    In particular, I assume that a spring is always unstretched when the particle is not in contact with it.

    We can readily calculate the work of friction on the distance x=0 to D,
    and hence V(D,t1).
    (If imaginary, the object stopped somewhere x<D)

    Along the rail, we may calculate the velocity loss, and find V(D+L,T1),
    just prior to production of new potential energy.

    Hence, the contraction C1 of the second spring should be given by:
    1/2mV(D+L,T1)^(2)=1/2KC1^(2)+(mu)*mgC1
    , where (mu)*mgC1 is the frictional work.
    With infinite static friction, this gives your answer.
     
  4. Apr 14, 2004 #3
    No, the total length of the rail is L, including the unstretched springs. Forget about the static friction too, let's just say we are looking to find where the object first stops. What else do you need to know?
     
  5. Apr 14, 2004 #4

    arildno

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    The problem is a bit harder if you only have a finite static friction;
    the stopping criterion (when a spring is stretched) then becomes K*Cn<=(mustat)*mg,
    where (mustat) is the coefficient of static friction, and Cn the n'th spring contraction...
     
  6. Apr 14, 2004 #5

    arildno

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    1. Velocity at x=D:
    V(D,t1)=sqrt(((K/M)D^(2)-2(mu)*gD)
    If nonzero and real:
    2. On the strecth x=D, L-D, only frictional force (that's what I've assumed), hence:
    1/2MV1(L-D.T1)^(2)=1/2MV1(D,t1)^(2)-(mu)*Mg(L-2D)
    If nonzero and real:
    3.
    1/2MV1(L-D.T1)^(2)=1/2KC1^(2)+(mu)*MgC1.

    Position of particle should then be L-D+C1.
     
  7. Apr 14, 2004 #6

    Doc Al

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    I'm not getting the difficulty of this problem. (I'm probably missing something.)

    First find out if the mass stops before the spring fully decompresses, that is at x < L. (Ignore the fact that infinite static friction would not allow the mass to start moving. :rolleyes: ) Use the energy method that you started. If the mass does stop before x = L, then you are done. If not, you have some remaining KE at x = L. Then find out if it stops before hitting the second spring. Etc, etc.
     
  8. Apr 15, 2004 #7
    The problems is that the object is allowed, and does, bounce between the two springs several times before stopping. So x is most definitely larger than L.
     
  9. Apr 15, 2004 #8

    Doc Al

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    Ah... that just means that your values for k, μ, L, and D are such to allow it. Since the contraints are different in the three regions (x = 0-D, D-(L-D), (L-D)-L), I doubt there's a simple algebraic way to write down the answer. But given a particular set of parameters, I'm sure you'd have no problem cranking out the answer. (But I bet you knew that! :smile: )
     
  10. Apr 15, 2004 #9
    I thought I should try finding the energy that the object loses in every cycle, and from there find the difference in the spring's contraction. The problem is that the energy loss depends on the spring's contraction, so it's like a loop... I will try doing it 'manually', one cycle at a time, and see where that gets me.
     
  11. Apr 15, 2004 #10

    Doc Al

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    On second thought, given your "infinite static friction" contraint, the mass will never make it to x = L. If it has the energy to make it to the second spring, that spring will surely stop it.
     
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