Question

Theorem of perpendicular axis is used in obtaining the moment of inertia of a

Answer 1

Answer:

Square lamina.

Explanation:

This theorem states that the moment of inertia of a plane laminar about an axis perpendicular to its plane is equal to the sum of the moment of inertia of the lamina about two axes mutually perpendicular to each other in its plane and intersecting each other at the point where perpendicular axis passes through it.

Iz = Ixx + Iyy = I’x + I’y

Note: Only applicable to plane laminar bodies.

Answer 2

Answer:

The answer is square lamina

Given:

Theorem of perpendicular axis: According to this theorem, "the moment of inertia of a planar body (lamina) about an axis OZ Perpendicular to the pane of the lamina (O being a point in this lamina) is the sum of the moments of inertia about any two mutually perpendicular axes OX and OY, both lying in the same plane",

• Let $I_{z}$ = moment of inertia of the lamina about OZ axis.
• $I_{x}$ = moment of inertial about Ox axis.
• $I_{y}$= moment of inertia about OY axis
• Then, $I_{z} =l_{x} + l_{y}$

Proof : Consider a particle of mass m of the lamina at point p distant r from O. Let (x, y) be point co-ordinates of the point P.(Fig).

The moment of inertia of the particle about z-axis = mr2. Moment of inertia of the whole lamina about z-axis is,

$\begin{aligned} \mathrm{I}_{z} &=\sum m r^{2} \\ \text { But, } r^{2} &=x^{2}=y^{2} \end{aligned}$

Hence $I_{z}=\sum m\left(x^{2}+y^{2}\right)=\Sigma m x^{2}+m y^{2}$

Now, $\sum_{m} x^{2}=I_{x} and \sum m y^{2}=I_{y}$

Thus, $I_{z}=I_{x}+I_{y} .$

It is a square lamina