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Math Help - Physics meets Geometry

  1. #1
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    Physics meets Geometry

    I could not find a solution to the following problem:

    UNTAMED (I am walking 8 foxes at the top of my head and trying to keep their tails apart ! (*))

    I am the one who posted the problem. I vaugely remembering seeing something similar somewhere years ago. But just couldn't solve it.
    Has anyone seen anything like that?
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  2. #2
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    Quote Originally Posted by oswaldo View Post
    I could not find a solution to the following problem:

    UNTAMED (I am walking 8 foxes at the top of my head and trying to keep their tails apart ! (*)) [Thumbnail below.]

    I am the one who posted the problem. I vaguely remembering seeing something similar somewhere years ago. But just couldn't solve it.
    Has anyone seen anything like that?
    Potential energy lost = kinetic energy gained: mgr(1-\sin\theta) = \tfrac12mv^2.

    Radial component of weight just sufficient to provide centripetal acceleration: mg\sin\theta = mv^2/r.

    Put those two equations together to get \sin\theta = 2(1-\sin\theta), from which \sin\theta = 2/3 (independent of m, r and g !).
    Attached Thumbnails Attached Thumbnails Physics meets Geometry-8foxes.com_35.png  
    Last edited by Opalg; September 14th 2009 at 11:47 PM. Reason: (See following comments.)
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  3. #3
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    Good work! Thanks Opalg.

    I had the conservation of Energy part before. For the second equation I was thinking of equation of the tangent line at point A, but yours makes more sense.

    BUT I still have a question:
    "Radial component of weight just sufficient to provide centripetal acceleration
    mg\cos\theta = mv^2/r "

    Radial component is towards the center, so the equation should be:

    mg\sin\theta = mv^2/r

    which should give a different answer!!! But 3/5 makes more sense!

    Would you please clarify?
    Thanks.

    -O
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  4. #4
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    Quote Originally Posted by oswaldo View Post
    "Radial component of weight just sufficient to provide centripetal acceleration
    mg\cos\theta = mv^2/r "

    Radial component is towards the center, so the equation should be:

    mg\sin\theta = mv^2/r

    which should give a different answer!!!
    You're absolutely right! It should be \sin\theta, not \cos\theta. That makes the equation \sin\theta = 2(1-\sin\theta), with solution \sin\theta=2/3. I have edited my previous comment to correct it. (I think I was misled by the fact that θ is decreasing from π/2 towards 0, rather than increasing.)
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  5. #5
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    Thank you for the correction Opalg!
     \sin\theta = 2/3 gives:
     \theta is about 42 degrees.

    To me, this is intriguing. I would expect to object remain on course, say until 30 degrees. But apparently my intuition is wrong.

    One more details:  mg\sin\theta = mv^2/r
    holds only at point A, when the object is airborne. Not before nor after. Interesting!

    Having the answer not depending on m, g, r makes it more beautiful.

    Thanks for a great solution!

    -O
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