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Rotational Motion

Rotational Motion — American System Physics Revision Notes & Topical Questions

5 min read 6 parts Lesson 1 of 5 1 quiz

📚 American System Physics Hub / Topical Revision
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Lesson summary

Rotational Motion: Description, Torque and Equilibrium — AMERICAN SYSTEM PHYSICS notes: Describe rotation with angular quantities, calculate torque and moment of inertia, and apply equilibrium and centre-of-mass ideas.The axis of rotation is the line about which an object rotates…

  • translational equilibrium
  • Stable equilibrium
  • Worked example 1
  • Worked example 2
  • Worked example 3
  • Worked example 4

Part 1 of 4Describing rotation

The axis of rotation is the line about which an object rotates; points on it do not move. 1 revolution = 2π rad = 360°, 1 rad ≈ 57.3°.

Describing rotation
Revolutions per minute (rpm) is a common non-SI rate: 1 rpm = 2π/60 rad/s ≈ 0.105 rad/s.
Angular velocity is ω = Δθ/Δt = 2πf = 2π/T, and angular acceleration is α = Δω/Δt.
For a point at radius r: s = rθ, linear speed v = rω (m/s) and tangential acceleration a = rα (m/s2).
LinearRotational
xθ
vω (v = rω)
aα (a = rα)
mI
F = maτ = Iα
p = mvL = Iω

To convert rpm to rim speed

  • divide rpm by 60 to get revolutions per second (f);
    • multiply by 2π to get ω in rad/s;
    • measure the radius r in metres;
    • multiply v = rω and give v in m/s to 2-3 significant figures.
  • Convert degrees to radians by × π/180 and revolutions by × 2π.

Part 2 of 4Torque, moment of inertia and moments

Torque is the turning effect of a force, τ = Fr sin θ (N m), where r sin θ is the perpendicular distance from the axis to the line of action.

Torque, moment of inertia and moments
Torque is increased by pushing harder (τ ∝ F), pushing farther from the pivot (τ ∝ r) and pushing at 90° to the lever (sin θ = 1); wrenches with longer handles give more torque,
  • and a door is easiest to turn by pushing at the handle perpendicular to it.
Newton's second law for rotation is α = τnet/I, so doubling net torque doubles α.

Moment of inertia I depends on mass and its distribution; the farther the mass from the axis, the larger I (tightrope walkers use poles to increase I). For point masses I = Σmr2.

ObjectAxisI
Point mass / thin hoopdistance r (hoop: through centre)mr2
Solid cylinder or discthrough centre½mr2
Solid spherethrough centre(2/5)mr2
Thin rodcentre, perpendicular(1/12)ml2
Thin rodone end, perpendicular(1/3)ml2
1
Choose the pivot (fulcrum, the fixed point about which the object turns) and label all forces.
2
Find each perpendicular distance r sin θ.
3
Calculate τ = Fr sin θ with sign (counterclockwise +, clockwise −).
4
Add to get the net torque; find I; use α = τnet/I (rad/s2).

For a seesaw or moments problemmark the pivot and each force's distance, classify moments as clockwise or counterclockwise, compute force × perpendicular distance, set total clockwise = total counterclockwise and solve (seesaws balance when m1r1 = m2r2).

Part 3 of 4Equilibrium and centre of mass

Equilibrium requires ΣF = 0 (translational equilibrium, constant linear velocity) and Στ = 0. Weight W = mg acts through the centre of mass; for a uniform beam it is at the midpoint. Stable equilibrium is when a small displacement raises the centre of mass, so the weight produces a torque that returns the object.

To find the centre of mass of a flat irregular lamina: hang it freely from a pin through a small hole near its edge, hang a plumb line from the same pin, let both settle and mark the plumb line on the card; repeat from a second and third hole well away from the first. The lines cross at the centre of mass; check by balancing the card on a pencil tip there.

ConditionResult
Low centre of mass, wide basestable
High centre of mass, narrow baseeasily toppled
Centre of mass outside basetopples

Why a tilted object topples

  • it is tilted about an edge of its base, its weight still acts vertically down through the centre of mass;
    • if this line is inside the base the weight's moment turns it back, but once outside the base the moment turns it further over.
  • The Coriolis effect is the apparent sideways deflection of a moving object seen from a rotating frame such as Earth.

Part 4 of 4Worked examples

  1. A wheel turns at 120 rpm and has radius 0.30 m. Find the rim speed. [3]3 marks

  2. A 200 N weight is 2.0 m from a pivot. What counterclockwise moment is needed for balance? [1]1 mark

  3. A point mass of 3.0 kg is 0.50 m from an axis. Find its moment of inertia. [2]2 marks

  4. A hoop and a solid disc have equal mass and radius. Which has the larger moment of inertia about its centre, and why? [2]2 marks

  5. Describe how to find the centre of mass of an irregular flat card. [4]4 marks

Key terms

TermMeaning
TorqueThe turning effect of a force, τ = Fr sin θ.
Axis of rotationThe line about which an object rotates; points on it do not move.
Revolutions per minute (rpm)A non-SI rotation rate; 1 rpm = 2π/60 rad/s.
Pivot (fulcrum)The fixed point or axis about which an object turns.
Translational equilibriumNet force is zero so linear velocity is constant.
Stable equilibriumA small displacement raises the centre of mass so the weight returns the object.
Coriolis effectApparent sideways deflection of a moving object viewed from a rotating frame such as Earth.

Avoid these mistakes

1
Using degrees in v = rω — angles must be in radians.
2
Forgetting to divide rpm by 60 before multiplying by 2π.
3
Using the distance along the force rather than the perpendicular distance in torque.
4
Thinking a higher centre of mass makes an object more stable — a lower one and wider base do.
5
Mixing up moment of inertia values — a hoop has mr2, a disc ½mr2.

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