11.2 Work Done by Torques

The Wrench on the Bolt

You tighten a bolt with a wrench. You grip the handle, apply a force, and the wrench rotates through an angle. You can feel the effort --- energy is being transferred from your muscles into the bolt's resistance. The harder you push and the further you turn, the more energy you transfer.

But how much, exactly?

In Chapter 7, you learned that translational work is force times displacement --- or more precisely, the component of force along the displacement times the distance traveled. The formula $W = Fd\cos\theta$ (or its integral form $W = \int \vec{F} \cdot d\vec{r}$) told you exactly how much energy a force transfers through a displacement.

Now we need the rotational version. A torque acts through an angular displacement. The energy transfer should depend on how large the torque is and how far the object rotates. The question is: what is the precise relationship?

Before you read on: You apply a constant torque of 10 N$\cdot$m to a wheel and rotate it through half a revolution. How much work is done?

Commit to a numerical answer before continuing. (Hint: be careful with units of angle.)

The Guiding Question

How does torque transfer energy into or out of rotational motion?

In Section 11.1, you learned that rotational kinetic energy is $KE_{\text{rot}} = \frac{1}{2}I\omega^2$. A spinning object stores energy. But how does that energy get put in or taken out? When you spin up a wheel from rest, something must do work on it --- and that something is a torque acting through an angular displacement. This section makes that connection precise.

Building the Formula from the Translational Case

Before turning to the interactive, let's see why the formula should take the form it does.

In translational dynamics, a force $F$ acting on an object that moves a small distance $dx$ does work:

$$dW = F\,dx$$

Now think about a force applied tangentially to a wheel of radius $r$. The force acts at the rim, and as the wheel rotates through a small angle $d\theta$, the point of application moves a distance $ds = r\,d\theta$ along the arc. The work done is:

$$dW = F\,ds = F \cdot r\,d\theta$$

But $Fr$ is the torque $\tau$ produced by that tangential force. So:

$$dW = \tau\,d\theta$$

That is the result: the work done by a torque in a small rotation is the torque times the angular displacement. The structure mirrors $dW = F\,dx$ exactly --- just with rotational quantities in place of translational ones.

Exploration: Turning the Bolt

[Interactive: Wrench-and-Bolt Work Accumulator. A top-down view shows a bolt at the center of the screen with a wrench attached. The student drags the wrench handle to rotate it. Two panels update in real time:

Left panel: The wrench and bolt. An arc shows the angle turned so far, labeled $\theta$ in radians (and degrees in parentheses). The applied torque is displayed numerically. The student can choose between "Constant torque" and "Variable torque" modes using a toggle.
Right panel: A torque-versus-angle graph ($\tau$ on the vertical axis, $\theta$ on the horizontal axis). As the wrench turns, the curve is drawn and the area under it fills in with blue shading. A numerical readout shows: current angle $\theta$, current torque $\tau(\theta)$, and accumulated work $W$ (the shaded area).

Mode 1 --- Constant torque: The torque is fixed at a value the student sets with a slider (default: 10 N$\cdot$m). The $\tau(\theta)$ graph is a horizontal line. The shaded area is a rectangle.

Guided prompts: - "Set the torque to 10 N$\cdot$m. Rotate the wrench through $\pi$ radians (half a revolution). Read the work from the display. Does it match your prediction from the opening question?" - "Now rotate through a full revolution ($2\pi$ radians). How does the work compare to the half-revolution case?" - "Double the torque to 20 N$\cdot$m and rotate through $\pi$ radians again. What happened to the work?"

Mode 2 --- Variable torque: The torque increases linearly with angle: $\tau(\theta) = k\theta$, representing a bolt that gets harder to turn as it tightens. The student sets the constant $k$ with a slider.

Guided prompts: - "In this mode, the torque-angle graph is a straight line from zero. What shape is the shaded area? How would you compute it geometrically?" - "Compare the work for turning from $\theta = 0$ to $\theta = \pi$ versus turning from $\theta = \pi$ to $\theta = 2\pi$. Which half-revolution involves more work? Why?" - "How does this compare to stretching a spring in Section 7.2, where the work was the area under the $F(x)$ curve?"]

What the Exploration Reveals

If you worked through both modes, here is what you should have found.

Constant torque. The $\tau(\theta)$ graph is a horizontal line. The area under it is a rectangle: height $\tau$, width $\Delta\theta$. So:

$$W = \tau \, \Delta\theta$$

For 10 N$\cdot$m through half a revolution: $W = 10 \times \pi = 31.4$ J. If you predicted $W = 10 \times 0.5 = 5$ J, you fell into the unit trap --- "half a revolution" is $\pi$ radians, not 0.5. More on this common error in Practice Layer 4.

Variable torque. When the torque depends on the angle, you cannot simply multiply. You must add up contributions from every tiny rotation $d\theta$. The total work is the area under the $\tau(\theta)$ curve:

$$W = \int_{\theta_i}^{\theta_f} \tau(\theta)\,d\theta$$

For the linear case $\tau(\theta) = k\theta$, the area is a triangle: $W = \frac{1}{2}k\theta_f^2$ (starting from $\theta_i = 0$). The second half-revolution always requires more work than the first, because the torque is larger throughout that portion. This is exactly the same idea as the spring in Section 7.2, where $W = \frac{1}{2}kx^2$ for a linearly increasing force.

The Concept: Work by a Torque

Here is the general result.

The work done by a torque $\tau$ on an object that rotates through an angular displacement from $\theta_i$ to $\theta_f$ is:

$$W = \int_{\theta_i}^{\theta_f} \tau\,d\theta$$

For a constant torque, this simplifies to:

$$W = \tau\,\Delta\theta$$

where $\Delta\theta = \theta_f - \theta_i$ is the angular displacement in radians.

The structural parallel to translational work is exact:

Translational	Rotational
Force $F$	Torque $\tau$
Displacement $dx$	Angular displacement $d\theta$
$W = \int F\,dx$	$W = \int \tau\,d\theta$
$W = F\,\Delta x$ (constant $F$)	$W = \tau\,\Delta\theta$ (constant $\tau$)

Every structural feature carries over. The integral accumulates contributions. For constant quantities, the integral collapses to a simple product. The sign of the work tells you whether energy flows into the object (positive) or out of it (negative) --- a braking torque opposing the rotation does negative work, just as friction opposing translational motion does negative work.

Units

Work is still measured in joules. Torque has units of N$\cdot$m. Angle in radians is dimensionless. So:

$$[\tau\,\Delta\theta] = \text{N}\cdot\text{m} \cdot \text{rad} = \text{N}\cdot\text{m} = \text{J}$$

This is why radians are essential here. Degrees are not dimensionless in the same way, and using them would give the wrong numerical answer.

The Rotational Work-Energy Theorem

In Chapter 7, you saw that the net work done on an object equals its change in translational kinetic energy:

$$W_{\text{net}} = \Delta KE_{\text{trans}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$

The rotational version follows the same logic. If a net torque $\tau_{\text{net}}$ acts on a rigid body rotating about a fixed axis, Newton's second law for rotation gives $\tau_{\text{net}} = I\alpha$. The net work is:

$$W_{\text{net}} = \int_{\theta_i}^{\theta_f} \tau_{\text{net}}\,d\theta = \int_{\theta_i}^{\theta_f} I\alpha\,d\theta$$

Using the chain rule ($\alpha = d\omega/dt$ and $d\theta = \omega\,dt$, so $\alpha\,d\theta = \omega\,d\omega$):

$$W_{\text{net}} = \int_{\omega_i}^{\omega_f} I\omega\,d\omega = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$$

This is the rotational work-energy theorem:

$$\boxed{W_{\text{net}} = \Delta KE_{\text{rot}} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2}$$

The net work done by all torques equals the change in rotational kinetic energy. The derivation is identical in structure to the translational case --- just replace $F$ with $\tau$, $m$ with $I$, $v$ with $\omega$, and $x$ with $\theta$.

Pause and think: A flywheel spinning at 100 rad/s is brought to rest by a braking torque. Is the net work positive or negative? Where does the kinetic energy go?

Check your answer

The net work is negative. The braking torque opposes the rotation, so $\tau$ and $d\theta$ point in opposite directions. The flywheel goes from $\omega_i = 100$ rad/s to $\omega_f = 0$, so $\Delta KE_{\text{rot}} = 0 - \frac{1}{2}I(100)^2 < 0$. The kinetic energy is converted to thermal energy (heat) in the braking mechanism, just as friction converts translational kinetic energy to heat.

Connection to Translational Work

It is worth pausing to see just how tight the parallel is.

In Section 7.1, you learned that $W = \int F\,dx$: work is force accumulated over displacement. The force causes a change in velocity, and the total work equals the change in $\frac{1}{2}mv^2$.

Here, $W = \int \tau\,d\theta$: work is torque accumulated over angular displacement. The torque causes a change in angular velocity, and the total work equals the change in $\frac{1}{2}I\omega^2$.

The mathematical structure is the same. The physical interpretation is the same. The underlying reason is the same: the work-energy theorem is a consequence of Newton's second law combined with a kinematic identity, and that derivation works in both the translational and rotational domains.

This is not a coincidence. It reflects the deep structural parallel between translational and rotational dynamics that has been building since Chapter 9. Every translational concept has a rotational counterpart, and the relationships between them carry over unchanged.

Practice

Layer 1: Concrete

Problem 1. A constant torque of 25 N$\cdot$m is applied to a wheel, rotating it through $3.0$ rad. How much work is done by the torque?

Check your answer

For constant torque: $$W = \tau\,\Delta\theta = (25)(3.0) = 75 \text{ J}$$

Problem 2. A motor applies a constant torque of 8.0 N$\cdot$m to a shaft that makes 20 complete revolutions. How much work does the motor do?

Check your answer

First, convert revolutions to radians: $$\Delta\theta = 20 \times 2\pi = 40\pi \text{ rad} \approx 125.7 \text{ rad}$$ Then: $$W = \tau\,\Delta\theta = (8.0)(40\pi) = 320\pi \approx 1005 \text{ J}$$ About 1.0 kJ of work. Note that the conversion to radians is essential --- using $\Delta\theta = 20$ would give a wildly wrong answer.

Problem 3. A grinding disk ($I = 0.40 \text{ kg}\cdot\text{m}^2$) is spinning at 50 rad/s. A braking torque brings it to rest. How much work does the braking torque do?

Check your answer

Use the rotational work-energy theorem: $$W_{\text{net}} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2 = \frac{1}{2}(0.40)(0)^2 - \frac{1}{2}(0.40)(50)^2 = 0 - 500 = -500 \text{ J}$$ The work is negative because the braking torque opposes the rotation. The 500 J of rotational kinetic energy is removed from the disk (converted to thermal energy in the brake).

Layer 2: Pattern

Problem 4. The graph below shows the torque applied to a wheel as a function of angle.

[Graph description: $\tau(\theta)$ is plotted from $\theta = 0$ to $\theta = 4$ rad. From $\theta = 0$ to $\theta = 2$ rad, the torque is constant at 15 N$\cdot$m. From $\theta = 2$ to $\theta = 4$ rad, the torque decreases linearly from 15 N$\cdot$m to 0.]

(a) Find the work done from $\theta = 0$ to $\theta = 2$ rad.

(b) Find the work done from $\theta = 2$ to $\theta = 4$ rad.

(c) Find the total work done from $\theta = 0$ to $\theta = 4$ rad.

Check your answer

The work is the area under the $\tau(\theta)$ curve --- exactly the same idea as finding work from a $F(x)$ graph in Chapter 7. **(a)** From $\theta = 0$ to $\theta = 2$ rad, the torque is constant at 15 N$\cdot$m. The area is a rectangle: $$W_1 = 15 \times 2 = 30 \text{ J}$$ **(b)** From $\theta = 2$ to $\theta = 4$ rad, the torque decreases linearly from 15 to 0. The area is a triangle: $$W_2 = \frac{1}{2}(2)(15) = 15 \text{ J}$$ **(c)** Total work: $$W_{\text{total}} = 30 + 15 = 45 \text{ J}$$ Notice that the second half of the rotation contributes only half as much work as the first half, even though the angular displacement is the same. The torque is weaker during the second half, so less energy is transferred per radian.

Problem 5. A torsional spring exerts a restoring torque $\tau = -\kappa\theta$, where $\kappa = 4.0 \text{ N}\cdot\text{m/rad}$. You twist the spring from $\theta = 0$ to $\theta = 1.5$ rad. How much work do you do against the spring?

Check your answer

You apply a torque $\tau_{\text{you}} = +\kappa\theta$ (equal and opposite to the spring's restoring torque). The work you do is: $$W = \int_0^{1.5} \kappa\theta\,d\theta = \kappa \cdot \frac{\theta^2}{2}\bigg|_0^{1.5} = (4.0)\frac{(1.5)^2}{2} = (4.0)(1.125) = 4.5 \text{ J}$$ This is stored as elastic potential energy in the spring. The structure is identical to the translational case: for a linear spring, $W = \frac{1}{2}kx^2$; for a torsional spring, $W = \frac{1}{2}\kappa\theta^2$.

Layer 3: Structure

Problem 6. Why is $W = \tau\,\Delta\theta$ the rotational analog of $W = F\,\Delta x$?

Don't just say "because the formulas look the same." Trace the reasoning from first principles: start with a force applied tangentially to a rotating object, and show how the translational work formula $dW = F\,ds$ becomes the rotational work formula $dW = \tau\,d\theta$.

Check your answer

Consider a force $F$ applied tangentially to an object at a distance $r$ from the axis of rotation. When the object rotates through a small angle $d\theta$, the point of application moves through an arc of length: $$ds = r\,d\theta$$ The translational work formula gives: $$dW = F\,ds = F \cdot r\,d\theta$$ But $Fr = \tau$ (the definition of torque for a tangential force at distance $r$). So: $$dW = \tau\,d\theta$$ Integrating over a finite rotation: $$W = \int_{\theta_i}^{\theta_f} \tau\,d\theta$$ The rotational formula is not an independent definition --- it is a *consequence* of the translational formula $dW = F\,ds$, rewritten using the relationship between arc length and angle ($ds = r\,d\theta$) and the definition of torque ($\tau = Fr$). The analogy is exact because the rotational variables are built from the translational ones. This is the same pattern we saw in Section 10.5, where $\sum \tau = I\alpha$ turned out to be $F = ma$ reorganized using rotational variables. The rotational work formula is $dW = F\,ds$ reorganized the same way.

Layer 4: Debug

Problem 7. A student solves the following problem:

A constant torque of 12 N$\cdot$m is applied to a wheel. The wheel rotates through 3 revolutions. Find the work done.

The student writes:

$$W = \tau \times (\text{number of revolutions}) = 12 \times 3 = 36 \text{ J}$$

The answer is wrong. What did the student do incorrectly, and what is the correct answer?

Check your answer

The student plugged in the number of revolutions (3) directly as the angle, without converting to radians. The formula $W = \tau\,\Delta\theta$ requires $\Delta\theta$ in radians. The correct calculation: $$\Delta\theta = 3 \text{ rev} \times 2\pi \text{ rad/rev} = 6\pi \text{ rad}$$ $$W = \tau\,\Delta\theta = (12)(6\pi) = 72\pi \approx 226 \text{ J}$$ The student's answer of 36 J is off by a factor of $2\pi \approx 6.28$. That is not a small rounding error --- it is a factor-of-six mistake. Why does this happen? The formula $W = \tau\,\Delta\theta$ is derived from $dW = F\,ds$ using $ds = r\,d\theta$, and that relationship $ds = r\,d\theta$ is only valid when $\theta$ is in radians. Using degrees or revolutions breaks the geometric relationship between arc length and angle. Radians are not optional here --- they are built into the derivation. This is the same reason that rotational kinematics formulas ($\omega = d\theta/dt$, $\alpha = d\omega/dt$, $v = r\omega$, $a_t = r\alpha$) all require radians. The radian is the natural unit of angle for any formula that connects rotational and translational quantities.

Spaced Retrieval

Before you move on, test your recall of earlier material.

Recall prompt 1: What is rotational kinetic energy, and how does it depend on $\omega$? (Section 11.1)

Recall prompt 2: What is the definition of torque? How does the distance from the axis affect the torque produced by a given force? (Section 10.3)

Recall prompt 3: What was the translational work-energy theorem? (Section 7.4)

Reflection

Think back over this section.

How do the translational and rotational work formulas fit into a single framework?

The translational formula is $W = \int F\,dx$: force accumulated over displacement gives work, which equals the change in $\frac{1}{2}mv^2$. The rotational formula is $W = \int \tau\,d\theta$: torque accumulated over angular displacement gives work, which equals the change in $\frac{1}{2}I\omega^2$. Same accumulation idea. Same work-energy theorem structure. Different variables, same physics.

This is not just a pattern to memorize --- it is a way of thinking. Whenever you encounter a new domain of motion, ask: "What plays the role of force? What plays the role of displacement? What quantity accumulates into work?" If you can answer those questions, you can write down the work formula without memorizing it separately.

Self-assessment: Can you derive $dW = \tau\,d\theta$ starting from $dW = F\,ds$? If you can do that derivation from memory, you understand this section at a structural level, not just a computational one.

Looking Ahead

You now know how torques do work and how that work changes rotational kinetic energy. Combined with Section 11.1, you have the rotational counterparts of kinetic energy and the work-energy theorem.

But energy and work are only half of the conservation story. In translational dynamics, there was a second conserved quantity --- momentum --- that proved just as powerful as energy and sometimes more useful. That quantity had its own version of Newton's second law ($F = dp/dt$), its own conservation condition (no external force), and its own problem-solving strategy.

In the next section, you will meet the rotational counterpart: angular momentum. You will define $L = I\omega$ for rigid bodies and $\vec{L} = \vec{r} \times m\vec{v}$ for particles, and you will begin to see why a figure skater who pulls her arms in spins faster without anyone pushing her. The conservation architecture is about to repeat itself once more.