A uniform rod of mass $m=2 \sqrt{10} \mathrm{~kg}$ and length $l$ can rotate in vertical plane about a smooth horizontal axis hinged at point $H$. Angular acceleration $a$ of the rod just after it released from initial position making an angle of $37^{\circ}$ with horizontal from rest.…

Question

A uniform rod of mass $m=2 \sqrt{10} \mathrm{~kg}$ and length $l$ can rotate in vertical plane about a smooth horizontal axis hinged at point $H$. Angular acceleration $a$ of the rod just after it released from initial position making an angle of $37^{\circ}$ with horizontal from rest. Find force (in $\mathrm{N}$) exerted by the hinge just after the rod is released from rest.

Aicharya · Accepted Answer

Torque about hing $=\tau H=I \alpha$ $M g \frac{\cos 37^{\circ} l}{2}=\frac{m l^{2}}{3} \alpha$ $\alpha 6 g / 5 l$ $\begin{array}{l} a_{t}=\alpha \frac{l}{2}=\frac{3 g}{5} \ M g \cos 37^{\circ}-N-1=m a_{1} \ N_{1}=\frac{m g}{5} \end{array}$ Angular velocity of rod is zero so $N-2=m g \sin 37^{\circ}=3 m g / 5$ $N=\sqrt{N-1^{2}+N_{2}^{2}}=\sqrt{\left(\frac{m g}{5}\right)^{2}+\left(\frac{3 m g}{5}\right)^{2}}=\frac{m g \sqrt{10}}{5}$

A uniform rod of mass \(m=2 \sqrt{10} \mathrm{~kg}\) and length \(l\) can rotate in vertical plane about a…

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