Consider an "L-shaped rod" of uniform mass density, hinged at point 'O' is held at rest initially. The mass of the rod is 'm = 1kg'. The earth's gravitational field is assumed to be uniform and directed downwards. Assume no friction at hinges. The rod is now released. Find the maximum magnitude of angular speed of the rod (in rad/s). (Take : L = 1m, \sqrt{10} = 19/6)

Engineering

Physics

Moment of Inertia

Question

Consider an "L-shaped rod" of uniform mass density, hinged at point 'O' is held at rest initially. The mass of the rod is 'm = 1kg'. The earth's gravitational field is assumed to be uniform and directed downwards. Assume no friction at hinges. The rod is now released. Find the maximum magnitude of angular speed of the rod (in rad/s). (Take : L = 1m, $\sqrt{10}$ = 19/6)

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$I_{P} = \frac{1}{3} \frac{m}{2} L^{2} + [\frac{1}{12} \frac{m}{2} L^{2} + \frac{m}{2} (\frac{L^{2}}{4} + L^{2})] = \frac{5}{6} {mL}^{2}$

‘ω’ will be max. when com is at its lowest position.

∴ vertical distance moved by com = $\frac{\sqrt{10} L}{4} + \frac{L}{4}$

$PC = \frac{\sqrt{10} L}{4}$

$mg [\sqrt{10} + 1] \frac{L}{4} = \frac{1}{2} I_{P} ω^{2} = \frac{5}{12} {mL}^{2} ω^{2}$

$\Rightarrow ω^{2} = \frac{3 g}{5 L} (\sqrt{10} + 1) ∴ ω = 5 \frac{r}{s}$

Please subscribe our Youtube channel to unlock this solution.