HomeConcept libraryPhysicsMechanicsRotational Mechanics

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.

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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.

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Why it behaves this way

Explanation

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.

Frozen walkthrough

Step through the frozen example

Frozen walkthrough

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.

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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

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?

Make a prediction before you reveal the next step.

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 against the live bench.

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?

Use the live bench to test the result before moving on.

Accessibility

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

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Support-region balanceMechanics

Torque

Rotational Mechanics

Explanation

Step through the frozen example

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 this idea moving

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?