A uniform rod of mass $m$ and length $\ell$ makes a constant angle $\theta$ with an axis of rotation which passes through one end of the rod. Find the moment of inertia about this axis.

A uniform rod of mass m and length ell makes a constant angle θ with an axis of rotation which passes through

V. Numerical Problems · Q4

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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. Set up the geometry. The rod of mass $m$ and length $\ell$ is pivoted at one end, and makes a constant angle $\theta$ with the axis of rotation (which also passes through that same end). A small element of the rod, at arc-length $x$ from the pivoted end ( $0\le x\le\ell$ ), is not lying directly on the axis — because the whole rod is tilted at angle $\theta$ to the axis, this element's actual perpendicular distance from the axis is

r=x\sin\theta.

Step 2. Write the mass of the small element. Since the rod is uniform, its linear mass density is $\lambda=m/\ell$ , so a small element of length $dx$ has mass

dm=\frac{m}{\ell}\,dx.

Step 3. Write its contribution to the moment of inertia. Using $dI=(dm)r^2$ :

dI=\frac{m}{\ell}\,dx\,(x\sin\theta)^2=\frac{m\sin^2\theta}{\ell}x^2\,dx.

Step 4. Integrate over the whole rod, from $x=0$ (the pivoted end) to $x=\ell$ (the free end):

I=\int_0^\ell\frac{m\sin^2\theta}{\ell}x^2\,dx=\frac{m\sin^2\theta}{\ell}\left[\frac{x^3}{3}\right]_0^\ell=\frac{m\sin^2\theta}{\ell}\cdot\frac{\ell^3}{3}.

Step 5. Simplify.

$I=\frac{1}{3}m\ell^2\sin^2\theta.$ …

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Q.A uniform rod of mass mmm and length ℓ\ellℓ makes a constant angle θ\thetaθ with an axis of rotation which passes through one end of the rod. Find the moment of inertia about this axis.

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Q.A uniform rod of mass $m$ and length $\ell$ makes a constant angle $\theta$ with an axis of rotation which passes through one end of the rod. Find the moment of inertia about this axis.

Unlock everything free for 14 days