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Study Notes

Unit 5 — Torque and Rotational Dynamics

Motion around an axis: angular position, velocity and acceleration, how they connect to linear motion, and how torque and rotational inertia govern rotation. Each section ends with a short practice set — click "Show answer" to check yourself.

5.1

Rotational Kinematics

Rotation is described with angular quantities, measured in radians. One full revolution is 2π rad = 360°.

QuantitySymbol / definitionUnits
Angular displacementΔθrad
Angular velocityω = Δθ / Δtrad/s
Angular accelerationα = Δω / Δtrad/s²

For constant angular acceleration, the equations mirror the linear ones:

ω  = ω₀ + α·t
Δθ = ω₀·t + ½·α·t²
ω² = ω₀² + 2·α·Δθ
  • Choose a positive rotation direction (usually counterclockwise) and use signs, just as with linear motion.
  • If ω and α have the same sign, the object spins faster; opposite signs, it slows down.
  • Period and frequency: T = 2π/ω and f = 1/T. To convert: 1 rpm = 2π/60 rad/s.
  • Graphs work as before: the slope of θ–t is ω, the slope of ω–t is α, and the area under ω–t is Δθ.
Tip: always convert degrees and rpm to radians and rad/s before using the equations.
Practice

1. A wheel starts from rest and reaches 20 rad/s in 4.0 s with constant angular acceleration. Find α and the angle it turns through.

Show answer
α = Δω/Δt = 20/4.0 = 5.0 rad/s². Δθ = ½αt² = ½(5.0)(4.0)² = 40 rad (about 6.4 revolutions).

2. A record player spins at 300 rpm. What is its angular velocity in rad/s?

Show answer
ω = 300 × 2π / 60 = 10π ≈ 31.4 rad/s.
5.2

Connecting Linear and Rotational Motion

Every point on a rigid rotating object has the same ω and α, but points farther from the axis move faster along a longer circle.

s   = r·θ            (arc length)
v_t = r·ω            (tangential speed)
a_t = r·α            (tangential acceleration)
a_c = ω²·r = v²/r    (centripetal acceleration, toward the axis)
  • The tangential quantities (vt, at) point along the circle; the centripetal acceleration points toward the axis. A point on a spinning object can have both.
  • These relations require angles in radians.
  • Rolling without slipping: the center moves at v = ωr, and the distance rolled equals the arc length turned, d = rθ.
Key idea: ω is shared by the whole object; speed v = ωr depends on how far the point is from the axis.
Practice

1. A point 0.50 m from the axis of a disk rotates at ω = 12 rad/s. Find its tangential speed and centripetal acceleration.

Show answer
v = ωr = 12 × 0.50 = 6.0 m/s. ac = ω²r = 144 × 0.50 = 72 m/s².

2. A bicycle wheel of radius 0.30 m rolls without slipping at 6.0 m/s. What is its angular velocity, and how far does the bike travel in one revolution?

Show answer
ω = v/r = 6.0/0.30 = 20 rad/s. One revolution: d = 2πr = 2π(0.30) ≈ 1.9 m.
5.3

Torque

Torque measures how effectively a force causes or changes rotation about a chosen axis (pivot).

τ = r · F · sin θ  =  r⊥ · F        units: N·m
  • r is the distance from the axis to where the force is applied; θ is the angle between the vector r (from the axis to the point of application) and the force.
  • Only the force component perpendicular to r produces torque. The perpendicular distance from the axis to the force's line of action is the lever arm, r⊥.
  • Torque is greatest when the force is perpendicular to r (θ = 90°). A force along the line through the axis (θ = 0° or 180°) gives zero torque.
  • Sign convention: counterclockwise torque is positive, clockwise is negative. The net torque is the signed sum Στ.
  • Torque depends on the axis chosen. Weight acts at the object's center of mass.
Why it matters: a longer wrench or a door handle far from the hinges gives more torque for the same force.
Practice

1. A 20 N force pushes perpendicular to a door 0.40 m from the hinges. What is the torque about the hinges?

Show answer
τ = rF sin 90° = 0.40 × 20 = 8.0 N·m.

2. A 30 N force acts on a 0.50 m wrench at 30° to the wrench. What is the torque on the bolt?

Show answer
τ = rF sin θ = 0.50 × 30 × sin 30° = 7.5 N·m.
5.4

Rotational Inertia

Rotational inertia (moment of inertia), I, is an object's resistance to changes in its rotation. It plays the role of mass in rotational motion.

I = Σ m·r²          units: kg·m²
  • Each piece of mass contributes m·r², where r is its distance from the axis of rotation.
  • I depends on the total mass, how the mass is distributed, and the choice of axis. The same object has different I about different axes.
  • Mass farther from the axis contributes far more (because of r²). Moving mass outward increases I.
  • A single point mass on a light rod or string: I = mr².
Object (mass M)AxisI
Thin hoop / ring (radius R)CenterMR²
Solid disk / cylinderCenter½MR²
Uniform rod (length L)Center(1/12)ML²
Uniform rod (length L)One end(1/3)ML²
Note: the table is for reference and comparison. For a system made of several masses, add up mr² for each.
Practice

1. Two 2.0 kg masses sit at the ends of a light rod, each 0.50 m from the axis through the rod's center. What is the rotational inertia?

Show answer
I = Σmr² = 2(2.0)(0.50)² = 1.0 kg·m².

2. A hoop and a solid disk have the same mass and radius. Which has the larger rotational inertia, and why?

Show answer
The hoop (I = MR² vs. ½MR²). All of its mass is at the maximum distance R from the axis, whereas the disk has mass closer to the center.
5.5

Rotational Equilibrium and Newton's First Law in Rotational Form

The rotational form of Newton's first law: an object keeps rotating at constant angular velocity (or stays non-rotating) unless a net external torque acts on it.

Rotational equilibrium:   Στ = 0     (α = 0)
Static equilibrium:       ΣF = 0  and  Στ = 0
  • In static equilibrium the object is not translating and not rotating, so both the net force and the net torque are zero.
  • You may choose any point as the pivot when writing Στ = 0. Pick the point where an unknown force acts, since that force then has zero torque and drops out.
  • A uniform object's weight acts at its center of mass, at its geometric center.
  • Balanced torques: τcw = τccw, which is the principle of a seesaw or balance.

Steps: draw an FBD, choose a pivot, write each torque with its sign, set Στ = 0 and ΣF = 0, then solve.

Practice

1. On a seesaw pivoted at its center, a 40 kg child sits 2.0 m from the pivot. Where must a 30 kg child sit on the other side to balance it?

Show answer
(40)(2.0) = (30)x → x ≈ 2.7 m from the pivot. The lighter child sits farther out.

2. A light meter stick is pivoted at its 50 cm mark. A 2.0 kg mass hangs at the 20 cm mark. At which mark must a 1.5 kg mass hang to balance it?

Show answer
Left lever arm = 30 cm. (2.0)(30) = (1.5)x → x = 40 cm from the pivot, so the 90 cm mark. (g cancels on both sides.)
5.6

Newton's Second Law in Rotational Form

A net torque on an object produces an angular acceleration that is proportional to the torque and inversely proportional to the rotational inertia.

α = Στ / I        or        Στ = I·α
LinearRotational
Position xAngle θ
Velocity vAngular velocity ω
Acceleration aAngular acceleration α
Mass mRotational inertia I
Force FTorque τ
ΣF = maΣτ = Iα
  • The same torque gives a smaller α to an object with a larger I.
  • Use the same axis for both τ and I, and use signs for direction (α has the sign of the net torque).
  • For problems that combine rotation and translation (e.g. a hanging mass on a pulley), write ΣF = ma for the mass, Στ = Iα for the pulley, and link them with a = αr.
Practice

1. A wheel has I = 2.0 kg·m² and a net torque of 6.0 N·m is applied. What is its angular acceleration?

Show answer
α = Στ/I = 6.0/2.0 = 3.0 rad/s².

2. A 10 N force acts tangentially on the rim of a disk of radius 0.20 m and I = 0.40 kg·m². Find the torque and the angular acceleration.

Show answer
τ = rF = 0.20 × 10 = 2.0 N·m. α = τ/I = 2.0/0.40 = 5.0 rad/s².