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're being tested on your ability 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. Master the conceptual "why" behind each calculation, and you'll handle both multiple-choice and FRQ scenarios with confidence.

---

The Torque Formula and Its Components

Torque measures how effectively a force causes rotation about an axis. The key insight is that only the perpendicular component of force contributes—force directed along the lever arm does nothing rotationally.

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

• Torque is a vector quantity—it has both magnitude and direction, with the direction indicating the axis and sense of rotation
• The position vector $r$ extends from the axis of rotation to the point where the force is applied, not just any distance
• Only the perpendicular component matters—this is why the cross product naturally extracts the rotational effect

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

• $\theta$ is the angle between $r$ and $F$—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 $180°$—force along the lever arm pushes toward or away from the pivot but causes no rotation

The Lever Arm (Moment Arm) Approach

• The lever arm is the perpendicular distance from the axis of rotation to the line of action of the force—an alternative way to visualize torque
• $\tau = r_{\perp}F$ 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 is faster than finding angles

---

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)

• 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 represents a different physical quantity—never write torque in "J"
• Dimensional analysis helps check work—if your answer isn't in N·m, something went wrong in your calculation

Sign Convention for Torque Direction

• Counterclockwise (CCW) is typically positive—this matches 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

---

Calculating Net Torque from Multiple Forces

Real systems have multiple forces acting simultaneously, so you must combine their rotational effects algebraically.

Summing Individual Torques

• Net torque is the algebraic sum $\Sigma\tau = \tau_1 + \tau_2 + \tau_3 + ...$, accounting for signs based on 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 axis of rotation for torque calculations—the physics works out the same, but some choices simplify math dramatically
• Place the pivot at an unknown force's location to eliminate that force from your torque equation (since $r = 0$ means $\tau = 0$)
• This technique is essential for FRQs—it reduces the number of unknowns and speeds up problem-solving

Torque from Gravity (Weight)

• Weight acts at the center of mass—treat the entire gravitational force as 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
• Horizontal distance matters for vertical weight—when weight points straight down, the lever arm is the horizontal distance to the pivot

---

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 constant rate while accelerating linearly, or vice versa
• Both conditions together define static equilibrium—$\Sigma F = 0$ AND $\Sigma\tau = 0$ for objects completely at rest

Solving Equilibrium Problems

• Draw a clear force diagram showing all forces, their points of application, and distances from the pivot
• Write $\Sigma\tau = 0$ with consistent signs—set CCW torques positive and CW torques negative (or vice versa)
• Solve for the unknown—equilibrium problems typically ask for an unknown force, distance, or mass

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—another reason to place your axis at support points
• Multiple supports require careful analysis—each support exerts an unknown force that contributes to equilibrium

---

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 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—mass distribution matters
• Direction of $\alpha$ matches 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—the constraint $a_{cm} = r\alpha$ connects translational and rotational acceleration
• Friction provides the torque for rolling—static friction at the contact point creates torque about the center of mass
• For rolling down an incline: $a = \frac{g\sin\theta}{1 + I/(MR^2)}$—objects with larger $I$ accelerate more slowly

Angular Impulse and Momentum Change

• Angular impulse = $\tau \Delta t$—the rotational analog of linear impulse
• $\tau \Delta t = \Delta L = I\Delta\omega$—torque applied over time changes angular momentum
• Graphs of $\tau$ vs. $t$—the area under the curve equals angular impulse delivered

---

Quick Reference Table

---

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 vs. a hoop 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.