A short problem.

Classical Mechanics Level 3

In the figure a stick of length $L = 1.85 m$ oscillates as a physical pendulum.What value of distance (in $m$ ) $x$ between the stick's centre of mass and its pivot point $O$ gives the least period

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The answer is 0.53.

1 solution

Purushottam Abhisheikh
Jan 31, 2015

Total torque acting on the rod about pivot must be equal to $I_{p}\alpha$ .

$\implies$ $mgxsin\theta =I_{p}\alpha$

Since this is a restoring torque so $-mgxsin\theta =I_{p}\alpha$

$\implies$ $\alpha = \frac{-mgxsin\theta}{I_{p}}$

Since for very small angular displacment $sin\theta\approx\theta$

So, $\alpha = \frac{-mgx\theta}{I_{p}}$

This is an equation of SHM whose angular frequency $w = \sqrt{\frac{mgx}{I_{p}}}$

Therefore its time period $T = \frac{2\pi}{w} = 2\pi \sqrt{\frac{I_{p}}{mgx}}$

$= 2\pi \sqrt{\frac{\frac{mL^2}{12} + {mx^2}}{mgx}}$

$= 2\pi\sqrt{\frac{L^2}{12gx} + \frac{x}{g}}$

Since A.M. is greater than or equal to G.M. and equality holds if and only if all the terms are equal. So for time period to be minimum $\frac{L^2}{12gx} = \frac{x}{g}$

$\implies$ $x = \boxed{\frac{L}{2\sqrt{3}}\approx 0.53m}$

I have a doubt . Shouldn't we also consider the torque due to the portion of rod above the pivot point?

Siddharth Yadav - 4 years, 2 months ago

If you consider the whole rod as a body with uniform density you only can considerate the torque due the center of mass. The case you're thinking on is when you considerate the upper and lour parts of the rod separetely, then you consider the center of mass of the lour part and the upper part and the result is right the same

Hjalmar Orellana Soto - 3 years, 3 months ago

A short problem.

The answer is 0.53.

1 solution

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