System of Particles and Rotational Motion · Lesson 8 of 9
Rotational Dynamics and Angular-Momentum Conservation
“Connect net axial torque, work, power and spin changes about a fixed axis.”
• Derive work, power and τz = Iα for fixed I. • Solve the source flywheel, hollow-cylinder and rotor problems. • Derive Lz = Iω and distinguish it from the full angular-momentum vector. • Explain changing-inertia spin rates and the source of rotational-energy changes.
Fixed-axis dynamics, work and power
Which torque changes the allowed rotation?
A fixed axle permits rotation about one axis while preventing the axis from tilting or translating. Only the external torque component along that axis enters the one-coordinate rotation equation. A force parallel to the axis has zero axial torque. A radial force in the rotation plane also has zero axial torque. Constraint reactions keep the axis fixed and can balance or supply perpendicular torques; these should not be confused with the driving axial torque.
Deriving work and power
A particle moves through a short tangential displacement ds = r⊥dθ. Only the tangential force component F_t does work, so dW = F_t r⊥dθ. The product F_t r⊥ is its signed axial torque. Every particle of the rigid body shares the same dθ, so summing over all applied forces gives the net axial torque times dθ. Dividing by dt gives instantaneous power.
For fixed I, differentiate K = Iω²/2: dK/dt = Iωα. Equate this to net torque power τ_zω to obtain τ_z = Iα where ω is nonzero; continuity or the angular-momentum derivation extends the equation through rest. This is the rotational counterpart of F = Ma. It assumes the axis stays fixed relative to the rigid body and the mass distribution, hence I, does not change.
Problem
A solid-disc flywheel has mass 20 kg and radius 0.20 m. A 25 N cord pull unwinds 2 m from rest on a frictionless axle.
- 1.I = MR²/2 = 0.40 kg m² and torque is FR = 5.0 N m. Thus α = 5.0/0.40 = 12.5 rad/s².
- 2.The unwound length gives Δθ = 2/0.20 = 10 rad. The work is Fℓ = 25(2) = 50 J, also τΔθ = 5(10) = 50 J.
- 3.The final angular speed satisfies ω² = 2αΔθ = 250, giving ω ≈ 15.81 rad/s.
- 4.Final kinetic energy is 0.40(250)/2 = 50 J, matching the work because there is no frictional loss.
Problem
A 3 kg thin-walled hollow cylinder of radius 0.40 m has a light rope wound round it. A 30 N pull acts on the rope without slipping; the axle is fixed.
- 1.I = MR² = 0.48 kg m² and τ = FR = 12 N m.
- 2.α = 12/0.48 = 25 rad/s². The rope’s linear acceleration is αR = 10 m/s².
- 3.The no-slip relation matches the rope displacement to rim arc displacement. The tangential rim acceleration, not its total radial-plus-tangential acceleration, matches the rope’s acceleration.
Problem
Equal torques act for equal times from rest on a thin hollow cylinder and a solid sphere of equal M and R.
- 1.I_cylinder = MR² while I_sphere = 2MR²/5. Their angular accelerations are torque divided by these values.
- 2.The sphere’s angular acceleration, and therefore its acquired angular speed from rest, is 5/2 times the cylinder’s.
- 3.The smaller moment of inertia makes the sphere easier to spin up about the stated symmetry axes.
Problem
A rotor turns at 200 rad/s while an engine supplies 180 N m to overcome friction. Find delivered power.
- 1.Power is τω = 180(200) = 36000 W = 36 kW.
- 2.Steady ω means net torque is zero, not that the driving torque or power is zero. The supplied 180 N m balances resisting torque.
- 3.The input work is dissipated by resistance rather than increasing rotational kinetic energy.
| Translation | Fixed-axis rotational analogue |
|---|---|
| Displacement x | Angular displacement θ |
| Velocity v | Angular velocity ω |
| Acceleration a | Angular acceleration α |
| Mass M | Moment of inertia I |
| Net force Ma | Net axial torque Iα |
| Work integral F · ds | Work integral τ_z dθ |
| Kinetic energy Mv²/2 | Kinetic energy Iω²/2 |
| Power F · v | Power τ_zω |
| Momentum Mv | Axial angular momentum Iω |
Axial angular momentum and its conservation
Start with one particle, then sum
Take the fixed axis as z and an origin O on it. For one particle P, split its position as OP = OC + CP, where C lies on the axis at the centre of its circular path. The CP × mv contribution is along z and has magnitude mr⊥v = mr⊥²ω. The OC × mv contribution is perpendicular to z. Hence even a particle moving in a circle need not have its full angular momentum along the axis when the reference origin is away from its circle’s centre.
For a body whose mass distribution has the appropriate symmetry about the chosen axis, equal opposite particle contributions cancel the transverse components. Then L is along the axis and the vector relation L = Iω is valid. A body driven around an axis without the required symmetry can have a nonparallel L. This is a significant limit to the simple analogy with p = mv, where p and v are always parallel for an ordinary particle of positive mass.
Torque changes the axial component
For an asymmetric body held to a fixed axis, support constraints may exert perpendicular torques to keep that axis from tilting. Projecting onto the allowed axis avoids assuming those transverse torques vanish. The complete vector equation remains dL/dt = τ_ext; the one-axis equation concerns only its z component. A zero axial torque guarantees constant L_z, while conservation of the full vector requires the full external torque to vanish.
Changing the mass distribution changes spin rate
On a freely turning swivel chair, extending the arms puts more mass farther from the axis and increases I. The angular speed falls. Bringing the arms inward reduces I and increases angular speed. Skaters, dancers, divers and acrobats use the same principle. For a diver in flight, changing body shape changes the distribution of mass about the spin axis; negligible external torque allows angular momentum to remain approximately constant.
Problem
A system turns at 40 rpm. Folding the arms reduces its total moment of inertia to 2/5 of its original value. Find final speed and energy ratio.
- 1.Angular-momentum conservation gives ω_f = (I_i/I_f)ω_i = (5/2)(40) = 100 rpm.
- 2.K_f/K_i = (I_f/I_i)(ω_f/ω_i)² = (2/5)(5/2)² = 5/2.
- 3.The rotational kinetic energy rises by 150% of its initial value. The child does internal work, converting chemical energy into mechanical energy while moving the arms inward.
- 4.If 2/5 refers only to the child’s I rather than the total including the turntable, add the unchanged turntable inertia before calculating the ratio. The simple 100 rpm answer assumes the stated factor applies to total rotating inertia or that platform inertia is negligible.
Angular-momentum conservation does not mean energy must be constant. It states the absence of an external angular impulse about the relevant axis. Internal work can change rotational energy, and friction can dissipate energy while transferring angular momentum to the surroundings. Always specify system boundaries before deciding which quantity is conserved.
Check your understanding
Quiz
Power from torque is what?
Power is work per time: τ(dθ/dt).
For a rigid body with fixed I, net torque equals what?
This is the fixed-axis dynamical law.
For a no-slip rope on radius R, tangential acceleration is what?
Differentiate rope speed v = ωR for constant R.
If total I halves with negligible axial torque, ω changes how?
Iω is conserved, so ω changes inversely with I.
A child pulls arms inward and gains rotational energy. The source is what?
Internal muscular work supplies the energy. The correct choice is the internal-work option.
Key Takeaways
• Use the net torque component about the permitted fixed axis. • Work is torque integrated over angle; power is torque times angular velocity. • Lz = Iω is an axial relation; the vector relation needs the appropriate symmetry. • With negligible external axial torque, Iω is conserved and internal work can change rotational energy.