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MechanicsIntroStarter track

Concept module

Torque

Push on one pivoted bar and see how lever arm distance, force direction, and turning effect stay tied to the same compact rotational bench.

Interactive lab

Keep the stage, graph, and immediate control feedback in one working view.

Time

0.00 s / 2.40 sLivePause to inspect a specific moment, then step or scrub through it.

0.00 s2.40 s

Torque

One pivoted bar, one applied force, and one fixed inertia are enough to keep lever arm, force direction, and turning response tied to the same honest bench.

Graphs

Switch graph views without breaking the live stage and time link.

Torque vs time

For a fixed setup on this bounded bench, the same lever-arm geometry keeps the torque level steady through the clip.

time (s): 0 to 2.4torque (N m): -4 to 4

tau

Hover or scrub to link the graph back to the stage.time (s) / torque (N m)

Controls

Adjust the physical parameters and watch the motion respond.

Force magnitude

F

2 N

Set the size of the push without changing where or how it is applied.

Force angle

ϕ

90°

Angle is measured relative to the bar. Positive angles twist counterclockwise, negative angles twist clockwise.

Application point

r

1.6 m

Move the same push closer to the pivot or farther toward the handle.

More tools

Secondary controls, alternate presets, and less-used toggles stay nearby without crowding the main bench.

Show

More presets

Presets

Predict -> manipulate -> observe

Keep the active prompt next to the controls so each change has an immediate visible consequence.

ObservationPrompt 1 of 1

Notice that the same perpendicular push twists much harder at the handle than near the pivot.

Try this

Switch between the handle right-angle push and the near-hinge version. The force arrow stays the same, but the torque and the end-of-clip angle should drop by about half.

Equation map

See each variable before you move it.

Select a symbol to highlight the matching control and the graph or overlay it most directly changes.

r

Lever arm distance

1.6 m

Moving the same push farther from the pivot increases the turning effect because the same perpendicular force now has a longer lever arm.

Graph: Torque vs timeGraph: Torque vs force angleOverlay: Moment armOverlay: Line of action

Equations in play

Choose an equation to sync the active symbol, control highlight, and related graph mapping.

More tools

Detailed noticing prompts, guided overlays, and challenge tasks stay available without taking over the main bench.

Hide

What to notice

Use the live prompt to keep the turning story focused on one thing at a time: where the force acts, how it is aimed, and what that does to the bar.

ObservationPrompt 1 of 1

Graph: Torque vs time

Notice that the same perpendicular push twists much harder at the handle than near the pivot.

Try this

Switch between the handle right-angle push and the near-hinge version. The force arrow stays the same, but the torque and the end-of-clip angle should drop by about half.

Why it matters

This is the cleanest lever-arm effect: same force direction, different distance, different turning effect.

Control: Application pointGraph: Torque vs timeGraph: Rotation angle vs timeOverlay: Moment armEquation

r

Guided overlays

Focus one overlay at a time to see what it represents and what to notice in the live motion.

2 visible

Overlay focus

Perpendicular component

Separates the twisting part of the force from the part that merely pushes along the bar.

What to notice

The amber arrow is the part of the force that actually creates torque.
A large parallel component can still look dramatic while contributing almost nothing to the turning effect.

Why it matters

Torque belongs to $F_\perp$, not to the full force blindly.

Control: Force magnitudeControl: Force angleGraph: Torque vs timeGraph: Torque vs force angleEquation

F

Equation

ϕ

Challenge mode

Use the same live bench to prove that you can reason about lever arm distance, force direction, and matched turning effects without hiding behind memorized formulas.

0/3 solved

ConditionCore

1 of 4 checks

Zero turn at the handle

Keep the push point near the handle but make the bar feel almost no turning effect.

Graph-linkedGuided start2 hints

Suggested start

Turn on the line of action and use the torque-angle map together.

Pending

Open the Torque vs force angle graph.

Torque vs time

Pending

Keep the Line of action visible.

Off

Matched

Keep the application point between

1.35

and

1.6 m

1.6 m

Pending

Make the torque land between

- 0.12

and

0.12 N m

3.2 N m

The checklist updates from the live simulation state, active graph, overlays, inspect time, and compare setup.

At t = 0 s, a 2 N force is applied 1.6 m from the pivot at 90°. The perpendicular component is 2 N, so the torque is 3.2 N m. The angular acceleration is 0.46 rad/s^2, the bar's angular speed is 0 rad/s, and its rotation is 0 rad. The positive torque keeps building counterclockwise rotation.

Equation detailsDeeper interpretation, notes, and worked variable context.

Torque from force geometry

Torque grows when the force is applied farther from the pivot and when more of the force points perpendicular to the lever arm.

τ = r F sin ϕ

r

Lever arm distance 1.6 m

F

Force magnitude 2 N

ϕ

Force angle 90°

Torque from the perpendicular part

Only the perpendicular component of the force contributes to turning.

τ = r F_{⊥}

r

Lever arm distance 1.6 m

F

Force magnitude 2 N

ϕ

Force angle 90°

Rotational response

For a fixed rotational inertia, larger torque produces larger angular acceleration.

α = \frac{τ}{I}

Bounded constant-torque motion

On this bench the bar starts from rest and keeps one constant torque for each fixed setup, so angular speed grows linearly while angle curves.

ω = α t, θ = \frac{1}{2} α t^{2}

Progress

Not startedMastery: NewLocal-first

Start exploring and Open Model Lab will keep this concept's progress on this browser first. Challenge mode has 3 compact tasks ready. No finished quick test, solved challenge, or completion mark is saved yet.

Let the live model runChange one real controlOpen What to notice

Try this setup

Copy the live bench state and reopen this concept with the same controls, graph, overlays, and compare context.

Stable links

Starter track

Step 1 of 50 / 5 complete

Rotational Mechanics

Next after this: Static Equilibrium / Centre of Mass.

1. Torque2. Static Equilibrium / Centre of Mass3. Rotational Inertia / Moment of Inertia4. Rolling Motion+1 more steps

This concept is the track start.

See next concept

Short explanation

What the system is doing

Torque is the turning effect of a force about a pivot. The same push can twist hard, twist gently, or barely twist at all depending on where you push and how much of the force points perpendicular to the lever arm.

This bench keeps one fixed bar and one fixed rotational inertia so the turning story stays honest without turning into a giant rigid-body engine. The force angle is defined relative to the bar itself, which lets the same live controls show lever arm distance, force direction, torque, and the resulting spin on one compact surface.

Key ideas

01Torque depends on both distance from the pivot and the perpendicular part of the force: $\tau = rF\sin\phi = rF_\perp$.

02A force aimed straight through the pivot can be large and still produce almost no torque because its line of action has almost no moment arm.

03Torque is the rotational side of force. Later it becomes the bridge to angular momentum change, and in static-equilibrium problems the same turning bookkeeping explains why weights acting at the centre of mass matter.

Live torque checks

Solve the exact state on screen.

Solve the turning effect you are actually watching. The first example reads the current geometry, and the second follows the same fixed-torque setup forward through the current inspected time.

Live valuesFollowing current parameters

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

r

Lever arm distance

1.6 m

F

Force magnitude

2 N

ϕ

Force angle

90 °

1. Start from the turning 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 part of the force

With

r = 1.6 m

F = 2 N

, and

ϕ = 9 0^{\circ}

, the perpendicular part is

F_{⊥} = 2 N

3. Compute torque and angular acceleration

τ = 3.2 N m

and

α = 0.46 rad/ s^{2}

Current 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.

Turning-effect checkpoint

Can you cut the lever arm roughly in half but keep the same turning effect?

Prediction prompt

Try shortening the application distance, then predict what has to happen to the perpendicular part of the force if torque is going to stay the same.

Check your reasoning

Yes. If

r

gets smaller, then

F_{⊥}

must get larger in just the right way so that the product

r F_{⊥}

stays unchanged.

This is the core torque trade. The turning effect is not owned by distance alone or force alone. It belongs to the product of the lever arm and the perpendicular component of the force.

Common misconception

The biggest force always creates the biggest turning effect.

Force size alone is not enough. A smaller force far from the pivot can out-twist a bigger force applied close to the pivot.

Direction matters just as much as distance. Only the perpendicular part of the force contributes to torque, so a large force aimed through the pivot can still give almost zero turning effect.

Quick test

Compare cases

Question 1 of 4

These checks ask whether you can reason with torque, not just repeat the formula. Use lever arm, force direction, and the live bar response together.

The same $2 N$ force is applied at $9 0^{\circ}$ in two places: once $1.6 m$ from the pivot and once $0.8 m$ from the pivot. Which statement is correct?

Choose one answer to reveal feedback, then test the idea in the live system if a guided example is available.

Accessible description

The simulation shows a single bar pivoted at its left end. A force arrow is attached somewhere along the bar, and the bar rotates from rest according to the torque from the current force magnitude, force angle, and application distance.

Optional overlays can separate the perpendicular force component, extend the line of action, and show the moment arm from the pivot. The linked readout and graphs report torque, angular speed, and rotation angle on the same fixed bench.

Graph summary

The torque graph stays flat for a fixed setup because the bounded bench keeps the same lever-arm geometry relative to the bar throughout the clip. The angular-speed graph is linear and the rotation-angle graph is curved because the bar starts from rest under constant angular acceleration.

The torque-versus-force-angle response graph shows a sine-like shape for the current force size and application distance, with the largest positive and negative turning effects near plus or minus ninety degrees and near-zero torque when the line of action passes through the pivot.

Carry turning ideas forward

Keep moving through the concept map.

These suggestions come from the concept registry, so the reason label reflects either curated guidance or the fallback progression logic.

Support-region balanceMechanics

Torque

See each variable before you move it.

Perpendicular component

Zero turn at the handle

Rotational Mechanics

What the system is doing

Solve the exact state on screen.

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

The same 2N force is applied at 90∘ in two places: once 1.6m from the pivot and once 0.8m from the pivot. Which statement is correct?

Keep moving through the concept map.

Static Equilibrium / Centre of Mass

Rotational Inertia / Moment of Inertia

Rolling Motion

The same $2 N$ force is applied at $9 0^{\circ}$ in two places: once $1.6 m$ from the pivot and once $0.8 m$ from the pivot. Which statement is correct?