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This problem can be solved using the centre of mass. We can just calculate the change of the potential energy and that'd be the work done. That way, $W = MgL/2$ (since the cm is at $L/2$). I understood this method. But then, when I approached this problem with torque, I got a different answer. At the time of raising it from the end, the tangential component of gravity will be $mg\cos\theta$ (angle with the horizontal). So, the torque will be $mg\cos\theta\cdot L\cdot\sin90$. Integrating, we can find that the work done is $MgL$.

What went wrong with this procedure?

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You seem to be mixing up the torque applied by you and the torque applied by gravity. You are right that the tangential component of gravity is $mg\cos\theta$, but remember that the force of gravity can be taken to act at the COM, so your torque by gravity is $\frac{mgL}{2}\cos\theta$. From here I will leave it to you to determine how this will match your first method.

Now, it also seemed like you were trying to look at the force you apply, but this is much more difficult to do in general. This is because, if you want the work you do to be equal to the change in potential energy, then it must be that the rod starts and stops at rest, since $$\Delta K=W_{net}=W_{me}+W_{grav}=W_{me}-\Delta U=0$$ This means that your applied torque has to have a time dependence in order to accelerate and decelerate the rod, so your work integral becomes more complex than the one above.

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    $\begingroup$ So, W(me) = -W(gravity). Since the force of gravity is working at the CM, so the moment arm is L/2. So we end up with W(gravity) = - MgL/2 which results in W(me) = MgL/2. Is this alright? $\endgroup$ – Tahsin Choudhury Nov 6 '18 at 6:24

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