JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 27)
A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity, ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is ____________ ms$$-$$1. (Take g = 10 ms$$-$$2)
Answer
6
Explanation
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by energy conservation $$mgl = {1 \over 2}I{\omega ^2} = {1 \over 2}{{m{l^2}{\omega ^2}} \over 3}$$
$$ \Rightarrow \omega = \sqrt {{{6g} \over l}} $$
As we know the relation between the linear speed and angular speed,
$$v = \omega r = \omega l = \sqrt {6gl} $$
$$v = \sqrt {6 \times 10 \times .6} $$ = 6 m/s
Hence, the speed of the free end of the rod when it passes through its lowest position is 6 m/s.
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