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Q7.26(b):
(b) Prove the theorem of parallel axes. (Hint : If the centre of mass of a system of $n$ particles is chosen to be the origin $\sum m_i r_i = 0$).

Solution :

Step 1: Theoretical Definition of the Parallel Axes Theorem

The Theorem of Parallel Axes states that the moment of inertia $I$ of a rigid body about any axis is equal to the sum of the moment of inertia $I_{cm}$ of the body about a parallel axis passing through its centre of mass and the product of the mass of the body $M$ and the square of the perpendicular distance $d$ between the two axes. Mathematically, this is expressed as:

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

Step 2: Coordinate System Setup

Let the centre of mass (CM) of a system of $n$ particles be at the origin $(0,0,0)$ of the coordinate system. By the definition of the centre of mass, the sum of the moments of mass about the origin is zero: $\sum m_i \vec{r}_i = 0$ [Definition of Centre of Mass].

Consider two parallel axes: one passing through the CM ($z$-axis) and another passing through a point $P$ at distance $d$ from the CM (let this be the $z'$-axis). Let the position vector of a particle $i$ be $\vec{r}_i$ with coordinates $(x_i, y_i, z_i)$.

Axis CM Axis P CM d m_i

Step 3: Algebraic Derivation

The moment of inertia about the axis passing through the CM (the $z$-axis) is:

$I_{cm} = \sum m_i (x_i^2 + y_i^2)$

The distance of the $i$-th particle from the parallel axis passing through point $P$ (let $P$ be at $x=d, y=0$) is given by $[(x_i - d)^2 + y_i^2]$. The moment of inertia $I$ about this new axis is:

$I = \sum m_i ((x_i - d)^2 + y_i^2)$

Step 4: Expansion and Simplification

Expand the quadratic term:

$I = \sum m_i (x_i^2 - 2x_id + d^2 + y_i^2)$

Distribute the summation index:

$I = \sum m_i (x_i^2 + y_i^2) + \sum m_i d^2 - 2d \sum m_i x_i$

Step 5: Applying Centre of Mass Constraints

1. The first term $\sum m_i (x_i^2 + y_i^2)$ is exactly $I_{cm}$ [By Definition].

2. The second term $\sum m_i d^2$ becomes $d^2 \sum m_i = Md^2$, where $M$ is the total mass of the system.

3. The third term $2d \sum m_i x_i$: Since the CM is at the origin, $\sum m_i x_i = 0$ [Property of Centre of Mass]. Thus, this term vanishes.

Conclusion

Substituting these values back into the expression for $I$:

$I = I_{cm} + Md^2 + 0$

Final Solution: $I = I_{cm} + Md^2$. This completes the proof of the Parallel Axes Theorem.


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