5.3 Torque

TLDR

Torque is the twisting effect a force has about an axis of rotation, and it depends on how hard you push, where you push, and the direction of your push. You calculate its magnitude with $\tau = rF\sin\theta = rF_{\perp}$, where only the perpendicular part of the force actually causes rotation. In AP Physics 1, you work with the size of the torque, not its direction.

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AP Physics Torque

In AP Physics 1, torque measures how strongly a force tends to rotate a rigid system about an axis. Torque depends on the force size, the distance from the axis to where the force is applied, and the angle between the force and that position vector.

The key idea is that only the perpendicular part of the force creates torque. Use $\tau = rF\sin\theta$ when you know the angle, or $\tau = rF_{\perp}$ when you already have the perpendicular force component. A longer lever arm or a more perpendicular force gives a larger torque.

Why This Matters for the AP Physics 1 Exam

Torque is the rotational version of force, and it sets up almost everything else in Unit 5, including rotational inertia, rotational equilibrium, and Newton's second law in rotational form. On the exam, you need to identify which forces produce torque, build force diagrams that show where each force acts relative to the axis, and calculate torque magnitudes.

This topic leans heavily on functional relationships, like predicting how torque changes when you double the force or move the pivot. That kind of reasoning shows up in the multiple-choice section and in free-response questions, including the Qualitative/Quantitative Translation question, where you explain how one quantity changes when another changes.

Key Takeaways

• Only the force component perpendicular to the position vector $r$ produces torque; the parallel part does nothing.
• The lever arm is the perpendicular distance from the axis of rotation to the line of action of the force.
• Calculate torque magnitude with $\tau = rF\sin\theta = rF_{\perp}$, measured in N·m.
• Torque is largest when the force is applied at $90^\circ$ to $r$, and zero when the force points along $r$.
• Force diagrams are like free-body diagrams but also show where each force acts relative to the axis.
• AP Physics 1 only deals with the magnitude of torque, not its direction.

How Torque Works

Only the perpendicular force component matters

Torque comes only from the force component that is perpendicular to the position vector $r$, which runs from the axis of rotation to the point where the force is applied.

• When you apply a force at an angle, split it into a piece perpendicular to $r$ and a piece parallel to $r$.
• The perpendicular piece causes rotation.
• The parallel piece pushes straight toward or away from the axis and produces no torque.

The lever arm

The lever arm is the perpendicular distance from the axis of rotation to the line of action of the force (the line the force points along, extended in both directions).

• A longer lever arm means more torque from the same force.
• Torque is maximized when the force is applied perpendicular to $r$.
• This is why wrench handles are long and why door handles sit far from the hinges. Pushing far from the axis gives more turning effect for the same push.

Force diagrams

Force diagrams are the main tool for analyzing rotation, and they work a lot like free-body diagrams.

• They show the relative magnitude and direction of each force on a rigid system.
• They also show where each force acts relative to the axis of rotation, which free-body diagrams usually leave out.
• That location matters because the same force can produce very different torques depending on how far from the axis it acts.

The torque equation

The magnitude of the torque from a single force is:

$\tau = rF\sin\theta = rF_{\perp}$

• $r$ is the distance from the axis of rotation to the point where the force is applied.
• $F$ is the magnitude of the applied force.
• $\theta$ is the angle between the force vector and the position vector $r$.
• When $\theta = 90^\circ$, the torque is at its maximum: $\tau_{max} = rF$.
• When $\theta = 0^\circ$ or $180^\circ$, the force points along $r$ and the torque is zero.

How to Use This on the AP Physics 1 Exam

Problem Solving

Practice Problem 1: Torque Calculation

Solution:

1. Identify the given values:

   • Force (F) = 15 N
   • Distance from axis of rotation (r) = 0.75 m
   • Angle (θ) = 90° (perpendicular force)

2. Apply the torque equation: $\tau = rF\sin\theta$ $\tau = (0.75 \text{ m})(15 \text{ N})(\sin 90°)$ $\tau = (0.75 \text{ m})(15 \text{ N})(1)$ $\tau = 11.25 \text{ N·m}$

3. The torque applied to the door is 11.25 N·m.

Practice Problem 2: Lever Arm Analysis

Solution:

1. For part (a):

   • Force (F) = 50 N
   • Distance (r) = 0.3 m
   • Angle (θ) = 30° $\tau = rF\sin\theta$ $\tau = (0.3 \text{ m})(50 \text{ N})(\sin 30°)$ $\tau = (0.3 \text{ m})(50 \text{ N})(0.5)$ $\tau = 7.5 \text{ N·m}$

2. For part (b), perpendicular force (θ = 90°): $\tau_{max} = rF$ $\tau_{max} = (0.3 \text{ m})(50 \text{ N})$ $\tau_{max} = 15 \text{ N·m}$

3. The increase in torque would be: $\Delta\tau = 15 \text{ N·m} - 7.5 \text{ N·m} = 7.5 \text{ N·m}$

4. The torque would double (increase by 100%) if applied perpendicularly.

Practice Problem 3: Identifying Torque Magnitude

Solution:

1. Identify the given values:

   • Force (F) = 20 N
   • Distance from axis (r) = 0.40 m
   • Angle (θ) = 60°

2. Apply the torque equation: $\tau = rF\sin\theta$ $\tau = (0.40\,\text{m})(20\,\text{N})(\sin 60°)$ $\tau = 8(0.866)\,\text{N·m}$ $\tau = 6.93\,\text{N·m}$

3. The magnitude of the torque is 6.93 N·m.

Common Trap

• Watch the angle in $\sin\theta$. It is the angle between the force and the position vector $r$, not the angle the force makes with the ground or some other reference. Drawing $r$ and the force on the same diagram helps you find the right angle.
• When a question asks how torque changes, reason with functional dependence. If $F$ doubles and everything else stays the same, $\tau$ doubles. If $\theta$ goes from $30^\circ$ to $90^\circ$, $\sin\theta$ goes from $0.5$ to $1$, so the torque doubles.

Common Misconceptions

• Torque is not the same as force. Force can cause linear motion, but torque is specifically the turning effect about an axis. A large force applied right at the axis produces zero torque.
• The full force does not always count. Only the component perpendicular to $r$ produces torque. If you forget to use $\sin\theta$ or the perpendicular component, you will overestimate the torque.
• A force pointing straight toward or away from the axis gives zero torque, because it lines up with $r$ and has no perpendicular component.
• Bigger force does not guarantee bigger torque. Where the force is applied and at what angle matter just as much. A smaller force with a long lever arm can out-twist a large force applied close to the axis.
• Direction of torque is not tested in AP Physics 1. You work with the magnitude only, so do not waste time assigning a 3D direction to torque on this exam.

Related AP Physics 1 Guides

• 5.1 Rotational Kinematics
• 5.2 Connecting Linear and Rotational Motion
• 5.4 Rotational Inertia
• 5.5 Rotational Equilibrium and Newton's First Law in Rotational Form
• 5.6 Newton's Second Law in Rotational Form