The question says "A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end (Figure is attached below). The rod is released from rest in the horizontal direction" (a) What is the angular speed of the block when it reaches its lowest position

I have tried the problem assuming the total mass to be concentrated at the center of mass so and the reduce in potential energy would give the increase in rotational energy $\frac{MgL}{2}$ must be equal to $\frac{Iω^2}{2}$ , but as I assumed the total mass is to be at center of mass which makes I as $\frac{ML^2}{4}$ but am ending with an incorrect solution. The textbook solution has taken moment of inertia of the straight rod as have assumed to reduce in potential energy of Centre of mass. When taking potential energy of COM then why is moment of inertia of rod considered than considering moment of inertia of COM with respect to the axis?

Please suggest me why am I wrong?

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1 Answer 1


I assumed the total mass is to be at center of mass which makes I as M(L/2)^2 but am ending with an incorrect solution.

Its very incorrect way of finding moment of inertia. Moment of inertia is found by:

$$I=\int r^2dm$$

So $I$ depends on the distribution of mass from the axis, and not the center of mass.

The standard value of $I$ for a rod (treated as 1 dimensional object), about center of mass is: $I=ml^2/12$. I am leaving you a link, where you can see the derivations of $I$ for standard objects:Link (Make sure you know basic calculus for derivations)

Here, the rod is hinged from one end, so by Parallel axis theorem, you can find $I$, about one end of rod ($I=ml^2/3$). Note that always moment of inertia is calculated about the axis about which rotation is taking place

Then you can proceed with your question.

Hope it helps!

  • $\begingroup$ I don't think u need to divide by integration dm in the denominator. Also how about gravitational potential energy for an extended object. Is it defined for as the mgh where h is the displacement of COM? $\endgroup$
    – user318937
    Commented Nov 11, 2021 at 5:35
  • 1
    $\begingroup$ My bad, I corrected the equation now. Yes in a uniform gravitational field, centre of gravity is same as centre of mass, so you can proceed with your idea. but in a non uniform gravitational field you would have to find centre of gravity manually. Objects near earth is treated to experience unifrom downward gravitational field. $\endgroup$ Commented Nov 11, 2021 at 7:16

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