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Math Help - Circular Motion

  1. #1
    Senior Member
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    Circular Motion

    i can do part a)
    and for b) i started with T-mgsin60-mw^2Lsin30=0

    but didnt end up with the answer.

    i got w^2 = (2-rt3)g/L
    instead of w^2 = 2(2-rt3)g/L

    thanks.Circular Motion-dsc_1043.jpg
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  2. #2
    MHF Contributor
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    Quote Originally Posted by BabyMilo View Post
    i can do part a)
    and for b) i started with T-mgsin60-mw^2Lsin30=0

    but didnt end up with the answer.

    i got w^2 = (2-rt3)g/L
    instead of w^2 = 2(2-rt3)g/L

    thanks.Click image for larger version. 

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    Firstly, note that T=mg.

    Then, draw the forces acting on the ball outside the cone(Reaction force, gravity, tension and the centripetal force).

    If the system is in equilibrium, R sin 30 + T cos 30 = mg

    0.5 R + mg (root3 /2)= mg

    R= (2- root(3)) mg

    For horizontal components, T sin 30 - R cos 30 = mrw^2

    Substitute T=mg and R.
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  3. #3
    Senior Member
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    Thanks for ur answer.
    but i want to understand where i went wrong with T-mgsin60-mw^2Lsin30=0.

    thanks
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