Question 529: Centre of Mass and Rotational Motion

Question

The moment of inertia of thin square plate PQRS of uniform thickness, about an axis passing through centre ' O ' and perpendicular to the plane of the plate is $\left(\mathrm{I}_1, \mathrm{I}_2, \mathrm{I}_3, \mathrm{I}_4\right.$ are respectively the moments of inertia about axis $1,2,3,4$ which are in the plane of the plate as shown in figure)

Accepted Answer

Axis of $I_1$ and $I_2$ and that of $I_3$ and $I_4$ are pependicular to each other.
By theorem of perpendicular axis,
$$\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2 \quad 	ext { or } \mathrm{I}=\mathrm{I}_3+\mathrm{I}_4\quad	ext{.... (i)}$$
As it is a square,
$$\mathrm{I}_1=\mathrm{I}_2 	ext { and } \mathrm{I}_3=\mathrm{I}_4$$
From (i),
$$\begin{array}{ll} 
& \mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{I}}{2} \
& \mathrm{I}_3=\mathrm{I}_4=\frac{\mathrm{I}}{2} \
	herefore \quad & \mathrm{I}_3=\mathrm{I}_1 \
	herefore \quad & 	ext { Momentof inertia of the plate }=\mathrm{I}=\mathrm{I}_1+\mathrm{I}_3
\end{array}$$

Answer

$\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3$

Answer

$\mathrm{I}_1+\mathrm{I}_3+\mathrm{I}_4$

Answer

$\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3+\mathrm{I}_4$

Answer

$ \mathrm{I}_1+\mathrm{I}_3$

Centre of Mass and Rotational Motion

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