A 2 kg steel rod of length 0.6 m is clamped on a table vertically at… | JEE Main 2021 (1 Sep, Shift 2)

Q41JEE Main 2021 · 1 SEP · SHIFT 2Medium

Rotational Kinetic Energy, Work & Power

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)

Numerical answer type. Work it out, then open the answer.

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Answer: 6

From vertically up to vertically down, the centre of mass falls by $L = 0.6$ m.
$mgL = \frac{1}{2}\cdot\frac{mL^2}{3}\omega^2 \Rightarrow \omega^2 = \frac{6g}{L} = \frac{60}{0.6} = 100$ , $\omega = 10$ rad/s
Speed of free end: $v = \omega L = 10 \times 0.6 = 6$ m/s
So the answer is 6 m/s.

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