Discussion
A thin uniform rigid bar of length L and mass M is hinged at point O, located at a distance of \(\frac{L}{3}\) from one of its ends. The bar is further supported using springs, each of stiffness k, located at the two ends. A particle of mass \(m = \frac{M}{4}\) is fixed at one end of the bar, as shown in the figure. For small rotations of the bar about O, the natural frequency of the system is
Answer: B
The natural frequency is \(\sqrt{\frac{5k}{2M}}\).
Take moments about O for a small rotation \(\theta\). The bar's centre is L/6 from O, the left spring is L/3 from O, and the right spring and the particle are 2L/3 from O.
\[ \begin{aligned} I_{O} &= \tfrac{ML^{2}}{12} + M\left(\tfrac{L}{6}\right)^{2} + \tfrac{M}{4}\left(\tfrac{2L}{3}\right)^{2} = \tfrac{2ML^{2}}{9} \\ k_{\theta} &= k\left(\tfrac{L}{3}\right)^{2} + k\left(\tfrac{2L}{3}\right)^{2} = \tfrac{5kL^{2}}{9} \end{aligned} \]
\[ \begin{aligned} \omega_{n} &= \sqrt{\frac{k_{\theta}}{I_{O}}} \\ &= \sqrt{\frac{5k}{2M}} \end{aligned} \]
Leaving out the particle's inertia gives \(\sqrt{5k/M}\); the 3k forms come from miscounting the spring arms.
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