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Rotational Mechanics — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsRotational Mechanics5 Marks
Question
A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of $60^\circ$ and then released. Find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes an angle of $37^\circ$ with the vertical.
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Answer


Let $l = $length of the rod, and $m =$ mass of the rod. Applying energy principle $\Big(\frac{1}{2}\Big)\text{l}\omega^2-0=\text{mg}\Big(\frac{1}{2}\Big)(\cos37^\circ-\cos60^\circ)$
$\Rightarrow\frac{1}{2}\times\frac{\text{ml}^2}{3}\omega^2$
$=\text{mg}\times\frac{1}{2}\Big(\frac{4}{5}-\frac{1}{2}\Big)\text{t}$
$\Rightarrow\omega^2=\frac{9\text{g}}{10\text{l}}=0.9\Big(\frac{\text{g}}{\text{l}}\Big)$
Again $\Big(\frac{\text{ml}^2}{3}\Big)\alpha=\text{mg}\Big(\frac{1}{2}\Big)\sin37^\circ=\text{mgl}\times\frac{3}{5}$
$\therefore\alpha=0.9\Big(\frac{\text{g}}{\text{l}}\Big)=$ angular acceleration.
So, to find out the force on the particle at the tip of the rod $F_i =$ centrifugal force $=(\text{dm})\omega^2\text{l}=0.9(\text{dm})\text{g} F_t =$ tangential force $=(\text{dm})\alpha\text{l}=0.9(\text{ dm})\text{g}$
So, total force $\text{F}=\sqrt{\big(\text{F}_{\text{i}}^2+\text{F}_{\text{t}}^2\big)}=0.9\sqrt2(\text{dm})\text{g}$

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