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    (Original post by Zacken)
    I really hope this isn't from an IAL paper, I'm saving those for mocks just after the new year.

    Resolving horizontally: \displaystyle T \sin \frac{\pi}{6} - N \sin \frac{\pi}{6} = m\omega^2 r \iff N = T - 2m\omega^2 r

    Resolving vertically: \displaystyle T \cos \frac{\pi}{6} + N \cos \frac{\pi}{6} = mg \iff T + N = \frac{2mg}{\sqrt{3}}

    Solving simultaneously yields (omitting the obvious algebra) \displaystyle T = \frac{mg}{\sqrt{3}} + m\omega^2 r

    And some trigonometry shows that \displaystyle \cos \frac{\pi}{3} = \frac{r}{\ell}

    So our final answer is \boxed{T = \frac{mg}{\sqrt{3}} + \frac{m\omega^2 \ell}{2}}

    Solving for the normal reaction gives us \displaystyle N = \frac{mg}{\sqrt{3}} - m \omega^2 r > 0 \iff \frac{g}{\sqrt{3}} - \frac{\omega^2 \ell}{2} > 0. So \displaystyle w < \sqrt{\frac{2g}{\ell \sqrt{3}}}. Using the famous \omega = \frac{2\pi}{T_p} yields that \displaystyle \boxed{T_p > 2\pi \sqrt{\frac{\ell \sqrt{3}}{2g}}} as required.

    Nice question, though. Cheers.
    Don't worry, it isn't from an IAL paper, haha.

    Very elegant solution. I find it interesting that you converted the angles to radians.
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    (Original post by aymanzayedmannan)
    Don't worry, it isn't from an IAL paper, haha.

    Very elegant solution. I find it interesting that you converted the angles to radians.
    Oh, no - I did it in degrees on paper (since the question used degrees), but I couldn't be bothered latexing the x^\circ on here, plus radians just look prettier.
 
 
 
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