Question

Give the statement and proof of theorem of perpendicular axes relating moment of inertia.

Answer 1

Suppose, we have a circular ring of radius R and mass M. We have to find its moment of inertia about axis YY' which is normal to the plane of the ring and passes through the centre O. This axis is also known as the geometric axis of the ring.

Circumference of the ring = 2πR

∴ Mass of the ring per unit length = \(\frac{M}{2\pi R}\)

∴ Mass of a small element of the circle of length dx,

dm  = \(\frac{M}{2\pi R}\ dx\)

∴ Moment of inertia of this element about the axis YY',

dI  = \(dmr^2 = \frac{M}{2\pi R}\ dx\ R^2\) 

or dI = \(\frac{MR}{2\pi} \ dx\)

∴ Moment of inertia of the whole ring about the axis YY',

1. Moment of inertia about the diameter of the ring:
In fig. two cross diameters AB and CD are shown. We have to find the moment of inertia of the ring about the diameter AB or CD. Moment of inertia of the ring about ZZ' axis

I_z = MR² .........(1)

Now according to the theorem of the perpendicular axes of moment of

\(I_z = I_{AB} + I_{CD} = I_d + I_d\)

\(or \ I_z = 2I_d\)

\(\therefore\ I_d = \frac{I_z}{2}\)

or \(I_d = \frac{1}{2}\ MR^2\)

2. Moment of inertia of the ring about the tangential axis lying in the plane of the ring:

Suppose we have to find the moment of inertia of the ring about the tangential axis PQ.

∴ According to the theorem of parallel axes of the moment of inertia,
I_PQ = I_cm + Mh²

Here I_cm = I_AB = \(\frac{1}{2}MR^2\)

and h = R

\(\therefore \ I_{PQ} = \frac{1}{2}MR^2 + MR^2\)

or \(I_{PQ} = \frac{3}{2}MR^2\)

3. Moment of inertia of the ring about the tangential axis at right angles to plane of the ring: We have to find the moment of inertia about the axis PQ which is tangent of the ring at right angles to plane of the ring. Again applying the theorem of parallel axes,

I_PQ = I_cm + Mh²

Here I_cm = I_zz' = MR²

and h = R

∴ I_PQ = MR² + MR²

\(or\ I_{PQ} = 2MR^2\)