A uniform rigid rod of mass M hinged at one end is released form rest in the position shown in the vertical plane. The magnitude of reaction force (in N) at hinge just after its release is ___
$(U s e M = 0.8 k g, θ = 45 °, g = 10 m / s^{2}, T a k e \sqrt{34} = 5.83)$

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answer is 5.8.

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Detailed Solution

$τ = 1 α m g \sin 45 ° \times \frac{l}{2} = \frac{m l^{2}}{3} α \Rightarrow α = \frac{3 g}{2 \sqrt{2} l}$

Tangential acceleration of centre of mass $(a_{t}) = α \times \frac{l}{2} = \frac{3 g}{4 \sqrt{2}}$

Radial acceleration of centre of mass $(a_{r}) = \frac{m ω^{2} l}{2} = 0 (i n i t i a l l y ω i s 0)$

$m g \sin 45 ° - N_{⊥} = m a_{t} = N_{⊥} = \frac{m g}{4 \sqrt{2}} - m g \cos 45 ° + N_{/ /} = m a_{r} \Rightarrow N_{/ /} = \frac{m g}{\sqrt{2}}$

Net hinge force $\sqrt{N_{⊥}^{2} + N_{/ /}^{2}} = m g \frac{\sqrt{17}}{4 \sqrt{2}} \Rightarrow \sqrt{34} = 5.83 N$

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