Short on time? This page is the recap. For explanations, the graph and worked examples, use the full study guide. This is the algebra-based course: no calculus is needed.
Recap
- Rotational inertia I measures how hard it is to change a rigid system’s rotation. It is the rotational partner of inertial mass.
- It depends on the mass and on the distribution of mass relative to the axis. Mass far from the axis counts much more than mass near it.
- One object has a different I for every axis. Always state the axis.
- You calculate I for up to five small objects in a flat layout. For rods, disks, hoops and spheres, the value is given.
- For any family of parallel axes, I is smallest through the centre of mass.
- Compare a hoop with a solid disk of equal mass and radius: the hoop’s I is larger, because all its mass is at the rim.
Key relationships
| Relationship | Symbols and units | Use it to |
|---|---|---|
| I = mr² | m in kg, r = perpendicular distance from axis in m; I in kg·m² | find I for one small object |
| I = Σmᵢrᵢ² | every rᵢ from the same axis | add up a system of small objects |
| I = I_cm + Md² | M = total mass; d = distance between the parallel axes | move from the centre-of-mass axis to a parallel one |
| I_new / I_old = (r_new / r_old)² | same mass | factor-of-change questions |
Assumptions behind the numbers
- The system is rigid: the masses stay at fixed distances from the axis.
- “Small” objects are treated as points at a single distance r.
- Connecting rods or frames described as light add nothing to I.
- Formulas for extended objects (such as ½MR² for a disk) are given, and apply only to the axis stated with them.
Mistakes to avoid
- Not squaring r. I = mr², not mr.
- Measuring r from the wrong point. Use the perpendicular distance from the stated axis.
- Assuming one I per object. Change the axis and I changes.
- Using part of the mass in Md². M is the total mass of the system.
- Starting the parallel axis theorem from the wrong axis. It must start from I_cm.
- Thinking more material always means more I. A hoop beats a disk of the same mass and radius.
- Putting the minimum at the geometric centre when the masses are unequal. It is at the centre of mass.
Quick self-check
- A 0.30 kg bead is 0.20 m from an axis. Find I. (0.012 kg·m²)
- A 2.0 kg object has I_cm = 0.050 kg·m². Find I about a parallel axis 0.10 m away. (0.070 kg·m²)
- A small mass is moved from 0.15 m to 0.30 m from the axis. By what factor does its I change? (4 times as large)
Next: practice questions.