Mini Physics
Part of Mini EducationSwitch subject

UY1: Torque & Angular Acceleration

Torque definitions, sign conventions, and practical use of the rotational dynamics relation between net torque and angular acceleration.

  • University Physics Year 1
On this page

Learning objectives

  • Use vector and differential or integral calculus to express physical change and accumulation.
  • Use complex numbers, linear algebra, and differential equations to solve coupled physical models.
  • Choose an efficient mathematical method and validate the result physically.
Cluster role + pathway links

This page is the core concept page for rotational dynamics in the UY1 mechanics pathway.

Scope and prerequisites

This page assumes you already know:

  • rotational kinematics variables (θ,ω,α),
  • moment-of-inertia lookup/theorem use,
  • free-body diagram habits from translational dynamics.

Use this page as your default for fixed-axis rotational dynamics before attempting rolling or angular-momentum transfer problems.

At a glance

  • Prerequisites: Newton’s second law, moment of inertia, vector basics.
  • Outcomes: compute torque correctly and apply Στ = Iα.
  • Key result:
vector τ = vector r × vector F, Στ = Iα
Common traps (direction + lever arm)
  • Only the component of vector F perpendicular to vector r produces torque: τ = rF sin φ.
  • r is measured from the chosen pivot/axis to the point of application, not to some “nice-looking” point.
  • In 2D problems, decide your sign convention once (e.g. CCW positive) and keep it everywhere.

Setup

Model and signs:

  • Rigid body rotates about a fixed axis through pivot O.
  • Positive torque is counterclockwise (or along + z hat by right-hand rule).

Equivalent scalar forms (about a fixed axis):

τ = rF sin φ = Fₜ r = Fl

where l is lever arm and Fₜ is tangential force component.

Geometry and direction (how to get the sign fast):

  • Lever arm l is the perpendicular distance from the pivot to the force line of action.
  • Right-hand rule for vector τ = vector r × vector F:
    • Point fingers along vector r (pivot → point of application), curl toward vector F.
    • Thumb gives the direction of vector τ (along the rotation axis).
  • In the xy plane, the z-component is
    τ_z = xF_y-yFₓ
    which is often the cleanest way to avoid “sin” sign mistakes.

Core method

Start with a particle at radius r:

Fₜ = maₜ = m(rα)

Multiply by r:

τ = rFₜ = mr²α

For an extended body, sum over mass elements:

Στ = ∑ᵢ mᵢ rᵢ²α = (∑ᵢ mᵢ rᵢ²)α = Iα

Continuous form gives same result:

Στ = α∫ r² dm = Iα

Non-trivial point: only external net torque appears in Στ for the system; internal torques cancel in pairs.

Quick checks (before you commit to the algebra):

  • If the line of action passes through the pivot, l = 0 so τ = 0.
  • If vector F∥ vector r, then φ = 0 and τ = 0.
  • Maximum |τ| happens when vector F⊥ vector r.

Worked example

A wheel has I = 0.80 kg m². A tangential force of 12 N is applied at radius 0.25 m, while friction produces an opposing torque of 0.40 N m.

Applied torque:

τₐₚₚ = Fr = (12)(0.25) = 3.0 N m

Net torque:

Στ = 3.0-0.40 = 2.6 N m

Angular acceleration:

α = Στ/I = 2.6/0.80 = 3.25 rad s⁻²

Checks:

  • Units check: N m/(kg m²) = s⁻² (radian is dimensionless), so α is rad s⁻².
  • Sanity check: if opposing friction rose to 3.0 N m, net torque would be zero and α would be zero.

Practice set

  1. A force 9 N acts tangentially at 0.40 m. Find torque. Hint: τ = Fr for tangential force. Answer: 3.6 N m.

  2. A body has I = 2.0 kg m² and net torque -5.0 N m. Find α. Hint: sign carries through. Answer: -2.5 rad s⁻².

  3. Why does a force through the pivot produce no rotation? Hint: lever arm. Answer: l = 0, so τ = Fl = 0.

Summary and next steps

Torque is rotational effectiveness of force, set by both magnitude and lever arm. With a fixed axis and rigid body, dynamics reduces to Στ = Iα.

Minimum exam loop:

  1. Mark pivot and sign convention.
  2. Convert each force into signed torque (or xF_y-yFₓ form).
  3. Sum torques first, then solve α from Στ = Iα.