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Kleppner and Kolenkow (block sliding in a ring)

  1. Dec 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A block of mass ##m## slides on a frictionless table. It is constrained to move inside a ring of radius ##l## which is fixed on the table. At ##t=0##, the block is moving along the inside of the ring with tangential velocity ##v_0##. The coefficient of friction between the ring and the block is##\mu##. Find the velocity and position of the block at later times.

    2. Relevant equations
    Newton's second law of motion.

    3. The attempt at a solution
    Since I'm not sure what is meant by later times, I consider a small interval of time where the block sticks to the ring, say ##I=[0,T]##, and the block is under circular motion in that interval.
    Since we want to know the speed which will have only tangential component in circular motion, we have to calculate ##\dot\theta##.

    As initial conditions, I get ##\dot\theta(0) = \frac{v(0)}{r(0)} = \frac{v_0}{l}## and ##\theta(0) = 0##

    Since the block is under ring reaction force radially, and under friction force tangentially, I get by Newton's second law that ##a_{\theta} = \mu a_r##, only for time ##t## in ##I##. Therefore I get the differential equation
    ## \frac{\ddot\theta}{\dot\theta} = -\mu \dot\theta ##
    which thanks to the initial condition gives
    ## \dot\theta =\frac{v_0}{l} e^{-\mu\theta}##
    That equation is equivalent to
    ##\frac{d}{dt} (e^{\mu\theta}) = \mu\frac{v_0}{l}##
    Solving it gives
    ## \theta(t) = \frac{1}{\mu} ln(1+\frac{\mu v_0}{l}t)##
    So that
    ## \dot\theta(t) = \frac{v_0}{l}\frac{1}{1+\frac{\mu v_0}{l}t}##
    Therefore
    ## v(t) = r(t) \dot\theta(t) =\frac{v_0}{1+\frac{\mu v_0}{l}t}##
    and
    ## \vec r(t) = l (cos(\theta) \vec i + sin(\theta) \vec j ) ##

    3. Question
    I begin in physics and I'm unsure about the solution, the question is not very clear either and I find the proof more complicated than other exercises in the book. Does it look good to you?
     
    Last edited: Dec 2, 2014
  2. jcsd
  3. Dec 2, 2014 #2
    The differential equation looks good. But the solution you found is not correct.
     
  4. Dec 2, 2014 #3
    Thanks for replying.
    What part do you disagree with ? I have done calculations again and I find the same result.
     
  5. Dec 2, 2014 #4
    Sorry, I misread your solution. Somehow it seemed to me that you had ##e^{-\mu t}##, whereas you had ##e^{-\mu \theta}##. I will check the rest of your work now.
     
  6. Dec 2, 2014 #5
    Yes, no problem. Let me know what you think afterwards.
     
  7. Dec 2, 2014 #6

    haruspex

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    I get the same answer, a slightly different way. I integrated ##\frac{d\omega}{\omega^2} = -\mu dt##.
     
  8. Dec 2, 2014 #7
    Your way is much better since you get to the result in a single pass while I have to do it in two passes to get to the same result.
    Thanks !
     
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