# NCERT Class 11 Physics Solutions: Chapter 7 – System of Particles and Rotation Motion-Part 11 (For CBSE, ICSE, IAS, NET, NRA 2022)

Doorsteptutor material for NCO Class-11 is prepared by world's top subject experts: fully solved questions with step-by-step explanation- practice your way to success.

**Question 7.26**:

(a) Prove the theorem of perpendicular axes.

(Hint: Square of the distance of a point () in the plane from an axis through the origin perpendicular to the plane is

(b) Prove the theorem of parallel axes.

(Hint: If the centre of mass is chosen to be the origin

**Answer**:

**(a)** Prove the theorem of perpendicular axes:

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 perpendicular axis and lying in the plane of the body. A physical body with centre and a point mass , in the plane at is shown in the following figure:

Moment of inertia about axis,

Moment of inertia about axis,

Moment of inertia about axis,

Hence, the theorem is proved.

**(b)** Prove the theorem of parallel axes:

The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

Suppose a rigid body is made up of particles, having masses , at perpendicular distances respectively from the centre of mass of the rigid body.

The moment of inertia about axis passing through the point

The perpendicular distance of mass , from the axis,

Hence, the moment of inertia about axis,

Now, at the centre of mass, the moment of inertia of all the particles about the axis passing through the centre of mass is zero, that is,

Also,

Where, total mass of the rigid body

Hence, the theorem is proved.