The two ends of a uniform thin rod of length \sqrt{2} R and of mass 2 \sqrt{2}kg can move without friction along a vertical circular path of radius R. The rod is released from the vertical position (ab). Find the force (in N) exerted by an end of the rod on the path when the rod passes the horizontal position(cd).

Engineering

Physics

Center of Mass

Question

The two ends of a uniform thin rod of length $\sqrt{2} R$ and of mass $2 \sqrt{2}$ kg can move without friction along a vertical circular path of radius R. The rod is released from the vertical position (ab).

Find the force (in N) exerted by an end of the rod on the path when the rod passes the horizontal position(cd).

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By energy conservation

$mg \frac{R}{\sqrt{2}} = \frac{1}{2} \frac{m (\sqrt{2} R)^{2} ω^{2}}{3}$

$\Rightarrow ω^{2} = \frac{3 g}{\sqrt{2} R}$

Now, 2Ncos 45° – mg = $m \times \frac{3 g}{\sqrt{2} R} \times \frac{R}{\sqrt{2}} \Rightarrow N = \frac{5 mg}{2 \sqrt{2}} = 50$

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