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A rod of length L is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod, when it is in vertical position, is

(A) √(2g/L)

(B) √(3g/L)

(C) √(g/2L)

(D) √(g/L)

This is what I understood from the question:

rotating rod

I consider the torque being applied by the force of gravity at the centre of mass of the rod (which is also the centre of gravity in this case). The angular displacement is $π/2$, which means the work done should be equal to $\tau\Delta\theta$ (torque multiplied by the angular displacement). Applying the work-energy theorem, work done equals the change in kinetic energy, so I write this:

$$\tau\Delta\theta=\frac12 I \omega^2$$

But I don't get the correct answer, and when I check the solution, this is what I find:

$$\frac{mgl}{2}=\frac 12 I \omega^2$$

By which it is meant that the loss in potential energy equals the gain in kinetic energy, which is technically correct. But is the formula that I've used wrong in some way? Is the work-energy theorem not supposed to be used here or am I missing something? There's no translational motion here, so what am I missing?

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closed as off-topic by John Rennie, stafusa, Kyle Kanos, sammy gerbil, Pulsar May 6 '18 at 21:34

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Work energy theorem, by definition of work and energy, is always applicable. What you're doing wrong is the work part. The work is not simply $\frac{\tau\pi}{2}$ Because the torque is changing every moment, (the moment arm is changing). So if you take that into account and integratethe entire work, you will get it right But instead you can also use energy conservation in the presence of conservative fields, in this case gravity. So total change in kinetic energy = - total change in potential energy

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