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Elastic Force over Semicircle

  1. Feb 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A variable force F_vec is maintained tangent to a frictionless, semicircular surface View Figure . By slowly varying the force, a block with weight w is moved through the angla theta, and the spring to which it is attached is stretched from position 1 to position 2. The spring has negligible mass and force constant k. The end of the spring moves in an arc of radius a.

    Calculate the work done by the force F_vec.

    2. Relevant equations
    K_1 + U_1 + Wother = K_2 + U_2


    3. The attempt at a solution
    I'm not sure if I'm getting this correctly. Besides the elastic force, there's also the varying force correct? So that should Wother?

    This is my equations so far:

    K_1 = 0
    U_1 = 0

    K_2 = 1/2mv^2
    U_2 = 1/2k(a*theta)^2

    But I'm stuck on finding the velocity for K_2 in terms of a and theta. Any help is greatly appreciated, thanks. ... Assuming I'm correct so far (which I sorta doubt).
     

    Attached Files:

    Last edited: Feb 18, 2007
  2. jcsd
  3. Feb 18, 2007 #2
    Actually, when the problem says it is moved very slowly, that means you can neglect the fact that it has a tiny kinetic energy. So, K_2 should be 0. The other parts of your work look correct. :)
     
    Last edited: Feb 18, 2007
  4. Feb 18, 2007 #3
    I entered that and it says it's incorrect. What else am I missing? What about the weight on top of the spring? Does that have something to do with it?
     
  5. Feb 18, 2007 #4
    Ah, yes, if the block is being raised vertically you will need to include potential energy of mgh after it is raised. (Of course you'll need to express mgh in terms of w and theta.)
     
  6. Feb 18, 2007 #5
    So I have to add the potential energy of the box to the potential energy of the spring (which is the elastic force right)? So mgh_box = (w*a*theta)?
     
  7. Feb 18, 2007 #6
    You do have to add the potential energy of the box to that of the spring, but be careful how you do it. Remember that h is the height the object has been raised, not the distance it has traveled along the semicircle.
     
  8. Feb 18, 2007 #7
    The height isn't theta, is it? Wait, shouldn't the height be the radius then? >< Sorry I'm so confused about this circle thing.
     
  9. Feb 18, 2007 #8
    It's okay, to see how much the block has been raised, draw a line on the diagram to represent the height the block has moved:

    [​IMG]

    Look familiar?
     
  10. Feb 18, 2007 #9
    Got it. So the total potential energy should be:

    deltaU = wasin(theta) + 1/2k(a*theta)^2

    Right? Just making sure because I only have one last attempt.
     
  11. Feb 18, 2007 #10
    That looks good to me. :)
     
  12. Feb 18, 2007 #11
    The answer is wrong. What else could we be missing? This is so frusterating. ><
     
  13. Feb 18, 2007 #12
    I'm not sure. I just worked the problem by integrating an infinitesimal amount of work dW = F*dx and I got the same answer.

    Net force acting on the block during its movement along the semicircle (mgsin(theta) and N cancel):
    [tex]
    \sum F_{net} = F - kx - wcos(\theta) = 0
    [/tex]

    So,

    [tex]
    F = wcos(\theta) + kx
    [/tex]

    (Remember x = a*theta and dx = a*dtheta)

    [tex]
    dW = F dx \\
    = (wcos(\theta) + kx) dx \\
    = (wcos(\theta) + k(a \theta)) a d\theta \\
    [/tex]

    [tex]
    W = \int_0^{\theta_2} w\,cos \theta \,a\,d\theta\,+\,\int_0^{\theta_2} k a \theta\,a\,d\theta \\
    = w\,a\,\int_0^{\theta_2} cos \theta \,d\theta\,+\,k a^2 \int_0^{\theta_2} \theta\,d\theta \\
    [/tex]
    [tex]
    W = w a sin(\theta_2) + \frac{1}{2}k a^2 \theta_2^2
    [/tex]

    where [itex]\theta_2[/itex] is the final angle (the initial angle is zero).

    ...which should be the correct answer...?

    Is the homework an online assignment where you have to fill out a textbox? Could the error be in the format you entered?
     
    Last edited: Feb 18, 2007
  14. Feb 18, 2007 #13
    Yeah, it's an online hw. Erm, I just checked again the answer I entered and I didn't make a mistake entering it. I don't know what's the answer since I've maxed out my attempts already. :x I'll ask my TA if it's possible. Thanks for helping me so much already, really appreciated it.
     
  15. Feb 18, 2007 #14
    Sure, I'd be interested in knowing the answer to this problem, so keep me posted when you find out.
     
  16. Feb 18, 2007 #15
    Peach. The answer is correct. Check the way you entered it. Make asin(theta), apparently, is not the same as a*sin(theta). The system probably screwed you over that way.
     
  17. Feb 19, 2007 #16
    I don't remember how I entered it, whether it was asin(theta) or a*sin(theta) but the answer that I entered is showing up as asin(theta).

    I can't copy it directly but this is what it's showing up as:

    (w*asin(theta)) + (1/2)*k*(a*theta)^2
     
  18. Feb 19, 2007 #17
    is there possibly a different way to write the solution? i've thought about this problem ALOT and although the answer is right, i know i'm missing something stupid.
     
  19. Feb 19, 2007 #18
    i had the same problem as you. i entered it with the asterisk and it was fine. i believe the system interprets "asin(theta)" as arcsin?
     
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