Question

A uniform rod pivoted at its upped 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.

Answer 1

Let l = length of the rod and m = mass of the rod.

Applying energy principle,

$\left(\frac{1}{2}\right)I{\omega }^2-0=mg\frac{1}{2}(cos{37}^{\circ }-cos{60}^{\circ })$

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

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

$\text{Again,}m\alpha =mg\left(1.2sin{37}^{\circ }\right)$

= $mg\left(1.2\right)\times \frac{3}{5}$

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

= angular acceleration

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

$F_1$ = centrifugal force

= $\left(dm\right){\omega }^{2}l=0.9\left(dm\right)g$

$F_t$ = (dm) $\alpha l$

= 0.9 (dm)g So, Total force,

F = $\sqrt{(F_1^2+F{t}^2)}$

$=0.9\sqrt{2}\left(dm\right)g$