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.
Rotational Kinematics
Rotation is described with angular quantities, measured in radians. One full revolution is 2π rad = 360°.
| Quantity | Symbol / definition | Units |
|---|---|---|
| Angular displacement | Δθ | rad |
| Angular velocity | ω = Δθ / Δt | rad/s |
| Angular acceleration | α = Δω / Δt | rad/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/ωandf = 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.
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.
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2. A record player spins at 300 rpm. What is its angular velocity in rad/s?
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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.
v = ωr depends on how far the point is from the axis.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.
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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?
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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
ris the distance from the axis to where the force is applied;θis the angle between the vectorr(from the axis to the point of application) and the force.- Only the force component perpendicular to
rproduces 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.
1. A 20 N force pushes perpendicular to a door 0.40 m from the hinges. What is the torque about the hinges?
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2. A 30 N force acts on a 0.50 m wrench at 30° to the wrench. What is the torque on the bolt?
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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, whereris 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) | Axis | I |
|---|---|---|
| Thin hoop / ring (radius R) | Center | MR² |
| Solid disk / cylinder | Center | ½MR² |
| Uniform rod (length L) | Center | (1/12)ML² |
| Uniform rod (length L) | One end | (1/3)ML² |
mr² for each.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?
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2. A hoop and a solid disk have the same mass and radius. Which has the larger rotational inertia, and why?
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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.
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?
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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?
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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
| Linear | Rotational |
|---|---|
Position x | Angle θ |
Velocity v | Angular velocity ω |
Acceleration a | Angular acceleration α |
Mass m | Rotational inertia I |
Force F | Torque τ |
ΣF = ma | Στ = 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 = mafor the mass,Στ = Iαfor the pulley, and link them witha = αr.
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?
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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.