Physics · Mechanics

Torque

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Wrap-up

What you learned

Recommended next: Open concept testCheck whether the core ideas are ready without leaving this concept.
Read next: Rotational Inertia / Moment of InertiaKeep the same twist and change where the mass sits.

Key takeaway

Torque depends on both lever arm distance and the perpendicular part of the force.
A large force aimed through the pivot can still create almost no turning effect.
For this fixed-inertia bench, larger torque means larger angular acceleration.

Common misconception

The biggest force always gives the biggest turning effect.

Not unless it also has a large moment arm. A smaller force farther from the pivot can create more torque than a bigger force applied close in.

Carry torque ideas forward

Keep this idea moving

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Support-region balanceMechanics

Static Equilibrium / Centre of Mass

Shift one support region under one loaded plank and see how centre of mass, support reactions, and torque balance decide whether the object stays stable or tips.

Mass-distribution responseMechanics

Rotational Inertia / Moment of Inertia

Keep the same total mass and torque, then slide equal masses inward or outward to see why moment of inertia makes some rotors much harder to spin up than others.

Rolling without slippingMechanics

Rolling Motion

Roll a sphere, cylinder, hoop, or custom mass distribution down one incline and see how rolling without slipping ties translation, rotation, and rotational inertia to the same honest run.

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Equation snapshotTorque from force geometry · Torque from the perpendicular partShow 3 equations

Torque from force geometry
$τ = r F sin ϕ$
Torque increases when the force acts farther from the pivot and when more of the force is perpendicular to the bar.
Torque from the perpendicular part
$τ = r F_{⊥}$
Only the perpendicular component twists the bar. The component along the bar does not add torque.
Rotational response
$α = \frac{τ}{I}$
With fixed rotational inertia, the sign and size of torque set the sign and size of angular acceleration.

Why it behaves this way

Explanation

Torque tells you how strongly a force tries to rotate an object about a pivot. It depends on both where the force is applied and how much of the force points perpendicular to the bar.

This bench keeps the bar and rotational inertia fixed so you can isolate the geometry of torque. Change the distance from the pivot, force size, or force angle relative to the bar, then compare the stage with the moment arm, line of action, and graphs. For one fixed setup, the torque stays constant, so the torque graph is flat while angular speed and rotation angle build up.

Key ideas

01Read torque as

τ = r F sin ϕ = r F_{⊥}

: a longer distance from the pivot or a larger perpendicular component gives a larger turning effect.

02A force can be large and still produce almost no torque if its line of action passes through the pivot, because the moment arm is then near zero.

03On this bench, a fixed setup gives constant torque, so angular acceleration is constant, angular speed changes linearly, and rotation angle curves.

Worked examples

Solve the live torque

Open examples when you want to see the same idea walked through step by step.

Frozen walkthrough

Step through the frozen example

Frozen walkthrough

Start from the geometry you can see. First calculate torque from distance and the perpendicular component of the force. Then use the same constant-torque setup to find angular speed and rotation angle at a chosen time.

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Example 1 of 2

Frozen valuesUsing frozen parameters

For the current setup, what torque does the force create and what angular acceleration follows on this fixed-inertia bench?

r

Distance from pivot

1.6 m

F

Force magnitude

2 N

ϕ

Angle relative to the bar

90 °

1. Write the torque relation

Use

τ = r F sin ϕ = r F_{⊥}

and then

α = τ / I

. On this bench the rotational inertia stays fixed at

I = 7 kg m^{2}

2. Find the perpendicular force component

With

r = 1.6 m

F = 2 N

, and

ϕ = 9 0^{\circ}

, the perpendicular part is

F_{⊥} = 2 N

3. Calculate torque and angular acceleration

τ = 3.2 N m

and

α = 0.46 rad/ s^{2}

Torque and angular acceleration

τ = 3.2 N m, α = 0.46 rad/ s^{2}

The perpendicular part of the push is positive here, so the same force geometry builds a counterclockwise twist.

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