Cycling Equipment · Public discussion

Re: Ground Impact Speed?

Started by Joe Riel · · Last activity · 2 posts · 357 views

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Cycling Equipment
Published
8 July 2005
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9 July 2005
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Joe Riel
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  1. [email hidden] writes:

    The time it takes for the rod to topple is indeterminate. Imagine
    that it were precisely balanced on its end---the duration would be
    infinite. So timing the toppling is useless. The relevant measure
    is the velocity at impact. In this case, the velocity at the tip
    at impact. That is readily computed from conservation of energy.

    Assume a uniform rod of length l, mass m. When standing on its end,
    its potential energy is m*g*l/2. Assume that as it topples, its bottom
    end does not slide (that is, acts as a pivot). When the tip strikes
    the ground, all the potential energy has been converted to kinetic
    energy. So

    m*g*l/2 = 1/2*I*w^2

    where

    I = moment of inertia of rod, about one end = m*l^2/3
    w = angular velocity of rod at impact

    The velocity at the tip is

    Vtip = w*l

    so

    Vtip = sqrt(3*g*l).

    The same rod held horizontally and dropped from a height of l
    has an impact velocity of

    V = sqrt(2*g*l).

    To achieve the same velocity as the toppling rod,
    the horizontal rod must start at a height of 3/2*l.

    Joe

  2. Joe Riel said:

    [email hidden] writes:

    The time it takes for the rod to topple is indeterminate. Imagine
    that it were precisely balanced on its end---the duration would be
    infinite. So timing the toppling is useless. The relevant measure
    is the velocity at impact. In this case, the velocity at the tip
    at impact. That is readily computed from conservation of energy.

    Assume a uniform rod of length l, mass m. When standing on its end,
    its potential energy is m*g*l/2. Assume that as it topples, its bottom
    end does not slide (that is, acts as a pivot). When the tip strikes
    the ground, all the potential energy has been converted to kinetic
    energy. So

    m*g*l/2 = 1/2*I*w^2

    where

    I = moment of inertia of rod, about one end = m*l^2/3
    w = angular velocity of rod at impact

    The velocity at the tip is

    Vtip = w*l

    so

    Vtip = sqrt(3*g*l).

    The same rod held horizontally and dropped from a height of l
    has an impact velocity of

    V = sqrt(2*g*l).

    To achieve the same velocity as the toppling rod,
    the horizontal rod must start at a height of 3/2*l.

    Joe

    Dear Joe,

    So the speed-at-impact for the idealized toppling beam is
    simpler than I expected? (As long as someone like you works
    it out for me.)

    And I had it backward? The toppling rod's impact speed when
    it reaches the ground is always faster than than the impact
    speed for free-fall?

    That is, a horizontal rod one meter long reaches the ground
    at 17.8 m/s when dropped from 1.5 meters:

    http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

    So the tip of the same 1-meter rod just toppling over would
    hit the ground at the same 17.8 m/s, even though it started
    out only 1.0 meters off the ground, not 1.5 meters?

    Sorry if I'm missing the obvious or confusing things.

    Thanks,

    Carl Fogel

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