Q.State and prove the parallel axis theorem.

Step 1. Statement of the theorem.

The moment of inertia of a rigid body about any axis equals the sum of (a) its moment of inertia about a PARALLEL axis passing through its own center of mass, and (b) the product of the body's total mass and the square of the perpendicular distance between the two axes.

Step 2. Setting up the proof.

Let $I_C$ be the (known) moment of inertia of the body about an axis $AB$ passing through its center of mass, and let $DE$ be a second axis, parallel to $AB$, at perpendicular distance $d$ from it. Call $I$ the (unknown) moment of inertia about $DE$, which is to be found.

Step 3. A representative point mass.

Consider a small point mass $m$ somewhere in the body, at perpendicular distance $x$ from the center-of-mass axis $AB$ (measuring $x$ as a signed distance, positive on one side of $AB$ and negative on the other). Since $DE$ is offset from $AB$ by $d$, this same point mass is at distance $(x+d)$ from $DE$. Its contribution to the moment of inertia about $DE$ is therefore

Step 4. Summing over the whole body.

Adding up this contribution over every point mass making up the body:

Step 5. Identifying each of the three sums.

• $\sum mx^2$ is, by definition, exactly $I_C$ — the moment of inertia about the center-of-mass axis $AB$.
• $\sum m$ is simply the total mass $M$ of the body. …

Direct summation over point masses of the body, using the defining property of the c …

• Forgetting the crucial step that $\sum mx=0$ — this term vanishes ONLY because $x$ is measured from an axis through the center of mass specifically; if the known axis $AB$ were NOT through the CM, this cross term would not vanish and the simple $I=I_C+Md^2$ form would not hold. …