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Consider the case here,

A uniform rod of mass $m$ and length $l$ is hanging vertically from the pivot $O$. A horizontal force $F$ acts at the lower end of the rod. If $F$ always remains horizontal then what is the maximum angular displacement of the rod? The image below show the rod's initial and final state

Rod hanging

The rod is a rigid body hence it will be at rotational equilibrium when the net external torque is zero

Hence,

$$\tau=F\ell\cos\theta - mg\frac{\ell}{2}\sin\theta = 0 $$ $$ \theta_{max} = \tan^{-1}\frac{2F}{mg}$$

The above solution is the one that i thought first, but the actual solution is twice that of $\theta_{max}$ which is obtained using work energy theorem,

The change in kinetic energy when the rod moves from it's initial to final state is zero hence the work done by total torque must be zero

Hence,

$$W_{ext} = \int_0^{\theta_{max}}\tau\,\mathrm d\theta = 0$$

By solving this integral we find that, $$ \theta_{max} = 2\tan^{-1}\frac{2F}{mg}$$

Obviously there is something fundamentally wrong with the first procedure, But if we think about a slightly different scenario where we keep this rod in a constant electric field(from left to right) and place a test charge at the bottom then too it seems intuitive that first procedure would give the correct result.

So my question is what is the difference between those two procedures and what is fundamental idea hidden in this problem...hope the community will give some intuitive way to understand what had gone wrong here, Thanks!

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When it will be in equilibrium, it will have some angular velocity and due to inertia it will continue to rotate till the work done is completely converted to potential energy.

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    $\begingroup$ yeah i just got it..the angular acceleration of the rod becomes zero at half the maximum but that does'nt mean that the angular velocity is zero....indeed the solution is simple but i was quite mad at it $\endgroup$ – user243016 Jul 15 '20 at 10:43

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