Learn Rigid-body rotation from scratch
Before this: Newton’s laws, Work & energy
1What rigid-body rotation is
A rigid body is an object that keeps its shape: a flywheel, a pulley, a door or a rolling wheel. In this topic the body turns about an axis.
Every point of the body turns through the same angle in the same time. So you describe the motion with angular quantities instead of distances:
- θ: the angle turned, in radians (rad). One revolution is 2π rad.
- ω: the angular velocity, in rad/s.
- α: the angular acceleration, in rad/s².
Machines give speeds in revolutions per minute (rpm). Convert them before you calculate:
ω = 2π·n / 60
A point at distance r from the axis moves with speed v = ω·r and has acceleration a = α·r along its path.
Everything you know about straight-line motion has a rotating twin: force becomes moment, mass becomes moment of inertia, and ΣF = m·a becomes ΣM = I·α.
Self-check A wheel turns at 600 rpm. What is its angular velocity in rad/s?
- 10 rad/s
- 3770 rad/s
- 600 rad/s
- 62.8 rad/s
Answer: 62.8 rad/s. ω = 2π·600/60 = 2π·10 ≈ 62.8 rad/s. 10 is revolutions per second (the 2π is missing), 3770 forgets to divide by 60, and 600 is still in rpm.
2The moment of inertia
Mass tells you how hard it is to speed a body up in a straight line. The mass moment of inertia I tells you how hard it is to set it spinning.
Mass far from the axis counts much more than mass close to it, because it counts with the square of its distance:
I = Σ m·r² (unit: kg·m²)
You do not need to add up the pieces yourself. Learn these standard results about the centre of mass G:
- Slender rod of length L, axis through the middle: m·L²/12.
- Solid disc or cylinder of radius R: ½·m·R².
- Thin ring or hoop: m·R².
- Solid sphere: (2/5)·m·R². Hollow sphere: (2/3)·m·R².
- Point mass at distance r: m·r².
When the axis O does not pass through G, use the parallel-axis theorem. Here d is the distance between O and G:
IO = IG + m·d²
A rod turning about its end has d = L/2, so IO = m·L²/12 + m·L²/4 = m·L²/3.
Some questions give a radius of gyration k instead of a shape. Then I = m·k². For a body made of parts, find each part’s I about the same axis and add them.
Self-check A uniform rod of mass 3 kg and length 2 m turns about one of its ends. What is its moment of inertia?
- 1 kg·m²
- 12 kg·m²
- 3 kg·m²
- 4 kg·m²
Answer: 4 kg·m². I = m·L²/3 = 3·4/3 = 4 kg·m², or IG + m·d² = 1 + 3·1² = 4. 1 kg·m² is the value about the centre, 3 kg·m² is only the m·d² part, and 12 kg·m² is m·L² without any factor.
3Moments, energy and angular momentum
Three tools solve almost every rotation problem. Pick the one that matches the question.
1. Moments. The sum of the moments about the axis equals I·α. Use it for accelerations and forces.
ΣM = I·α
With a constant torque the angular motion has constant α, and the kinematics equations have rotating twins: ω = ω₀ + α·t and ω² = ω₀² + 2·α·θ.
2. Energy. A spinning body stores kinetic energy. A torque T turning through an angle θ does work, and the power of a turning shaft is torque times angular velocity:
E = ½·I·ω² W = T·θ P = T·ω
A body that rolls without slipping moves and spins at the same time, with v = ω·r. It has both kinds of kinetic energy:
E = ½·m·v² + ½·I·ω² = ½·m·v²·(1 + c) c = I/(m·r²)
The number c depends only on the shape: 0.4 for a solid sphere, 0.5 for a solid cylinder, 2/3 for a hollow sphere and 1 for a hoop. On a slope, a = g·sin θ / (1 + c).
3. Angular momentum. L = I·ω. When no outside moment acts about the axis, L stays the same, even when bodies stick together and kinetic energy is lost:
I1·ω1 = I2·ω2
Self-check A motor delivers a torque of 100 N·m at 600 rpm. What is its power?
- 1 kW
- 377 kW
- 6.28 kW
- 60 kW
Answer: 6.28 kW. ω = 2π·600/60 = 62.8 rad/s, so P = T·ω = 100·62.8 ≈ 6283 W = 6.28 kW. 60 kW uses rpm as ω, 1 kW uses revolutions per second, and 377 kW forgets to divide by 60.
4The method step by step
- Draw the body and choose the axis: the fixed pivot or shaft, or the centre for a rolling body.
- Find the moment of inertia about that axis. Use the standard results and add m·d² when the axis is shifted.
- Convert every speed in rpm to rad/s with ω = 2π·n/60.
- Choose the tool. Accelerations and forces: ΣM = I·α (with ΣF = m·a for parts that move in a line). Speeds after a distance or angle: energy. Impacts, sudden grips and bodies that pull in their mass: angular momentum.
- Link the straight and the turning motion where a cord or a rolling contact joins them: v = ω·r and a = α·r.
- Solve, and check the result with a second tool if you can.
For a body hanging from a cord around a pulley, draw two free-body diagrams: one for the block (forces) and one for the pulley (moments). The cord tension appears in both.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked exampleA 2 kg block hangs from a light cord wrapped around a pulley. The pulley is a uniform solid disc of mass 4 kg and radius 0.2 m on a frictionless axle. The block is released from rest. Use g = 9.81 m/s². Find the acceleration of the block, the tension in the cord and the speed of the block after it has fallen 1 m.
1The pulley is a solid disc turning about its centre.
I = ½·M·R² = ½ · 4 · 0.2² = 0.08 kg·m²
2Take downwards as positive for the block. Gravity pulls it down and the cord tension T pulls it up.
m·g − T = m·a ⇒ 19.62 − T = 2·a
3Only the tension has a moment about the axle, with arm R. The cord does not slip, so α = a/R.
T·R = I·α = I·a/R ⇒ T = I·a/R² = 0.08·a/0.04 = 2·a
4Put T = 2·a into the block equation.
19.62 − 2·a = 2·a ⇒ a = 19.62/4 = 4.905 m/s²
T = 2 · 4.905 = 9.81 N
The pulley turns with α = a/R = 4.905/0.2 ≈ 24.5 rad/s².
5The acceleration is constant and the block starts from rest, so use v² = 2·a·s.
v = √(2 · 4.905 · 1) = √9.81 ≈ 3.13 m/s
6Use energy instead. The lost potential energy becomes kinetic energy of the block and of the pulley, with ω = v/R.
m·g·h = ½·m·v² + ½·I·(v/R)² ⇒ 19.62 = v² + v²
So v² = 9.81 and v ≈ 3.13 m/s, the same as before.
Answer a ≈ 4.91 m/s², T ≈ 9.81 N and v ≈ 3.13 m/s after 1 m.
6Common mistakes
- Leaving speeds in rpm in ω, E = ½·I·ω² or P = T·ω. Convert to rad/s first.
- Using the moment of inertia about the centre for a body that turns about its end. Add m·d².
- Forgetting the rotational kinetic energy of a rolling body or a pulley.
- Treating a sphere or disc on the end of a rod as a point mass. Add its own I about its centre as well.
- Assuming kinetic energy is conserved when bodies grip or stick together. Use angular momentum instead.
- Counting revolutions in radians: N = θ/(2π).
Self-check A solid cylinder of mass 2 kg rolls without slipping at 3 m/s. What is its total kinetic energy?
- 13.5 J
- 18 J
- 4.5 J
- 9 J
Answer: 13.5 J. E = ½·m·v²·(1 + c) = ½·2·3²·1.5 = 13.5 J, with c = 0.5 for a solid cylinder. 9 J is only the moving part, 4.5 J is only the spinning part, and 18 J uses c = 1, the value for a hoop.
7How exam questions look
- Easy: moment of inertia. A rod about a shifted axis, a pendulum made of a rod and a sphere or disc, the kinetic energy of a flywheel, or the torque or time needed to spin a flywheel up.
- Medium: ΣM = I·α, work and power. A rod released from horizontal, a block on a pulley, the revolutions a braked flywheel makes, the torque or power of a shaft, or the speed of a swinging rod.
- Hard: rolling and angular momentum. A sphere, cylinder or hoop rolling down a slope, the friction needed to roll, the share of rotational energy, a ring dropped on a spinning disc, a person pulling in dumbbells, or a bullet hitting a hanging rod.
For a braked flywheel, the brake does work equal to the stored energy: T·θ = ½·I·ω². Divide θ by 2π to get the revolutions.
A rod pinned at one end and released from horizontal has α = 3·g/(2·L) just after release. Its free end accelerates at 1.5·g.
For a bullet that sticks in a rod, the angular momentum about the pivot is kept: m·v·d = (Irod + m·d²)·ω.
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