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Prove the Theorem of Perpendicular Axes Square of the Distance of a Point (X, Y) in The X–Y Plane from an Axis Through the Origin Perpendicular to the Plane - Physics

Prove the theorem of perpendicular axes.

(Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is x+ y2).

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Solution 1

The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body.

A physical body with centre O and a point mass m,in the xy plane at (xy) is shown in the following figure.

Moment of inertia about x-axis, Ix = mx2

Moment of inertia about y-axis, Iy = my2

Moment of inertia about z-axis, Iz = `m(sqrt(x^2 + y^2))^2`

Ix + Iy = mx2 + my2

= m(x2 + y2)

`= m(sqrt(x^2 + y^2))`

`I_x + I_y = I_z`

Hence the theorem is proved

Solution 2

The theorem of perpendicular axes: According to this theorem, the moment of inertia of a plane lamina (i.e., a two dimensional body of any shape/size) about any axis OZ perpendicular to the plane of the lamina is equal to sum of the moments of inertia of the lamina about any two mutually perpendicular axes OX and OY in the plane of lamina, meeting at a point where the given axis OZ passes through the lamina. Suppose at the point ‘R’ m{ particle is situated moment of inertia about Z axis of lamina

= moment of inertia of body about r-axis

= moment of inertia of the body about y-axis.

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 7 System of Particles and Rotational Motion
Q 26.1 | Page 180
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