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 rod makes an angle of 37${}^{\circ }$ with the vertical.

• A. 0.9(dm)g
• B.
• C.
• D.

Answer 1

The correct option is C

Let I = length of the rod, and m = mass of the rod.

Applying energy principle

$\left(\frac{1}{2}\right)l{\omega }^2-O=mg\left(\frac{1}{2}\right)(cos{37}^{\circ }-cos{60}^{\circ })$

$\Rightarrow \frac{1}{2}\times \frac{m{l}^2}{3}{\omega }^2=mg\times \frac{1}{2}(\frac{4}{5}-\frac{1}{2})$

$\Rightarrow {\omega }^2=\frac{9g}{10l}=0.9\left(\frac{g}{l}\right)$

Again $\left(\frac{\frac{m}{2}}{3}\right)\propto =mg\left(\frac{1}{2}\right)sin{37}^{\circ }=\frac{mgl}{2}\times \frac{3}{5}$

$\therefore \alpha =0.9\left(\frac{g}{l}\right)=$ angular acceleration.

So, to find out the force on the particle at the tip of the rod

$F_1-$centrifugal force=(dm) $\omega 2l=0.9\left(dm\right)g$

$F_t-$tangential force=(dm) $\omega l=0.9\left(dm\right)g$

So, total force $F=\sqrt{(F_1^2+F_t^2)}=0.9\sqrt{2}\left(dm\right)g$