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    Hi guys.

    So i didn't have a clue how to approach part b of this question. Having consulted the solution bank i have seen and understand the logic behind their working out.

    However i have a few questions. What if the particle loses contact at a previous point on the chute but then comes back into contact with the chute before R? Would the solution banks method still work?

    Is it safe for me to assume that if there is a reaction at R then the particle never loses contact with the chute?



    Any input is appreciated.
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    (Original post by Shaanv)
    Hi guys.

    So i didn't have a clue how to approach part b of this question. Having consulted the solution bank i have seen and understand the logic behind their working out.

    However i have a few questions. What if the particle loses contact at a previous point on the chute but then comes back into contact with the chute before R? Would the solution banks method still work?

    Is it safe for me to assume that if there is a reaction at R then the particle never loses contact with the chute?



    Any input is appreciated.
    An interesting question.

    From the point of view of the given question, you're only interested in whether the particle leaves the surface at all. Doesn't matter whether it comes into contact with the surface again or not.

    And NO, it's not save to assume that as long as there is a reaction at R, then the particle never loses contact.

    The surface is part of a circle in outline. The line followed by the particle once it leaves the surface will be part of a parabola. Whether they will intersect again or not is going to depend on where the particle left the surface, it's velocity, and the curvature of the surface. It might be interesting to investigate....
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    (Original post by ghostwalker)
    An interesting question.

    From the point of view of the given question, you're only interested in whether the particle leaves the surface at all. Doesn't matter whether it comes into contact with the surface again or not.

    And NO, it's not save to assume that as long as there is a reaction at R, then the particle never loses contact.

    The surface is part of a circle in outline. The line followed by the particle once it leaves the surface will be part of a parabola. Whether they will intersect again or not is going to depend on where the particle left the surface, it's velocity, and the curvature of the surface. It might be interesting to investigate....
    Guess i will have to be extra cautious when approaching a question like this, thanks for the input
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    (Original post by ghostwalker)
    An interesting question.

    The surface is part of a circle in outline. The line followed by the particle once it leaves the surface will be part of a parabola. Whether they will intersect again or not is going to depend on where the particle left the surface, it's velocity, and the curvature of the surface. It might be interesting to investigate....
    The particle will lose contact with the circle when the component of mg acting towards the centre of rotation is no longer sufficient to maintain circular motion. This component can only reduce further as the particle falls freely under the influence of gravity, due to the changing angle above the horizontal. At the same time, the force needed to maintain circular motion at the original radius would increase with the increasing speed of the particle, so this is a double whammy - more radial force needed but less available. I think this means that once the particle has departed from circular motion is stays departed.

    Turning this round, I also think this also means that if the particle is in contact with the circle at point R, it has not previously departed from circular motion.

    (This is not a mathematical proof by any means, so I'm open to other thoughts).
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    (Original post by old_engineer)
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    Nice.
 
 
 
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