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Torque and Work

Lesson ~10 min read 8 MCQs

In simple terms: In simple terms, rotational work is about how a twist or torque, applied over a certain angle, transfers energy to a spinning object, making it speed up or slow down.

Why this matters

Have you ever tried to loosen a stubborn lug nut on a car tire? You grab a wrench, fit it on the nut, and push. If you push with a tiny force, nothing happens. If you use a short wrench, it’s tough. But if you use a long wrench and apply a solid, steady force, the nut begins to turn. That turning action is you doing work. You’re not just pushing in a straight line; you’re applying a torque over an angle.

This is the same physics that spins a merry-go-round, opens a heavy castle door, or even gets a wind turbine moving. In this lesson, we’ll connect the familiar idea of work to the world of rotation. We’ll see how torque does the job that force does for linear motion.

Diagram

Work Done by a Constant Torque A two-part diagram. On the left, a disk of radius r=0.5m has a tangential force F=10N applied, causing it to rotate by Δθ=2.0 rad. On the right, a graph of Torque vs. Angular Position shows a constant torque of 5.0 N·m. The area under the graph from 0 to 2.0 rad is shaded, representing 10 J of work. Physical Scenario r = 0.5 m F = 10 N Δθ = 2.0 rad Torque vs. Angle Graph θ (rad) Torque, τ (N·m) 0 2.0 5.0 Work = Area = τΔθ = (5.0 N·m)(2.0 rad) = 10 J
This diagram shows the relationship between rotational work, torque, and angular displacement. On the left, a physical diagram shows a 10N force applied to a disk of 0.5m radius, causing it to rotate 2.0 radians. On the right, a graph plots the resulting constant 5.0 N·m torque against the angular position, with the area under the curve shaded to show the calculated work of 10 Joules.

Concept map

flowchart TD
    A[Force F applied at radius r] --> B{Creates Torque τ = rF};
    B --> C[Torque applied over an angular displacement Δθ];
    C --> D{Does Rotational Work W = τΔθ};
    D --> E[Changes Rotational Kinetic Energy ΔK_rot];
    subgraph Legend
        direction LR
        F1(Start) --> F2(Process) --> F3(Result);
    end

Core explanation

Hey everyone, it’s Saavi. Let's dive into how energy works in rotating systems.

You already have a solid foundation for the concept of work from linear physics. Remember the work-energy theorem? When you apply a net force to an object over a distance, you do work on it, and that work changes its kinetic energy.

The equation you know and love is: Work = Force × distance W = Fd

Now, let's switch our thinking from straight lines to circles. In rotation, every linear concept has a rotational twin.

  • Linear displacement (d) becomes angular displacement (Δθ).
  • Linear velocity (v) becomes angular velocity (ω).
  • Force (F) becomes torque (τ).

So, if Work = Force × distance, what do you think the equation for rotational work will be? We just swap the linear variables for their rotational twins.

Rotational Work = Torque × angular displacement W = τΔθ

This simple-looking equation is incredibly powerful. It tells us that a torque does work only when it causes a rotation. If you apply a massive torque to a bolt that’s rusted shut and it doesn't move (Δθ = 0), you’ve done zero physics work, no matter how tired you get!

How Torque Transfers Energy

Work is a measure of energy transfer.

  • If you apply a torque in the same direction that an object is already spinning, you do positive work. You are adding energy to the system, making it spin faster. Think of pushing a friend on a merry-go-round to make it go faster.
  • If you apply a torque opposite to the direction of rotation, you do negative work. You are removing energy from the system, making it spin slower. This is like dragging your foot to slow that same merry-go-round down.

This directly connects to the work-energy theorem for rotation: the net work done by torques on an object equals the change in its rotational kinetic energy. W_net = ΔK_rot = (1/2)Iω_f² - (1/2)Iω_i² (We'll explore this more in the next topic, but it's good to see the connection now!)

The "Area Under the Curve" Connection

Here’s a concept that you'll see again and again in AP Physics. Just like work is the area under a Force vs. Position graph, rotational work is the area under a Torque vs. Angular Position graph.

Let's imagine a simple scenario, like the one in our main diagram. A constant force of 10 Newtons is applied to the edge of a disk with a radius of 0.5 meters.

First, let's find the torque. Since the force is tangential, the angle is 90 degrees, and sin(90) = 1. τ = rF_⊥ = (0.5 m)(10 N) = 5.0 N·m

Now, let's say this constant torque causes the disk to rotate through an angle of 2.0 radians.

We can calculate the work done with our new formula: W = τΔθ = (5.0 N·m)(2.0 rad) = 10 J

Let's visualize this on a graph of Torque vs. Angular Position (θ).

  • The y-axis is Torque (τ).
  • The x-axis is Angular Position (θ).

Since our torque is constant at 5.0 N·m, the graph is just a horizontal line at y = 5.0. The rotation happens from θ = 0 to θ = 2.0 rad. The area under this "curve" is just a rectangle.

Area = height × width = τ × Δθ = (5.0 N·m)(2.0 rad) = 10 J

It's the exact same result! This graphical method is essential. The AP exam might not give you a constant torque. They might give you a graph where the torque changes, and you'll have to find the area of the shape (like a triangle or a trapezoid) to find the total work done.

This is where many students slip up: They see a graph and try to plug the final value of torque into the W = τΔθ formula. That formula only works if the torque is constant. If the torque changes, you must find the area under the graph.

And one final, crucial point: the angular displacement Δθ in this formula must be in radians. If you're given degrees, your first step is always to convert. The entire framework of rotational physics is built on radians, and using degrees will give you the wrong answer every time.

Worked examples

Let's walk through a couple of problems together to make sure this clicks.

Example 1

Constant Torque on a Flywheel

Problem: A motor applies a constant torque of 50 N·m to a large flywheel. How much work does the motor do on the flywheel as it rotates through 4.0 full revolutions?

Solution:

  1. 1
    Identify the Goal
    We need to find the work done (W). We are given a constant torque (τ) and a rotational displacement.
  2. 2
    Recall the Formula
    The formula for work done by a constant torque is W = τΔθ.
  3. 3
    Check the Units
    The torque is τ = 50 N·m, which is great. The angular displacement is given in revolutions. This is the trap! The formula requires Δθ to be in radians.
  4. 4
    Convert Angular Displacement
    We need to convert 4.0 revolutions into radians. We know that one full revolution is 2π radians. Δθ = 4.0 rev × (2π rad / 1 rev) = 8π rad
  5. 5
    Calculate the Work
    Now we can plug our values into the formula. W = τΔθ W = (50 N·m)(8π rad) W ≈ (50)(25.13) J W ≈ 1257 J or 1.26 kJ

Why this matters: The most common mistake here is forgetting to convert revolutions to radians. If you had used Δθ = 4, your answer would be 200 J, which is incorrect and would definitely be a distractor choice on a multiple-choice question. Always check your units for Δθ.

Example 2

Work from a Graph

Problem: The graph below shows the net torque applied to a propeller as it rotates. What is the total work done on the propeller as it rotates from θ = 0 to θ = 5.0 rad?

(Imagine a graph where the y-axis is Torque (N·m) and the x-axis is Angular Position (rad). The line on the graph starts at (0, 20), stays horizontal until (3, 20), then slopes down linearly to (5, 0).)

Solution:

  1. 1
    Identify the Goal
    We need to find the total work done from a graph of torque vs. angular position.
  2. 2
    Recall the Concept
    Work is the area under the τ-θ graph.
  3. 3
    Break the Area into Shapes
    The torque is not constant, so we can't just use W = τΔθ. We need to find the area. The shape under the graph from θ = 0 to θ = 5 is a trapezoid. A simpler way is to break it into a rectangle and a triangle.
    • Shape 1 (Rectangle)
      From θ = 0 to θ = 3.0 rad.
      • Height = τ = 20 N·m
      • Width = Δθ = 3.0 rad
      • Area₁ = height × width = (20 N·m)(3.0 rad) = 60 J
    • Shape 2 (Triangle)
      From θ = 3.0 to θ = 5.0 rad.
      • Height = τ = 20 N·m
      • Base = Δθ = 5.0 rad - 3.0 rad = 2.0 rad
      • Area₂ = (1/2) × base × height = (1/2)(2.0 rad)(20 N·m) = 20 J
  4. 4
    Calculate Total Work
    The total work is the sum of the areas. W_total = Area₁ + Area₂ = 60 J + 20 J = 80 J

Why this matters: This shows how to handle non-constant torques. You can't just pick one value of torque. You must find the total area, which represents the total work done over the entire angular displacement.

Try it yourself

Time to try a couple on your own. Don't worry about getting it perfect, focus on setting up the problem correctly.

Problem 1: Priya is using a wrench to tighten a bolt on her bike. She applies a constant torque of 15 N·m. The bolt turns 90 degrees before it is fully tightened. How much work did Priya do on the bolt?

Problem 2: A spinning potter's wheel is slowing down due to friction. The frictional torque is not constant; it increases as the wheel's lubricant warms up. A graph of the frictional torque vs. angle shows a straight line going from τ = -0.5 N·m at θ = 0 to τ = -1.5 N·m at θ = 10 rad. How much work did friction do on the wheel?

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