Question

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$\large\sf{Question \: :}$
A uniform rod of length L and mass M is held vertical, with its bottom and pivoted to the floor. The rod falls under gravity, freely turning about the pivot. If acceleration due to gravity is g. what is the instantaneous angular speed of the rod when it makes angle 60° with the vertical. ​

Answer 1

GIVEN :

• Length of rod = L.
• Mass is held with vertical and pivoted to the floor.

TO CALCULATE :

• what is the instantaneous angular speed of the rod when it makes angle 60° with the vertical.

$\Large\red {\sf Solution \: :}$

The fall of centre of gravity is given by :-

$\sf\dfrac{\big[\frac{L}{2}-h\big]}{\big[{\frac{L}{2}\big]}} = cos60^{ \circ}$

$\therefore$ Decrease in potential energy = $\sf Mgh \: = \: Mg \dfrac {L}{2} (1 - cos 60^{ \circ})$

Kinetic energy of rotation = $\sf \dfrac {1}{2} I \omega^{2} \: = \: \dfrac {1}{2} \: \times \dfrac {ML^{2}}{3} \omega^{2}$

Now,

According to the law of conservation of energy,

$\implies \sf Mg \dfrac{L}{2} (1 \: - \: cos 60^{ \circ})$

$\implies \sf \dfrac{ML^{2}}{6} \omega^{2}$

$\implies \sf \omega \: = \: \sqrt \dfrac {3g}{2L}$

$\large {\boxed {\sf \omega \: = \: \sqrt \dfrac {3g}{2L}}}$