Cycling Equipment · Public discussion

Re: Ground Impact Speed?

Started by Dave Lehnen · · Last activity · 1 post · 375 views

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Cycling Equipment
Published
8 July 2005
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8 July 2005
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Dave Lehnen
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  1. [email hidden] wrote:
    < extensive snippage >

    Quoted message said:

    I think that the end of a rigid, uniform 62"-long object
    topples over and whomps into the ground significantly more
    slowly than the same object held horizontal and dropped from
    62", but I'm hoping that someone either has a simple
    equation that I can follow, a more complicated equation that
    I can't follow but will take on faith, or even an
    explanation that the end of the toppling beam doesn't go
    more slowly than it would in free fall.

    . . - - - .
    | / . - - - .
    |/_ _ . - - - .

    That is, how fast does the dot hit the ground in the two
    figures above if it starts out 62" above the ground? I think
    that it's going slower when it topples, but I want
    reassurance.

    Along similar lines, does anyone know of a sneaky
    demonstration that would slow these two paths down enough to
    make them clear, something like Galileo's trick of rolling
    balls down inclines to slow the effect of gravity and show
    that large and small balls accelerate at the same rate?

    For anyone interested, here's a page with a miniature
    falling chimney (two sections topple at different speeds)
    and the short-armed cup and ball trick (a short beam
    accelerates faster):

    http://www.physics.umd.edu/lecdem/outreach/QOTW/arch3/a043.htm

    Thanks,

    Carl Fogel

    There is no simple answer for the time it takes a toppling pole
    to topple, since it depends on how big the initial imbalance
    or disturbance was that caused it to fall. If the pole was very
    well balanced, and undisturbed, it could take an arbitrarily
    long time to fall, even though unstable.

    If the disturbance that makes it fall is small enough not to
    impart significant kinetic energy to the pole, it is fairly
    straightforward to compute its speed of groundwhompage, at least
    in the case of pivoting without its base slipping, and the usual
    ignoring of air resistance. The potential energy of the center
    of gravity of the pole falling to ground level is converted to
    kinetic energy of the pole rotating about its pivot point.
    Knowing the change in potential energy, and the moment of inertia
    of the pole about its pivot, you can solve for its angular velocity,
    and its speed of whompage of any point on the pole.

    Dave Lehnen

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