Question

A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60° 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° with the vertical.

Answer 1

Let the length of the rod be l.

Mass of the rod be m.

Let the angular velocity of the rod be ω when it makes an angle of 37° with the vertical.

On applying the law of conservation of energy, we get

\[\frac{1}{2}I \omega^2  - 0 = mg\frac{l}{2}\left( \cos37^\circ - \cos60^\circ \right)\]

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

\[ \Rightarrow  \omega^2  = \frac{9g}{10l}\]

Let the angular acceleration of the rod be α when it makes an angle of 37° with the vertical.

Using  \[\tau = I\alpha,\] we  get 

\[I\alpha = mg\frac{l}{2}\sin37^\circ\] 

\[ \Rightarrow \frac{m l^2}{3}\alpha = mg\frac{l}{2} \times \frac{3}{5}\] 

\[ \Rightarrow \alpha = 0 . 9\left( \frac{g}{l} \right)\]

Force on the particle of mass dm at the tip of the rod

\[F_c  =\text{ centrifugal force}\]

\[= \left( dm \right) \omega^2 l = \left( dm \right)\frac{9g}{10l}l\]

\[ \Rightarrow  F_c  = 0 . 9g\left( dm \right)\]

\[ F_t  =\text{ tangential force}\]

\[       = \left( dm \right)\alpha l\]

\[ \Rightarrow  F_t  = 0 . 9g\left( dm \right)\]

So, total force on the particle of mass dm at the tip of the rod will be the resultant of F_cand F_t.

\[\therefore   F = \sqrt{\left( {F_c}^2 + {F_t}^2 \right)}\]

\[= 0 . 9\sqrt{2}g\left( dm \right)\]