Discussion
A uniform stick of mass m and length l is pivoted about a horizontal axis through its lower end at O (as shown in the figure). Initially, it is held vertically and is allowed to fall freely downwards (under the action of gravity, with no air resistance). Its angular velocity at the instant when it makes an angle of 60° with the vertical is (given that the moment of inertial I = ml2/3 and g is acceleration due to gravity) :
Answer: A
The angular velocity is \(\sqrt{3g/2l}\).
Energy is conserved. The centre of mass, at \(l/2\) from O, falls by \(\frac{l}{2}(1 - \cos 60^{\circ}) = \frac{l}{4}\) as the stick turns through 60 degrees:
\[ \begin{aligned} mg\frac{l}{4} &= \frac{1}{2}\cdot\frac{ml^{2}}{3}\omega^{2} \\ \omega^{2} &= \frac{3g}{2l} \end{aligned} \]
\(\sqrt{2g/3l}\) inverts the factor \(\frac{3}{2}\), and \(\sqrt{2g/l}\) and \(\sqrt{g/2l}\) do not satisfy the energy equation for a 60 degree turn.
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