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    Mechanics problem, TeeEm style.

    A mass m is placed on the surface of a smooth hemisphere, radius R, at a height R \frac{\sqrt{3}}{2} from the base of the hemisphere and is given an initial horizontal velocity of U, tangential to the sphere.

    Find an expression for the normal reaction N(\theta) of the hemisphere on the mass when it has fallen to a height R\cos\theta, where \theta is the polar angle measured from the vertical. (So that \theta=0 corresponds to the top of the hemisphere)

    (I think that I've got the correct result for this - it was messier than I'd anticipated)
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    :afraid:
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    (Original post by atsruser)
    Mechanics problem, TeeEm style.

    A mass m is placed on the surface of a smooth hemisphere, radius R, at a height R \frac{\sqrt{3}}{2} from the base of the hemisphere and is given an initial horizontal velocity of U.

    Find an expression for the normal reaction N(\theta) of the hemisphere on the mass when it has fallen to a height R\cos\theta, where \theta is the polar angle measured from the vertical. (So that \theta=0 corresponds to the top of the hemisphere)

    (I think that I've got the correct result for this - it was messier than I'd anticipated)
    sorry teaching today
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    (Original post by TeeEm)
    sorry teaching today
    I'm happy to wait.
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    (Original post by atsruser)
    I'm happy to wait.
    He likes his excuses
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    (Original post by atsruser)
    Mechanics problem, TeeEm style.

    A mass m is placed on the surface of a smooth hemisphere, radius R, at a height R \frac{\sqrt{3}}{2} from the base of the hemisphere and is given an initial horizontal velocity of U.

    Find an expression for the normal reaction N(\theta) of the hemisphere on the mass when it has fallen to a height R\cos\theta, where \theta is the polar angle measured from the vertical. (So that \theta=0 corresponds to the top of the hemisphere)

    (I think that I've got the correct result for this - it was messier than I'd anticipated)
    I got

    N = mU2/R +mg( √3-cosθ )
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    (Original post by TeeEm)
    I got

    N = mU2/R +mg( √3-cosθ )
    Correct
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    (Original post by GeologyMaths)
    Correct
    reassuring ....
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    (Original post by GeologyMaths)
    He likes his excuses
    He didn't wimp out - Teeem's answer has been provided.
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    (Original post by TeeEm)
    I got

    N = mU2/R +mg( √3-cosθ )
    Oh dear. I get something quite different:

    \displaystyle N = 2mg \cos\theta -\frac{mg\sqrt{3}}{2}+\frac{mU^2}  {R}(\frac{3}{4\sin^2\theta}-1)
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    (Original post by atsruser)
    Oh dear. I get something quite different:

    \displaystyle N = 2mg \cos\theta -\frac{mg\sqrt{3}}{2}+\frac{mU^2}  {R}(\frac{3}{4\sin^2\theta}-1)
    we may be solving a very different problem ...

    what I see being described here is a very simple problem and I suspect you mean something a lot more complicated which I interpreted in a simplistic way
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    (Original post by TeeEm)
    we may be solving a very different problem ...

    what I see being described here is a very simple problem and I suspect you mean something a lot more complicated which I interpreted in a simplistic way
    Did you notice that the initial velocity is horizontal i.e. initially parallel to the base?
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    (Original post by atsruser)
    Did you notice that the initial velocity is horizontal i.e. initially parallel to the base?
    I did not ...
    then it leaves the surface?
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    (Original post by TeeEm)
    I did not ...
    then it leaves the surface?
    Not sure that I follow. It will leave the surface for a sufficiently large value of U, I guess, otherwise it'll spiral round the sphere until it falls off, no?
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    (Original post by atsruser)
    Not sure that I follow. It will leave the surface for a sufficiently large value of U, I guess, otherwise it'll spiral round the sphere until it falls off, no?

    Now I see what you are describing
    The velocity is horizontal and tangential.
    will try it again if not tonight tomorrow when I am off
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    (Original post by TeeEm)
    Now I see what you are describing
    The velocity is horizontal and tangential.
    will try it again if not tonight tomorrow when I am off
    Right. That is a key point that I didn't make clear - it is also tangential. [I've edited the original post]
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    (Original post by atsruser)
    Right. That is a key point that I didn't make clear - it is also tangential. [I've edited the original post]
    I spent over 2 hours this afternoon
    I feel I got something (I can post my attempt) but the parameterization or connection between z and r is a total mess
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    (Original post by TeeEm)
    I spent over 2 hours this afternoon
    I feel I got something (I can post my attempt) but the parameterization or connection between z and r is a total mess
    Sure, put it up. I'll post my attempt at a solution tomorrow - I took a Lagrangian approach again.
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    (Original post by atsruser)
    Not sure that I follow. It will leave the surface for a sufficiently large value of U, I guess, otherwise it'll spiral round the sphere until it falls off, no?
    Someone *really* likes this thread...
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    (Original post by atsruser)
    Sure, put it up. I'll post my attempt at a solution tomorrow - I took a Lagrangian approach again.
    Yeah, OK, Gnomes&Knights, I get the message - I'll put up my solution when I get a chance to make an electronic copy.
 
 
 
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