System of Particles and Rotational Motion · Lesson 11 of 15
Moment of Inertia and Rotational Kinetic Energy
“Relate rotational resistance and energy to the mass distribution about a specified axis.”
• Derive rotational kinetic energy from particle energies. • Calculate I for a ring and a two-mass system. • Use all eight standard moments of inertia with their axes. • Interpret radius of gyration and translational-plus-rotational energy.
Why the distribution of mass matters
Two bodies with equal mass can respond differently to the same torque because their mass lies at different distances from the rotation axis. In a rigid body rotating about a fixed axis, each particle has speed vᵢ = ωrᵢ⊥. Its kinetic energy is mᵢω²rᵢ⊥²/2. Sum all particle energies and factor out the common ω²/2. The remaining mass-weighted squared-distance sum is the moment of inertia I.
A particle twice as far from the axis contributes four times as much I if its mass is unchanged. Moving the axis can change I without changing the body’s mass or shape. For a given fixed mass distribution about a fixed body axis, I is independent of angular speed. It is not a universal scalar property of a body detached from its axis.
Problem
Find I for a thin ring of mass M and radius R about its central normal axis, and for two masses M/2 separated by length L on a light rod about their midpoint.
- 1.Every ring element lies at distance R, so I = ΣmᵢR² = MR². Its energy is therefore MR²ω²/2.
- 2.Each end mass of the light rod lies at L/2: I = 2(M/2)(L/2)² = ML²/4.
- 3.A solid uniform rod with the same total mass is different: its mass is spread along the rod, giving the smaller midpoint value ML²/12.
Standard values and the axes they require
| Body | Specified axis | Moment of inertia |
|---|---|---|
| Thin circular ring | Through centre, perpendicular to plane | MR² |
| Thin circular ring | Any diameter in its plane | MR²/2 |
| Uniform thin rod of length L | Through midpoint, perpendicular to rod | ML²/12 |
| Uniform thin disc | Through centre, perpendicular to plane | MR²/2 |
| Uniform thin disc | Any diameter in its plane | MR²/4 |
| Thin-walled hollow cylinder | Longitudinal symmetry axis | MR² |
| Uniform solid cylinder | Longitudinal symmetry axis | MR²/2 |
| Uniform solid sphere | Any diameter | 2MR²/5 |
The hollow-cylinder value assumes a thin wall so essentially all mass is at radius R. A thick-walled cylinder needs its inner and outer radii. In the supplied textbook, most standard-value derivations are stated to be beyond scope. Here the ring and two-mass derivations show how the definition works; the table supplies the remaining reference values. Do not use a disc’s normal-axis value about a diameter or a rod’s midpoint value about an end.
Radius of gyration
This definition does not assert that the body’s mass is actually concentrated there. A ring about its normal axis has k = R. A uniform rod about its perpendicular midpoint axis has k = L/√12. A disc about a diameter has k = R/2; about its central normal axis it has k = R/√2. A flywheel uses substantial inertia to resist rapid changes in speed for a given fluctuating torque, smoothing the rotational motion.
Problem
A solid cylinder has mass 20 kg, radius 0.25 m and angular speed 100 rad/s.
- 1.Its symmetry-axis moment is I = 20(0.25²)/2 = 0.625 kg m².
- 2.Rotational energy is Iω²/2 = 3125 J. Its axial angular momentum, using the relation developed fully later, is Iω = 62.5 kg m²/s.
- 3.The cylinder rotates about its symmetry axis. Different axes require different I values.
Motion about the CM and motion of the CM
When the body translates as well as rotates, apply the previously derived energy decomposition. For a rigid body rotating about its CM, K′ = I_CMω²/2 and total kinetic energy is MV_CM²/2 + I_CMω²/2. Pure fixed-axis energy is I_axisω²/2. Avoid counting the same kinetic energy twice by adding CM translation to an I_axis expression that already describes every particle’s complete fixed-axis speed.
Problem
An oxygen molecule has mass 5.30 × 10⁻²⁶ kg, I = 1.94 × 10⁻⁴⁶ kg m² and translational speed 500 m/s. Take rotational energy to be two thirds of its translational energy.
- 1.Write Iω²/2 = (2/3)(Mv²/2), so ω = √[2Mv²/(3I)].
- 2.Substitution gives approximately 6.75 × 10¹² rad/s.
- 3.This is the representative angular speed corresponding to the stipulated energy. In a real distribution of molecular speeds, an average-energy calculation relates to a mean-square speed, not automatically an arithmetic mean.
Quiz
A particle’s contribution to I is what?
The distance is squared and measured perpendicularly to the axis.
For equal M and R, which has larger symmetry-axis I?
MR² exceeds MR²/2.
Radius of gyration equals what?
It is defined by I = Mk².
Does increasing ω change I for a rigid distribution and fixed axis?
I depends on the mass distribution about the axis.
Which value applies to a solid sphere about a diameter?
This is the sphere’s standard diameter-axis value.
Key Takeaways
• I measures the mass distribution about a particular axis. • Particle-energy summation gives K_rot = Iω²/2. • Standard values must remain attached to their axis specifications. • k = √(I/M) describes an equivalent mass distance.