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Math Help - A physics problem

  1. #1
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    A physics problem

    This physics problem appeared in my Calculus book:
    Given a rod with length a it is hindged thus it can freely rotate. An object is attached to the end of the rod and it is brough all the way vertically up. Then a slight disturbance and it flys down (from equilibrium). How long does it take for it to come down to the minimum point? Just remember it travels in a circular motion.
    My book gave the answer:
    \sqrt{\frac{a}{g}}\ln{(1+\sqrt{2})}
    Where g I am guessing it the gravitational acceleration.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    This physics problem appeared in my Calculus book:
    Given a rod with length a it is hindged thus it can freely rotate. An object is attached to the end of the rod and it is brough all the way vertically up. Then a slight disturbance and it flys down (from equilibrium). How long does it take for it to come down to the minimum point? Just remember it travels in a circular motion.
    My book gave the answer:
    \sqrt{\frac{a}{g}}\ln{(1+\sqrt{2})}
    Where g I am guessing it the gravitational acceleration.
    It seems to me that there is something wrong here. It looks to me that
    as the disturbance goes to zero the time to fall goes to \infty.

    RonL
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    This is just like a free-falling problem from physics. Except this is one is in a circular path? Did you try to do the problem and got some unusual result?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    This is just like a free-falling problem from physics. Except this is one is in a circular path? Did you try to do the problem and got some unusual result?

    What I did is derived the relevant DE for this problem (I was overcome
    with curiosity because physical intuition was that it should take a arbitrarily
    long time to fall from an unstable equilibrium as the initial disturbance becomes
    arbitrarily small).

    The DE can be solved but gives an answer in terms of for the (quarter)
    period in terms of an elliptic integral (not necessarily a problem in itself as
    we want a special case which could have in principle reduced to something
    more easily manageable).

    Examining the integral showed that there was a singularity at one end of the
    range of integration for the given initial condition, and that near the
    singularity the integral behaved like 1/x, and so the integral
    should be divergent.

    So I did some research, and as far as I can tell (the research did not turn
    up this exact problem but something similar) it indicates that my initial
    feelings about this may be correct.

    RonL

    The nearest reference to this problem is in Morris Klein's

    "Calculus : An Intuitive and Physical Approach"
    http://www.amazon.com/gp/product/048...lance&n=283155
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  5. #5
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    But this problem cannot have some kind of extremely complicated solution (elliptic integral) it appeared in my Calculus book for univeristies.

    Perhaps, I am thinking, this is somehow connected with the swinging pendulum?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    But this problem cannot have some kind of extremely complicated solution (elliptic integral) it appeared in my Calculus book for univeristies.

    Perhaps, I am thinking, this is somehow connected with the swinging pendulum?
    An elliptic integral is not complicated in principal, its just that in
    general it does not have a closed form representation in terms
    of elementary functions.

    Morris Klein's book I referred to is an undergraduate calculus
    text.

    And yes it is a rigid pendulum (being rigid allows the thing to be
    started with the weight above the pivot without it just falling
    straight down initially).

    RonL
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  7. #7
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    Do you know how to solve it?
    Because all my attempts failed.
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  8. #8
    Grand Panjandrum
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    There are two methods of obtaining the Differential Equation governing
    the motion of the bob of a (rigid)pendulum.

    (I will assume that the rod is "light" that is its mass is negligible compared
    to that of the bob)

    One is to just consider the energy of the bob. The total energy
    is:

    TE=PE+KE,

    where TE is the total energy, PE is the potential energy and KE
    is the kinetic energy.

    (another is to resolve the force due to gravity on the bob into components
    along the rod and perpendicularly to the rod, the perp component gives the
    torque which gives rise to angular acceleration, and the other component
    is cancelled by the tension/compression forces in the rod and the
    reaction at the pivot)

    Now we will describe the position of the bob by a single angle \theta (see the diagram).

    Take the reference height for the zero of potential energy to be the height
    of the bob when \theta=0, then:

    PE=m.g.(a-a\cos(\theta)),

    and:

    KE=\frac{1}{2}I\dot \theta ^2,

    where I=a^2m is the moment of inertia (assuming a light rod)

    So:

    m.g.a(1-\cos(\theta_0))=m.g.a(1-\cos(\theta))+\frac{1}{2}m.a^2.\dot \theta^2,

    where \theta_0 is the maximum angular amplitude. Rearranging:

     \dot \theta ^2 = \frac{2g}{a} (\cos (\theta)-\cos (\theta_0)) .

    Which is a DE of separable type which may be integrated to give:

    \pm \int_{\theta_0}^{\phi} \frac{1}{(\cos (\theta)-\cos (\theta_0))^{\frac{1}{2}}}d\theta=t_{\phi}\sqrt{ \frac{2g}{a} },

    where t_{\phi} is the time that the bob is a angle \phi
    (after having been released from \theta=\theta_0 at t=0).

    Now consider just the first swing the we want the signs in front
    of the integral to be -ve, and the bob reaches the bottom of its swing when \phi=0,
    so the time we are looking for is:

    \tau=\sqrt{ \frac{a}{2g} } \int_{0}^{\theta_0} \frac{1}{(\cos (\theta)-\cos (\theta_0))^{\frac{1}{2}}}d\theta.

    Now I think that the integral on the RHS diverges as \theta_0 \rightarrow \pi.

    RonL

    (apologies in advance for any errors that may have crept into my
    LaTeX).
    Attached Thumbnails Attached Thumbnails A physics problem-pendulum.png  
    Last edited by CaptainBlack; January 12th 2006 at 01:09 PM.
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  9. #9
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    Thank you
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