Rotation · AP · 5 min
Torque is not just force times distance; the angle matters.
01 The trap
Wrong path
Students use tau = FL even when the force is not perpendicular to the handle or beam.
Why it feels right
Most simple torque examples place the force at 90 degrees, hiding the sine factor.
02 Correct model
Torque magnitude is rF sin θ, or force times the perpendicular lever arm.
03 Mini-example
Prompt
A force F is applied at 30 degrees to a wrench of length L. What is torque?
Common wrong answer
tau = FL.
Correct reasoning
tau = FL sin30 = 0.5 FL.
Diagnostic cue
If the force is angled, the full force is probably not the torque-producing component.
04 Guided practice
Warm-up isolates the principle. Bridge changes the context. Contest-style requires a complete setup on less familiar geometry.
A 20 N force is applied at 30° to a 0.25 m wrench. Find the torque magnitude about the bolt.
Use the component perpendicular to the wrench.
τ = rF sin 30° = (0.25)(20)(0.5) = 2.5 N·m.
Trap: Using τ = rF without the angle factor.
A uniform beam of weight W and length L is pivoted at the left end and held by a vertical rope at the right end. Find the rope tension.
Take torque about the pivot.
T L = W(L/2), so T = W/2. The pivot force does not appear in the torque equation about the pivot.
Trap: Balancing vertical forces first and assuming T = W.
A ladder rests against a frictionless wall and rough floor. Why do you need both net force and net torque equations?
Rigid-body equilibrium has translational and rotational balance.
Net force zero keeps the center of mass from accelerating; net torque zero keeps the ladder from angularly accelerating. The wall normal and floor friction are linked through torque balance.
Trap: Solving only force balance and missing the rotational condition.
A uniform rod of length L and mass M pivots without friction about one end. It is released from rest at angle θ above the vertical. (a) Find its angular speed when vertical. (b) Explain why constant-angular-acceleration kinematics is not valid.
The center of mass drops (L/2)(1 - cos θ), and I_end = (1/3)ML².
Energy gives Mg(L/2)(1 - cos θ) = (1/2)(1/3 ML²)ω². Therefore ω = √(3g(1 - cos θ)/L). Constant-angular-acceleration kinematics fails because the gravitational torque is Mg(L/2)sin(θ), which changes as the rod falls.
Trap: Using τ = Iα at the starting angle and treating α as constant.
A 45 N force acts 0.32 m from a pivot at an angle of 55° to the lever arm. What is the torque magnitude about the pivot?
τ = rF sin θ.
τ = (0.32)(45)sin55° = 11.8 N·m.
Trap: Using rF without the sine factor.
A uniform disk of mass M and radius R rotates freely at angular speed Ω₀. A small puck of mass m moving tangentially at speed v sticks to the rim in the same rotational sense. (a) Find the final angular speed. (b) Find the mechanical energy converted to internal energy. (c) State why angular momentum is conserved about the axle even though linear momentum of the disk-puck system is not.
Use I_disk = ½MR² and add mR² after the puck sticks.
With I_d = ½MR², angular momentum gives Ωf = (I_dΩ₀ + mvR)/(I_d + mR²). The converted energy is [½I_dΩ₀² + ½mv²] - ½(I_d + mR²)Ωf². The axle can exert external linear impulse but has zero lever arm about itself, so its angular impulse about the axle is zero.
Trap: Conserving linear momentum despite the external axle impulse, or conserving kinetic energy in a sticking collision.