moment of inertia - 7

[logistic_guy]

Find the moment of inertia of a thin uniform rectangular plate of length $\displaystyle l$, width $\displaystyle w$, and mass $\displaystyle M$ when the rotation is through its center.

 

[khansaheb]

Find the moment of inertia of a thin uniform rectangular plate of length $\displaystyle l$, width $\displaystyle w$, and mass $\displaystyle M$ when the rotation is through its center.

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[logistic_guy]

$\displaystyle I_z = \int R^2 \ dm = \int (x^2 + y^2) \ dm = \sigma\int_{-l/2}^{l/2} \int_{-w/2}^{w/2} (x^2 + y^2) \ dx \ dy$

Double the integrals and start from zero. A trick that will live with you for a life time!

$\displaystyle I_z = 4\sigma\int_{0}^{l/2} \int_{0}^{w/2} (x^2 + y^2) \ dx \ dy = 4\sigma\int_{0}^{l/2} \left(\frac{x^3}{3} + xy^2\right)\bigg|_{0}^{w/2} \ dy$

Substitute.

$\displaystyle I_z = 4\sigma\int_{0}^{l/2} \left(\frac{w^3}{(8)3} + \frac{w}{2}y^2\right) \ dy = 4\sigma \left(\frac{w^3}{(8)3}y + \frac{w}{2}\frac{y^3}{3}\right)\bigg|_{0}^{l/2}$

Substitute.

$\displaystyle I_z = 4\sigma \left(\frac{w^3}{(8)3}\frac{l}{2} + \frac{w}{2}\frac{l^3}{3(8)}\right) = 4\frac{M}{lw}\left(\frac{w^3}{(8)3}\frac{l}{2} + \frac{w}{2}\frac{l^3}{3(8)}\right)$

Simplify.

$\displaystyle I_z = 4M\left(\frac{w^2}{48} + \frac{l^2}{48}\right) = M\left(\frac{w^2}{12} + \frac{l^2}{12}\right) = \textcolor{blue}{\frac{1}{12}M(w^2 + l^2)}$