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प्रश्न
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.
उत्तर
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 Fcand Ft.
\[\therefore F = \sqrt{\left( {F_c}^2 + {F_t}^2 \right)}\]
\[= 0 . 9\sqrt{2}g\left( dm \right)\]
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