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: $$\frac{1}{3}ML^2$$
Can the equation $\frac{1}{3}ML^2 $ be used to calculate the moment of inertia in a non-uniform rod?
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2 Answers
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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 $I = \frac{1}{3}ML^2$.
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$\begingroup$ \begin{equation} \rho(x)= \left \{ \begin{array}{lll} \rho_1, x\in (0,6)~\cup~(7,10), \\ \rho_2, x\in (6,7). \end{array} \right. \end{equation} $\endgroup$– nwolijinCommented Dec 10, 2020 at 19:21
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$\begingroup$ @nwolijin Can I interpret that as taking three segments and finding their moment of inertia about their center of mass and then using the parallel axis theorem to a point about which the whole body rotates about? $\endgroup$– LinkinCommented Dec 10, 2020 at 19:28
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$\begingroup$ @DuckTales You can just do piecewise integration over the three segments. The integral from 0 to 10 is the sum of the integrals from 0 to 6, 6 to 7, and 7 to 10. The axis of rotation is the same for each segment, not simply parallel. $\endgroup$– G. SmithCommented Dec 10, 2020 at 19:47
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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 Dec 10, 2020 at 19:03
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$\begingroup$ (well of course for some special density distributions it would hold :)) $\endgroup$ Commented Dec 10, 2020 at 19:04
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