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Math Help - Circular Motion Question

  1. #1
    Super Member craig's Avatar
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    Circular Motion Question

    2 Particles, having mass of m and 2m respectively are connected by a light inextensible string, threaded through a fixed smooth ring. The lighter particle moves uniformly around a horizontal circle radius of radius r, while the other particle remains at rest.

    Find the speed of the lighter particle?

    No idea how start this, I've dealt with a bit of circular motion but not sure how to relate it to the mass of the particles?

    Thanks in advance for any help
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  2. #2
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    In order for the lighter particle to be moving in a circle, it has to be experiencing acceleration towards the center of the circle with magnitude mv^2/r. That acceleration is supplied by the heavier particle trying to respond to gravity, i.e., (2m)g. These 2 accelerations are equal.
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    Super Member craig's Avatar
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    Quote Originally Posted by qmech View Post
    In order for the lighter particle to be moving in a circle, it has to be experiencing acceleration towards the center of the circle with magnitude mv^2/r. That acceleration is supplied by the heavier particle trying to respond to gravity, i.e., (2m)g. These 2 accelerations are equal.
    Thank you, that's surprisingly easy...

    Accelerations are equal, therefore \frac{mv^2}{r} = 2mg

    Cancelling out the masses and multiplying through by r, we get v^2 = 2gr, so v = \sqrt{2gr}?

    Thanks again for the quick reply
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  4. #4
    Super Member craig's Avatar
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    Would the larger particle have an acceleration though if it remains stationary? Would have thought that this would be zero?
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