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Math Help - Circular Motion - Mechanics Question

  1. #1
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    Circular Motion - Mechanics Question

    This question has been bugging me for ages.... I can't seem to draw the diagram correctly.

    A light inextensible string of length 3 metres is threaded through a smooth vertical ring which is free to turn. The spring carries a particle at each end. One particle, P, of mass m is at rest at a distance of 1 metre below the ring, while the other particle B of mass M is rotating in a horizontal circle whose centre is P.

    Show that the velocity, v, of B in terms of m, and M is given by:


    \sqrt {\frac{{(M+m)g }} {M}}
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shinn View Post
    This question has been bugging me for ages.... I can't seem to draw the diagram correctly.

    A light inextensible string of length 3 metres is threaded through a smooth vertical ring which is free to turn. The spring carries a particle at each end. One particle, P, of mass m is at rest at a distance of 1 metre below the ring, while the other particle B of mass M is rotating in a horizontal circle whose centre is P.

    Show that the velocity, v, of B in terms of m, and M is given by:


    \sqrt {\frac{{(M+m)g }} {M}}
    I apologize, my ability to visualize can be rather bad at times. Can you post a diagram of some kind? I'm just not seeing the physical setup here.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shinn View Post
    This question has been bugging me for ages.... I can't seem to draw the diagram correctly.

    A light inextensible string of length 3 metres is threaded through a smooth vertical ring which is free to turn. The spring carries a particle at each end. One particle, P, of mass m is at rest at a distance of 1 metre below the ring, while the other particle B of mass M is rotating in a horizontal circle whose centre is P.

    Show that the velocity, v, of B in terms of m, and M is given by:


    \sqrt {\frac{{(M+m)g }} {M}}
    Perhaps I still have the physical situation wrong. I'm getting
    v = \sqrt{2g \left ( \frac{m^2 - M^2}{Mm} \right ) }

    -Dan
    Attached Thumbnails Attached Thumbnails Circular Motion  - Mechanics Question-setup.jpg  
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