Question

Find the moment of inertia of a uniform square plate of mass m and edge a about one of its diagonals.

Answer 1

Answer:

$i = \frac{ {ma}^{2} }{12}$

Explanation:

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Answer 2

The moment of inertia of a uniform square plate is given by, $2 I=\frac{m a^{2}}{6}$  .

Explanation:

Step 1 :

Assume that in a specific cross sectional area the distance is x.

Now it would be mass of that cross-sectional area,

                           $\frac{m}{a^{2}} \times a x d x$

Step 2 :

Hence the moment of this axis inertia,

                        $\begin{aligned}&I=\int_{0}^{\frac{a}{2}} \frac{m}{a^{2}} \times a x d x \times x^{2}\\&I=\left[2 \times \frac{m}{a} \times \frac{x^{3}}{3}\right]^{\frac{a}{2}}\\&I=\frac{m a^{2}}{12}\end{aligned}$

Step 3 :

Now for the other side in the same way,

                        $\begin{aligned}&\mathrm{I}^{\prime}=2 \times \frac{\mathrm{ma}^{2}}{12}\\&I^{\prime}=\frac{m a^{2}}{6}\end{aligned}$

And by the theorem of the perpendicular pole, the moment of inertia is as follows :    

                          $I+I=I^{'}$      

                         $2 I=\frac{m a^{2}}{6}$      

Therefore, the moment of inertia is, $2 I=\frac{m a^{2}}{6}$  .