System of Particles and Rotational Motion · Lesson 13 of 15
Rotational Dynamics, Work and Power
“Derive the fixed-axis torque law and connect it to energy transfer and practical rotors.”
• Identify the torque component relevant to the axis. • Derive work and power relations. • Derive τ = Iα under constant-I conditions. • Solve flywheel and unwinding-rope problems.
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 |
Quiz
Power from torque is what?
Power is work per time: τ(dθ/dt).
At constant angular speed with friction, driving torque is what?
The net torque vanishes while individual torques can remain nonzero.
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.
Which quantity belongs in α = τ/I?
All driving and resisting axial torques must be included.
Key Takeaways
• Only axial torque enters the fixed-axis one-coordinate dynamics. • Work is torque integrated over angle; power is torque times angular velocity. • τ = Iα requires constant I and net torque. • Driving power can be nonzero even at steady speed.