Torque Calculations

Why This Matters

Torque is the rotational equivalent of force. It's what makes things spin, tip, or stay balanced. In AP Physics 1, you need to analyze why objects rotate (or don't), which means understanding how force, distance, and angle combine to create rotational effects. This connects directly to Newton's Second Law in rotational form ($\tau_{net} = I\alpha$), rotational equilibrium ($\Sigma\tau = 0$), and the broader theme of extending linear mechanics into the rotational world.

Don't just memorize $\tau = rF\sin\theta$. Know when each variable matters and how to set up problems strategically. The exam loves asking you to compare torques from different forces, choose clever pivot points to simplify calculations, and explain why an object accelerates rotationally or stays in equilibrium.

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The Torque Formula and Its Components

Torque measures how effectively a force causes rotation about an axis. The key idea: only the component of force perpendicular to the lever arm contributes to rotation. Force directed along the lever arm just pushes toward or away from the pivot and does nothing rotationally.

The Cross Product Definition: $\tau = r \times F$

• Torque is a vector quantity with both magnitude and direction. The direction tells you the axis and sense of rotation.
• The position vector $r$ extends from the axis of rotation to the point where the force is applied. It's not just any distance in the problem.
• Only the perpendicular component matters. The cross product naturally extracts just the part of the force that causes rotation.

The Scalar Calculation: $\tau = rF\sin\theta$

• $\theta$ is the angle between $r$ and $F$ when they're placed tail-to-tail. This determines what fraction of the force actually contributes to rotation.
• Maximum torque occurs at $\theta = 90°$ because $\sin(90°) = 1$, meaning the entire force is perpendicular to the lever arm.
• Zero torque occurs at $\theta = 0°$ or $180°$. In these cases the force points along the lever arm, producing no rotational effect at all.

The Lever Arm (Moment Arm) Approach

• The lever arm $r_{\perp}$ is the perpendicular distance from the axis of rotation to the line of action of the force (the infinite line extending along the force vector).
• $\tau = r_{\perp}F$ gives the same result as $rF\sin\theta$ but uses geometry instead of trigonometry.
• Choose whichever method fits the problem. Sometimes drawing the perpendicular distance on a diagram is faster than finding angles, especially when the geometry is simple.

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Units and Sign Conventions

Keeping track of units and directions prevents careless errors and ensures your answers make physical sense.

Units of Torque: Newton-meters (N·m)

• The SI unit is the Newton-meter (N·m): force (N) multiplied by distance (m), reflecting torque's dependence on both.
• N·m is dimensionally identical to Joules, but torque and energy are different physical quantities. Never write torque in "J." The exam will count that as an error.
• Use dimensional analysis to check your work. If your answer doesn't come out in N·m, something went wrong.

Sign Convention for Torque Direction

• Counterclockwise (CCW) is typically positive, matching the standard mathematical convention for angles.
• Clockwise (CW) is typically negative, but always check what convention your problem establishes.
• Consistency is critical. Once you choose a sign convention, apply it to every torque in the problem without switching partway through.

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Calculating Net Torque from Multiple Forces

Real systems have multiple forces acting at once, so you need to combine their rotational effects algebraically.

Summing Individual Torques

• Net torque is the algebraic sum: $\Sigma\tau = \tau_1 + \tau_2 + \tau_3 + ...$ with signs based on each torque's rotation direction.
• Each torque requires its own $r$, $F$, and $\theta$. Don't assume all forces act at the same point or angle.
• Torques can partially or fully cancel. Two equal-magnitude torques in opposite directions produce zero net torque.

Strategic Pivot Point Selection

You can choose any point as your axis of rotation for torque calculations. The physics works out the same regardless, but some choices simplify the math dramatically.

The strategy: place the pivot at the location of an unknown force. Since $r = 0$ at that point, that force's torque vanishes from your equation entirely. This reduces the number of unknowns and is especially useful on FRQs where you're solving for a specific force or mass.

Torque from Gravity (Weight)

• Weight acts at the center of mass. You can treat the entire gravitational force as if it's applied at this single point.
• $\tau_{gravity} = r_{cm} \cdot mg \cdot \sin\theta$, where $r_{cm}$ is the distance from the pivot to the center of mass and $\theta$ is the angle between $r_{cm}$ and the weight vector.
• When weight points straight down (as it usually does) and the object is horizontal, the lever arm is simply the horizontal distance from the pivot to the center of mass. This is a common source of errors on FRQs, so draw it out carefully.

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Rotational Equilibrium: $\Sigma\tau = 0$

When net torque is zero, an object either doesn't rotate or rotates at constant angular velocity. This is the rotational analog of Newton's First Law.

The Equilibrium Condition

• $\Sigma\tau = 0$ means no angular acceleration. The object maintains its current rotational state (at rest or constant $\omega$).
• Rotational equilibrium is independent of translational equilibrium. An object can spin at a constant rate while accelerating linearly, or vice versa.
• Static equilibrium requires both conditions: $\Sigma F = 0$ AND $\Sigma\tau = 0$ for objects that are completely at rest.

Solving Equilibrium Problems

Follow this process:

1. Draw a clear force diagram showing all forces, their points of application, and distances from the pivot.
2. Choose a pivot point strategically, ideally at the location of an unknown force you don't need to solve for.
3. Calculate each torque using $\tau = rF\sin\theta$ or $\tau = r_{\perp}F$, assigning positive to CCW and negative to CW (or vice versa, as long as you're consistent).
4. Write $\Sigma\tau = 0$ and solve for the unknown force, distance, or mass.
5. If needed, use $\Sigma F = 0$ as a second equation to find remaining unknowns.

Balancing Beams and Levers

• Classic setup: forces on opposite sides of a pivot. The principle of moments states $r_1F_1 = r_2F_2$ for balance.
• Reaction forces at the pivot create zero torque (since $r = 0$). This is another reason to place your axis at support points.
• Multiple supports require careful analysis. Each support exerts an unknown force, and you may need both $\Sigma F = 0$ and $\Sigma\tau = 0$ to solve for all of them.

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Newton's Second Law in Rotational Form: $\tau_{net} = I\alpha$

When net torque isn't zero, the object undergoes angular acceleration proportional to the torque and inversely proportional to the rotational inertia.

The Fundamental Relationship

• $\alpha = \frac{\Sigma\tau}{I}$ directly parallels $a = \frac{\Sigma F}{m}$. Torque plays the role of force; moment of inertia plays the role of mass.
• Larger moment of inertia means smaller angular acceleration for the same torque. How mass is distributed relative to the axis matters, not just total mass.
• The direction of $\alpha$ matches the direction of net torque. If net torque is CCW, angular acceleration is CCW.

Connecting to Linear Motion

• Rolling without slipping links linear and angular motion through the constraint $a_{cm} = R\alpha$, connecting translational and rotational acceleration.
• Static friction provides the torque for rolling. It acts at the contact point and creates torque about the center of mass. It's static friction, not kinetic, because the contact point doesn't slide against the surface.
• For rolling down an incline: $a = \frac{g\sin\theta}{1 + I/(MR^2)}$. Objects with larger $I$ (relative to $MR^2$) accelerate more slowly. For example, a hoop ($I = MR^2$) rolls slower than a solid sphere ($I = \frac{2}{5}MR^2$) down the same ramp because more of the hoop's mass sits far from the axis.

Angular Impulse and Momentum Change

• Angular impulse $= \tau \Delta t$, the rotational analog of linear impulse ($F \Delta t$).
• $\tau \Delta t = \Delta L = I\Delta\omega$: torque applied over time changes angular momentum.
• On $\tau$ vs. $t$ graphs, the area under the curve equals the angular impulse delivered, just as the area under an $F$ vs. $t$ graph gives linear impulse.

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Quick Reference Table

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Self-Check Questions

1. A force is applied to a door at an angle of 30° to the door's surface. How does the torque compare to the same force applied perpendicular to the door, and why?

2. Two students sit on opposite sides of a seesaw. Student A (mass 40 kg) sits 2 m from the pivot. Where must Student B (mass 60 kg) sit for rotational equilibrium? Which equation and principle did you use?

3. Compare the angular acceleration of a solid disk ($I = \frac{1}{2}MR^2$) vs. a hoop ($I = MR^2$) of the same mass and radius when identical torques are applied. Which accelerates faster, and what property explains the difference?

4. When solving for an unknown support force on a beam, why is it strategically useful to place your pivot point at a different support location?

5. A constant torque of 8 N·m is applied to a wheel for 3 seconds. If the wheel starts at rest and has moment of inertia $I = 4 \text{ kg·m}^2$, what is its final angular velocity? Identify which form of Newton's Second Law you used.