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Parallel Axis Theorem

ABCD is a squ...

Question

ABCD is a square plate with centre O. The moments of inertia of the plate about the perpendicular axis through O is $I_{0}$ and about the axes 1, 2, 3 and 4 are $I_{1}, I_{2}, I_{3}$ & $I_{4}$ respectively. It follows that :

I_{2} = I_{3}

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I_{0} = I_{1} + I_{4}

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I_{0} = I_{2} + I_{4}

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I_{1} = I_{3}

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Solution

The correct options are
A $I_{0} = I_{1} + I_{4}$
B $I_{1} = I_{3}$
C $I_{0} = I_{2} + I_{4}$
D $I_{2} = I_{3}$
According to Theorem of Perpendicular Axes, Moment of Inertia of the plate about an axis passing through the center O and perpendicular to the plane is,
$I_{0} = I_{1} + I_{2} = I_{3} + I_{4}$ ................(1) (Since, diagonals are mutually perpendicular to each other)
But, by symmetry rule,
$I_{1} = I_{2} = I$ (say) and $I_{3} = I_{4} = I^{'}$ (say)
Hence, $I_{0} = 2 I = 2 I^{'}$ ...................(2)
Now, for square plate,
$I = \frac{M L^{2}}{12}$
Therefore,
$I_{0} = 2 (\frac{M L^{2}}{12})$
$I_{0} = \frac{M L^{2}}{6}$
Hence, from (1),
$I_{1} + I_{2} = I_{3} + I_{4} = \frac{M L^{2}}{6}$
$I = I^{'} = \frac{M L^{2}}{12}$ ...............from(2)
$\Rightarrow I_{1} = I_{2} = I_{3} = I_{4} = \frac{M L^{2}}{12}$
$\Rightarrow I_{0} = I_{1} + I_{4}$
$\Rightarrow I_{0} = I_{2} + I_{4}$

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ABCD is a square plate with centre O. The moments of inertia of the plate about the perpendicular axis through O is I0 and about the axes 1, 2, 3 and 4 are I1,I2,I3 & I4 respectively. It follows that :

ABCD is a square plate with centre O. The moments of inertia of the plate about the perpendicular axis through O is $I_{0}$ and about the axes 1, 2, 3 and 4 are $I_{1}, I_{2}, I_{3}$ & $I_{4}$ respectively. It follows that :