---
title: "Mass moment of inertia derivations"
subject: "Engineering Mechanics"
chapter: 6
key: "semester_3/engineering_mechanics:mass-moment-inertia-parallel-axis"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/mass-moment-inertia-parallel-axis
site: BCT Engineering
---

# Mass moment of inertia derivations

Engineering Mechanics · Chapter 6: Plane Kinetics of Rigid Bodies: Force, Mass and Acceleration

- Asked in: year not given, 8 marks
- Video: [How to Find Mass Moment of Inertia | Mechanics Statics | (Solved Examples)](https://www.youtube.com/watch?v=zmzUdFFCFkc) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/mass-moment-inertia-parallel-axis
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-6

## Question

Define mass moment of inertia and radius of gyration. State and prove the parallel axis theorem. Derive the mass moment of inertia of (a) a uniform slender rod of length L about a transverse axis through its centre and about one end, and (b) a solid circular cylinder of radius R about its geometric axis.

## Short answer

Mass moment of inertia I = ∫ r² dm measures rotational resistance; radius of gyration k = √(I/m). Parallel axis theorem: I = I_G + md², proved by shifting origin to centre of mass. Derived: slender rod I_G = mL²/12, I_end = mL²/3; solid cylinder I = ½ mR².

## Answer

#### Definitions
**Mass moment of inertia** of a body about an axis is $I = \int r^2\,dm$, where $r$ is the perpendicular distance of the element $dm$ from the axis (unit kg·m²). It measures resistance to angular acceleration, as mass does for linear acceleration.

**Radius of gyration** $k$ is the distance from the axis at which the whole mass could be concentrated to give the same moment of inertia: $I = mk^2$, so $k = \sqrt{I/m}$.

#### Parallel axis theorem
**Statement:** the moment of inertia about any axis equals the moment of inertia about a parallel axis through the mass centre plus $md^2$:

$$
I = I_G + md^2
$$

**Proof:** take the mass centre G as origin of x-y axes, with the axis G along z and the parallel axis passing through the point $(d, 0)$. For an element $dm$ at $(x,y)$:

$$
I = \int [(x - d)^2 + y^2]\,dm = \int (x^2 + y^2)\,dm - 2d\int x\,dm + d^2\int dm
$$

The first term is $I_G$. The second is zero because $\int x\,dm = m\bar x = 0$ (G is the origin). The third is $md^2$. Hence $I = I_G + md^2$.

#### (a) Slender rod, length L, mass m
Mass per unit length $= m/L$. About the transverse axis through the centre:

$$
I_G = \int_{-L/2}^{L/2} x^2 \frac{m}{L}\,dx = \frac{m}{L}\cdot\frac{2}{3}\left(\frac{L}{2}\right)^3 = \frac{mL^2}{12}
$$

About one end (parallel axis theorem, $d = L/2$):

$$
I_{end} = \frac{mL^2}{12} + m\left(\frac{L}{2}\right)^2 = \frac{mL^2}{3}
$$

#### (b) Solid cylinder, radius R, length h, density $\rho$
Take a thin cylindrical shell of radius $r$ and thickness $dr$: $dm = \rho\,(2\pi r\,h)\,dr$.

$$
I = \int_0^R r^2\,\rho\,2\pi h\,r\,dr = 2\pi\rho h\,\frac{R^4}{4} = \frac{1}{2}(\rho\pi R^2 h)R^2 = \frac{1}{2}mR^2
$$

(independent of the length h).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does the radius of gyration k represent?
   A. The distance of the centre of mass from the axis
   B. The distance from the axis at which the whole mass could be concentrated to give the same I, so k = sqrt(I/m)
   C. The average of all particle distances from the axis
   D. The largest distance of any particle from the axis
   Answer: B. k is defined by I = mk^2, the single distance at which all the mass would give the same moment of inertia. (Key point: Radius of gyration)

2. In the proof of the parallel axis theorem, why does the term -2d times the integral of x dm vanish?
   A. Because d is very small
   B. Because the integral of x dm equals m times x-bar, and x-bar = 0 as G is the origin
   C. Because x^2 and y^2 cancel each other
   D. Because dm is zero at the axis
   Answer: B. With G as origin, the first moment of mass about G is zero, so only I_G and md^2 remain. (Key point: Why the cross term vanishes)

3. Using I_G = mL^2/12 for a slender rod, what is I about one end, and how is it obtained?
   A. mL^2/12, since the axis is parallel
   B. mL^2/6, by doubling I_G
   C. mL^2/3, from I_G + m(L/2)^2
   D. mL^2/4, from I_G + m(L)^2
   Answer: C. The end is L/2 from G, so I = mL^2/12 + mL^2/4 = mL^2/3. (Key point: Rod about its end)

4. For a solid cylinder, integrating thin shells dm = rho(2 pi r h)dr from 0 to R gives which result?
   A. mR^2
   B. (1/2)mR^2, independent of length h
   C. (1/2)mR^2, but depends on length h
   D. (1/3)mR^2
   Answer: B. The integral gives 2 pi rho h R^4/4 = (1/2)mR^2; length only enters through the mass. (Key point: Solid cylinder I)

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