Q.State and prove the theorem of parallel axes about moment of inertia.

Sum the moment of inertia of every mass element about the centroidal axis and about the new parallel axis, using the geometry that the centre of mass term vanishes by definition.

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

$I=I_{cm}+Md^2$

Proof: Let the body have mass $M$, and let $I_{cm}$ be its moment of inertia about an axis through its centre of mass G (perpendicular to the plane containing the axis of interest). Let $AB$ be another axis parallel to the one through G, at a perpendicular distance $d$ from it, and let $I$ be the moment of inertia of the body about $AB$.

Consider a small mass element $dm$ of the body, at perpendicular distance $r$ from G (measured in the plane perpendicular to the axes) and at perpendicular distance $r'$ from $AB$. Set up coordinates with G at the origin, the axis $AB$ passing through a point at distance $d$ from G along, say, the x-axis. If the element's coordinates relative to G are $(x,y)$, then

$r^2=x^2+y^2$

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

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