A small mass m is attached to a point O by an inextensible string of length a. The mass is held with the string taut, at the same level as O, and released. Determine the angular velocity of the mass when the string makes an angle [tex]\theta[/tex]with the downwards vertical. From this relationship, determine the period of one complete swing. How does this time compare with the value of the period of a small pendulum of length a?(adsbygoogle = window.adsbygoogle || []).push({});

My solution is

[tex]mga = \frac{1}{2}m\omega^2a^2 + mga(1-\cos\theta)[\tex]

[tex]\omega^2 = \frac{2g\cos\theta}{a}[/tex]

But the answer says

[tex]\omega^2=\frac{2g\sin\theta}{a}[/tex]

For the second part on the period. Assuming the answer given is right.

[tex]T^2=\frac{4\pi^2}{\omega^2}[\tex]

[tex]T = \pi\sqrt\frac{2a}{g\sin\theta}[\tex]

But instead the answer says

[tex]T=7.04\sqrt\frac{a}{g}[/tex]

Can anyone please tell me what I've done wrong?

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# Homework Help: Pedulum Problem

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