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  • A uniform stick of length '  ' is pivoted at one end on a horizontal table. The stick is free to rotate in the ve

A uniform stick of length ' L ' is pivoted at one end on a horizontal table. The stick is free to rotate in the vertical plane. A block of mass ' m ' is placed on the other end of the stick. What should be the angle of the stick from the horizontal so that block leaves the contact just after the release. [Given, g=10 \mathrm{~m} / \mathrm{s}^2 ]

Option: 1

\cos \theta=\sqrt{\frac{3}{2}} \\


Option: 2

\cos \theta=\sqrt{\frac{2}{3}} \\


Option: 3

\cos \theta=\frac{\sqrt{2}}{3} \\


Option: 4

\cos \theta=\frac{\sqrt{3}}{2 \sqrt{2}}


Answers (1)

best_answer

The block will loose the contact with the stick after the release if the acceleration of the stick in the vertical direction is more than 'g'.

\begin{aligned} \Psi&=m g\left(\frac{L}{2}\right) \cos \theta \\ \alpha&=\frac{m g\left(\frac{L}{2}\right) \cos \theta }{\frac{mL^2}{3} }=\frac{3 g}{2 L} \cos \theta \\ a_t&=L \alpha=\frac{3 g}{2} \cos \theta \\ a_t \cos \theta&=\frac{3 g}{2} \cos ^2 \theta \end{aligned}\\\\ For ~N=0\\\\ \begin{aligned} \frac{3 g}{2} \cos ^2 \theta & \geqslant g \\ \cos ^2 \theta & \geqslant 2 g \frac{2}{3} \\ \cos \theta & \geqslant \sqrt{\frac{2}{3}} \end{aligned}

Posted by

avinash.dongre

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