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Rolling Coin, Kleppner and Kolenkow, off by a factor of 3

  1. Jan 12, 2013 #1
    1. The problem statement, all variables and given/known data
    If you start a coin rolling on a table with care, you can make it roll in a circle. The coin "leans" inward, with its axis tilted. The radius of the coin is b. The radius of the circle traced by the coin's center of mass is R, and the velocity of its center of mass is v. The coin rolls without slipping. Find the angle [itex]\phi[/itex] that the coin's axis makes with the horizontal.


    2. Relevant equations



    3. The attempt at a solution
    Here is my attempt at a solution: (attached)

    A correct solution that I understand and agree with is given here: https://www.physicsforums.com/showthread.php?t=365994. The only difference between that solution and my own is that instead of choosing the point of contact to be the center of my coordinate system, I chose the center of mass of the coin. I cannot figure out why my answer is 1/3 of his/hers and I would be very happy if you could point out the flaw in my reasoning.
     

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    Last edited: Jan 12, 2013
  2. jcsd
  3. Jan 12, 2013 #2

    rude man

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    Too hard to read!
     
  4. Jan 12, 2013 #3
    Apologies - hopefully this is better. Thank you for your help!
     

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  5. Jan 12, 2013 #4
    I uploaded it right-side up, but now it's upside-down. I will now upload upside-down and hope it comes out right-side up...
     

    Attached Files:

  6. Jan 12, 2013 #5
    Well this is absolutely ridiculous... I'd very much appreciate it if you could still look at it, rotating a photo isn't hard in a photo-viewing program.
     
  7. Jan 12, 2013 #6

    rude man

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    The problem is not rotation. The problem is legibility.
     
  8. Jan 12, 2013 #7
    Apologies, here is my solution in text form:

    We only consider the component of angular momentum due to rotation of the coin (and not procession around the circle) since the latter does not change with time. Then |L| = |Iω| = 0.5 mb^2 ω = mvb^2/2b =mvb/2. Then L undergoes circular motion, with radius Lcos[itex]\phi[/itex] and angular velocity v/R. Then |dL/dt| = Lcos[itex]\phi[/itex]v/R. On the other hand, the magnitude of the torque about the center of mass is bmg sin[itex]\phi[/itex], equating the two gives tan[itex]\phi[/itex] = Lv/Rmgb = (0.5mbv^2)/(Rmgb) = v^2/2Rg
     
  9. Jan 12, 2013 #8

    TSny

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    There is another force acting on the coin. Recall that the acceleration of the center of mass point of the coin is due to the net external force. The center of mass of the coin is moving in a circle and therefore has an acceleration. What force provides that acceleration?

    Another thing is that the tilt of the coin makes the center of mass move in a circle of radius R that is smaller than the radius of the circle on the table that the coin is rolling around. If ##v## is the speed of the center of mass, then I don't think that ##\omega b## is equal to ##v##. However, I believe you can show that if ##(b/R)sin\phi << 1## then to a good approximation ##\omega b \approx v##
     
  10. Jan 12, 2013 #9
    Thank you so much! I am a high school student self-studying, so it is extremely helpful to receive your guidance. I really should've seen that!

    Using the other person's solution (shifting the frame of reference to the point of contact) makes the answer more exact, whereas I had to use an approximation (alongside the one you mentioned).
     
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