Q.(a) State and prove the parallel axis theorem. OR

The parallel axis theorem, I = I_cm + Md^2, relates the moment of inertia about any axis to the moment of inertia about a parallel axis through the centre of mass, proved by expanding the defining integral and showing the cross term vanishes.

Statement: The moment of inertia (I) of a rigid body about any axis is equal to the sum of its moment of inertia (I_cm) about a parallel axis passing through its centre of mass, and the product of the mass (M) of the body and the square of the perpendicular distance (d) between the two parallel axes:

I = I_cm + Md^2

Proof:

Consider a rigid body of mass M, and let AB be an axis through its centre of mass C, with moment of inertia I_cm about AB. Let A'B' be another axis, parallel to AB, at a perpendicular distance d from it, about which we want to find the moment of inertia I.

Set up coordinates with the centre of mass C at the origin, and the axis AB along the z-direction through the origin. The parallel axis A'B' is then a line parallel to AB but displaced by distance d in the x-direction.

Consider a small mass element dm of the body, located at perpendicular distance r from axis AB (through the centre of mass), with coordinates (x, y) in the plane perpendicular to AB (so r^2 = x^2 + y^2).

The perpendicular distance of this same mass element from the new axis A'B' (displaced by d along x) is r', where, by the geometry of the displaced axis,

r'^2 = (x-d)^2 + y^2 = x^2 - 2xd + d^2 + y^2 = r^2 - 2xd + d^2

The moment of inertia about the new axis A'B' is:

I = integral of r'^2 dm = integral of (r^2 - 2xd + d^2) dm

I = integral of r^2 dm - 2d*(integral of x dm) + d^2*(integral of dm)

Now examine each term:

1. integral of r^2 dm = I_cm, the moment of inertia about the axis through the centre of mass (by definition). …