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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

A 2 kg steel rod of length 0.6 m is clamped ....

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 _ _ _ _ $m s^{-1}$.

Solution

Decrease in gravitational potential energy = Increase in kinetic energy

So, $mgl = \frac {1}{2} I \omega ^2 = \frac {1}{2} .\frac {1}{3} ml^2 \omega ^2 $

$\Rightarrow 6g = l \omega ^2 $

Using $v = l \omega $, we have

$6 g = l. \frac {v^2}{l^2}$

$\Rightarrow v^2 = 6gl $

$\Rightarrow v = \sqrt {6gl} = \sqrt {6\times 10 \times 0.6 } = 6ms^{-1} $

Answer: 6