Derive the expression for the moment of inertia of a rod about its center and perpendicular to the rod.

III. Long Answer Questions · Q4

Q. Derive the expression for the moment of inertia of a rod about its center and perpendicular to the rod.

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Concept understanding — Moment of Inertia

Moment of inertia ( $I$ ) is the rotational analogue of mass: just as mass measures a body's inherent resistance to a change in its state of linear motion (inertia), moment of inertia measures a body's resistance to a change in its state of rotational motion, for rotation about a specific axis. Its SI unit is kg m $^2$ , and its dimension is $[ML^2]$ .

For a single point mass $m_i$ at perpendicular distance $r_i$ from a fixed axis, the moment of inertia about that axis is $I=m_ir_i^2$ . For a rigid body made up of many such point masses, the moment of inertia about the axis is the sum over all of them:

I=\sum_i m_ir_i^2.

Key contrast with mass. Ordinary mass is (to excellent approximation, ignoring relativistic effects) a fixed, invariant property of a body. Moment of inertia is emphatically not invariant — it depends not only on how much mass a body has, but crucially on exactly how that mass is distributed relative to the chosen axis. The same rigid body has a different moment of inertia for every different axis of rotation one might choose, even axes lying entirely outside the physical body itself.

Computing $I$ for a continuous body. Treating an infinitesimally small mass element $dm$ , at perpendicular distance $r$ from the axis, as a point mass gives $dI=(dm)r^2$ ; integrating over the whole body gives the moment of inertia of any continuous, uniformly-distributed bulk object:

I=\int dI=\int r^2\,dm.

This integral, carried out with the mass element $dm$ expressed via the object's linear density $\lambda=M/\ell$ (for a rod), surface density $\sigma=M/(\pi R^2)$ (for a disc), or similar, yields the standard results for common shapes:

Uniform rod (mass $M$ , length $\ell$ ), axis through the center, perpendicular to the rod: $I=\dfrac{1}{12}M\ell^2$ ; axis through one end: $I=\dfrac{1}{3}M\ell^2$ .
Uniform ring (mass $M$ , radius $R$ ), axis through the center, perpendicular to the plane: $I=MR^2$ (every mass element is at the same distance $R$ from the axis, so the integral is trivial). …

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Step 1. Setting up the geometry.

Consider a uniform rod of mass $M$ and length $\ell$ , and find its moment of inertia about the axis passing through its own center, perpendicular to the rod. Place the origin at the rod's midpoint (which, for a uniform rod, is also its center of mass), with the rod lying along the x-axis. Consider a thin element of the rod of width $dx$ , located at distance $x$ from this origin.

Step 2. Mass of the element.

Since the rod is uniform, its linear mass density (mass per unit length) is constant:

\lambda=\frac{M}{\ell}.

The mass of the small element is therefore

dm=\lambda\,dx=\frac{M}{\ell}\,dx.

Step 3. Contribution of the element to the moment of inertia.

Treating this element as a point mass at perpendicular distance $x$ from the axis, its contribution to the total moment of inertia is

dI=(dm)\,x^2=\frac{M}{\ell}x^2\,dx.

Step 4. Integrating over the whole rod.

Because the axis passes through the center, the rod extends symmetrically from $x=-\ell/2$ to $x=+\ell/2$ , so these are the correct integration limits:

I=\int_{-\ell/2}^{\ell/2}\frac{M}{\ell}x^2\,dx=\frac{M}{\ell}\left[\frac{x^3}{3}\right]_{-\ell/2}^{\ell/2}.

Step 5. Evaluating the limits.

At the upper limit, $\dfrac{(\ell/2)^3}{3}=\dfrac{\ell^3}{24}$ ; at the lower limit, $\dfrac{(-\ell/2)^3}{3}=-\dfrac{\ell^3}{24}$ . Subtracting the lower from the upper:

I=\frac{M}{\ell}\left(\frac{\ell^3}{24}-\left(-\frac{\ell^3}{24}\right)\right)=\frac{M}{\ell}\cdot\frac{\ell^3}{12}.

Step 6. Simplifying. …

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Q.Derive the expression for the moment of inertia of a rod about its center and perpendicular to the rod.

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