Derive the expression for the moment of inertia of a uniform ring about an axis passing through the center and

III. Long Answer Questions · Q5

Q. Derive the expression for the moment of inertia of a uniform ring about an axis passing through the center and perpendicular to the plane.

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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 ring of mass $M$ and radius $R$ , and find its moment of inertia about the axis through its center, perpendicular to the plane of the ring. Take a thin element of the ring — a tiny arc of length $dx$ along its circumference — as the mass element $dm$ .

Step 2. The key simplifying feature of a ring.

Unlike a rod or a disc, every single point of a ring lies at exactly the same perpendicular distance $R$ from the central axis, since $R$ is simply the ring's own radius. This means the element's contribution to the moment of inertia is

dI=(dm)\,R^2,

with $R^2$ a plain constant, not a variable that must be tracked through the integration.

Step 3. Mass of the element.

The full length of the ring is its circumference, $2\pi R$ . Since the mass is uniformly distributed, the linear mass density is

\lambda=\frac{M}{2\pi R},

so the mass of the arc element is $dm=\lambda\,dx=\dfrac{M}{2\pi R}\,dx$ .

Step 4. Integrating around the full circumference.

Summing (integrating) the contributions of every such element all the way around the ring, from $x=0$ to $x=2\pi R$ :

I=\int_0^{2\pi R}\frac{M}{2\pi R}R^2\,dx=\frac{MR^2}{2\pi R}\int_0^{2\pi R}dx=\frac{MR}{2\pi}\Big[x\Big]_0^{2\pi R}.

Step 5. Evaluating.

$I=\frac{MR}{2\pi}(2\pi R)=MR^2.$ …

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Q.Derive the expression for the moment of inertia of a uniform ring about an axis passing through the center and perpendicular to the plane.

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