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m3 question exam tomorow help! watch

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    hey exams tomorow and i just cant solve this question of an old m2 past paper!

    ive attached my very crap representation of the diagram included with the q.

    Two small spheres, A and B, of mass 0.8kg and 0.1kg respectively are attached to the ends of a light inextensible string which passes through a small smooth ring at the fixed point O. The ring is free to turn about a vertical axis. Sphere A describes horizontal circles with a constant angular speed of w rad s^-1. The portion of the string between O and A makes a constant angle (PHETA) with the vertical, and the portion OB is vertical. Sphere B remains in equilibrium.

    a) find the value of cos (PHETA)

    if someone could help me with this first part of the question i would love it!

    i resolved vertically and get Tcos(PHETA)=1.8g
    but no idea what to do from there
    and resolving horizontally is useless because it brings the other unknowns that you gota solve later..

    thanks a LOT!
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    i know mechanics is craaaaaap but pwetty pwease :tsr:
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    The masses of A and B are wrong. Sphere A can't be heavier than Sphere B.

    Let mA and mB be the real masses of A and B.

    Resolving vertically at A, T cos(theta) = mA g.
    Resolving vertically at B, T = mB g.

    So cos(theta) = mA g / (mB g) = mA/mB
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    sphere A is heavier than B, i just double checked, the sphere A is moving in horizontal circles which is providing some of the extra tension needed to keep the sphere B hanging in equilibrium.
    anyone please?
 
 
 
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Updated: January 12, 2005

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