Physics Asked by DuckTales on December 14, 2020
I am wondering whether the moment of inertia about a pivot at one of its ends of a non-uniform rod can be calculated using the following equation:
$$frac{1}{3}ML^2$$
No. When deriving this equation from the moment of inertia definition, the density $rho$ is assumed to be uniform. If the density is not uniform, the $dfrac{1}{3}ML^2$ equation does not hold.
You would have to use $$I = int r^2 dm$$ equation to find the moment of inertia of your non uniformly dense rod.
EDIT: as noted by user @Andrew Steane, for some 'special' density distributions it would hold.
But as a general rule, no (except in certain cases).
Answered by user256872 on December 14, 2020
No you cannot use this formula. Imagine that all mass of the rod is concentrated at one end, the the moment of inertia will be either $0$ or $M L^2$. The correct formula for the moment of inertia in your case is begin{equation} I = int_0^L dx rho(x) x^2, end{equation} where $rho(x)$ is the linear mass density of the rod. For a uniform rod $rho(x)=frac{M}{L}$, and then integration leads to $I = frac{1}{3}ML^2$.
Answered by nwolijin on December 14, 2020
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