moment of inertia - 8

[logistic_guy]

Find the moment of inertia of a long uniform rod of length $\displaystyle l$ and mass $\displaystyle M$ when the rotation is through its center.

 

[khansaheb]

Find the moment of inertia of a long uniform rod of length $\displaystyle l$ and mass $\displaystyle M$ when the rotation is through its center.

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[logistic_guy]

Let us assume that we placed the rod on the $\displaystyle x$-axis on the $\displaystyle x$-$\displaystyle y$ plane with its center at the origin. We will rotate it around the $\displaystyle y$-axis.

$\displaystyle I_y = \int R^2 \ dm = \int x^2 \ dm = \int_{-l/2}^{l/2} x^2 \lambda \ dx$

Because of symmetry, we can double the integral and start from zero to make the calculations easier.

$\displaystyle I_y = 2\int_{0}^{l/2} x^2 \lambda \ dx = 2\lambda \frac{x^3}{3}\bigg|_{0}^{l/2} = 2\lambda\frac{l^3}{(8)3}$

We know that the length density $\displaystyle \lambda = \frac{M}{l}$, then

$\displaystyle I_y = 2\frac{M}{l}\frac{l^3}{(8)3} = \textcolor{blue}{\frac{1}{12}Ml^2}$