---
title: "Composite body moment of inertia"
subject: "Engineering Mechanics"
chapter: 6
key: "semester_3/engineering_mechanics:rod-sphere-composite-inertia"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/rod-sphere-composite-inertia
site: BCT Engineering
---

# Composite body moment of inertia

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

- Asked in: year not given, 6 marks
- Video: [How to find the moment of inertia for composite shapes](https://www.youtube.com/watch?v=3wIi4O6a6_Q) — Engineer4Free
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/rod-sphere-composite-inertia
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-6

## Question

A uniform slender rod of mass 3 kg and length 1.2 m carries at one end a solid sphere of mass 2 kg and radius 0.1 m, the surface of the sphere just touching the rod end so that the sphere centre lies on the axis of the rod. Find the mass moment of inertia of the composite body about a transverse axis through the free end of the rod, the position of its centre of mass and its radius of gyration about that axis.

## Short answer

Compute I_O by summing rod about end (mL²/3 = 1.44 kg·m²) and sphere using parallel axis (2/5 mR² + md² = 3.388 kg·m²), total 4.828 kg·m². Centre of mass from free end: 0.88 m; radius of gyration k_O = 0.983 m.

## Answer

**Given:** rod $m_r = 3$ kg, $L = 1.2$ m; sphere $m_s = 2$ kg, $R = 0.1$ m, centre on the rod axis at $1.2 + 0.1 = 1.3$ m from the free end O.

```text
 O                               ___
 |============================( o )
 |<------- 1.2 m ------------>|<R>
 |<------ 1.3 m to sphere centre
```

#### Moment of inertia about O (transverse axis)
Rod about its end: $I_r = \dfrac{m_rL^2}{3} = \dfrac{3(1.2)^2}{3} = 1.440$ kg·m²

Sphere about its own centre: $\tfrac25 m_sR^2 = 0.4 \times 2 \times 0.01 = 0.008$ kg·m²

Parallel axis shift: $m_s d^2 = 2(1.3)^2 = 3.380$ kg·m²

| Part | Calculation | I about O (kg·m²) |
|---|---|---|
| Rod | $mL^2/3$ | 1.440 |
| Sphere | $\tfrac25 mR^2 + md^2$ | 3.388 |
| Total | | 4.828 |

#### Centre of mass from O

$$
\bar x = \frac{3(0.6) + 2(1.3)}{5} = \frac{1.8 + 2.6}{5} = 0.88\ \text{m}
$$

#### Radius of gyration about O

$$
k_O = \sqrt{\frac{I_O}{m}} = \sqrt{\frac{4.828}{5}} = 0.983\ \text{m}
$$

(As a check, the moment of inertia about the centre of mass is $I_G = 4.828 - 5(0.88)^2 = 0.956$ kg·m².)

**Answer: I_O = 4.828 kg·m²; centre of mass 0.88 m from the free end; k_O = 0.983 m.**

## Check yourself

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

1. How far is the sphere centre from the free end O of the rod?
   A. 1.2 m
   B. 1.3 m
   C. 1.1 m
   D. 1.4 m
   Answer: B. The sphere touches the rod end, so its centre is 1.2 + R = 1.3 m from O. (Key point: Distance to sphere centre)

2. How is the sphere's moment of inertia about O found?
   A. (2/5)mR^2 only
   B. (2/5)mR^2 + m(1.3)^2 using the parallel axis theorem
   C. m(1.3)^2 only
   D. (2/5)mR^2 + m(1.2)^2
   Answer: B. Its own I about the centre (0.008) plus md^2 (3.380) gives 3.388 kg·m^2. (Key point: Sphere inertia about O)

3. Which expression gives the centre of mass from O?
   A. (3(1.2) + 2(1.3))/5
   B. (3(0.6) + 2(1.3))/5 = 0.88 m
   C. (3(0.6) + 2(0.1))/5
   D. (0.6 + 1.3)/2
   Answer: B. The rod's mass centre is at 0.6 m and the sphere's at 1.3 m, weighted by mass. (Key point: Centre of mass)

4. The total I_O is 4.828 kg·m^2 for a total mass of 5 kg. What is k_O?
   A. 0.966 m
   B. 0.983 m
   C. 0.88 m
   D. 1.00 m
   Answer: B. k = sqrt(4.828/5) = 0.983 m. (Key point: Radius of gyration)

## Related in Engineering Mechanics

- [Mass moment of inertia derivations](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Amass-moment-inertia-parallel-axis) — Engineering Mechanics, chapter 6
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