6.2 - Torque and Work

Introduction: The Rotational Version of Work

Forces do work when they cause a linear displacement, and in the same way torques do work when they cause an angular displacement.

The key parallel: torque is the rotational version of force, so work done by torque follows the same pattern as work done by force.

Linear Work (Review)

For translational motion:

W = \int \vec{F} \cdot d\vec{s}

For a constant force:

W = Fd\cos\theta

Rotational Work

For rotational motion, the work done by a torque is:

W = \int \tau \, d\theta

where: -

W

= work (J). -

\tau

= torque (N·m). -

\theta

= angular displacement (rad). - You integrate over the angular displacement.

What it means: energy moves into or out of a rotating system whenever a torque acts over some angular displacement.

Into or out of the system: the sign of the work tells you which way the energy goes. If the torque points the same way the object turns (

\tau

and

\Delta\theta

have the same sign),

W > 0

and the torque transfers energy into the object (a motor spinning up a fan). If the torque opposes the rotation (a brake, friction in a bearing),

W < 0

and the torque transfers energy out. A torque on an object that doesn’t turn (

\Delta\theta = 0

) does no work at all.

A collection of torques: when several torques act, each does its own work,

W_i = \int \tau_i\,d\theta

. The net work is the sum, which equals the work done by the net torque, and that’s what changes the rotational kinetic energy (

\Delta K = \sum W_i

on the equation sheet):

W_{net} = \sum W_i = \int \tau_{net}\,d\theta = \Delta K_{rot}

Constant vs. Variable Torque

Constant Torque

When the torque holds steady:

W = \tau \Delta\theta

where: -

\tau

= the constant torque (N·m). -

\Delta\theta

= angular displacement (rad).

This is the twin of

W = Fd

for a constant force.

Variable Torque

When the torque changes with angle:

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

Now you need calculus to evaluate it.

Connection to the Linear Work Formula

Start from the definitions.

Linear quantities: - Force:

F

. - Displacement:

d

. - Work:

W = Fd

Rotational quantities: - Torque:

\tau = rF

. - Angular displacement:

\theta

, where

d = r\theta

Put them together:

W = Fd = F(r\theta) = (rF)\theta = \tau\theta

W = \tau\theta

for a constant torque.

Work-Energy Theorem for Rotation

Just like in linear motion, work equals the change in kinetic energy.

Work-Energy Theorem (rotational):

W = \Delta K_{\text{rot}}

W = K_{\text{rot},f} - K_{\text{rot},i}

W = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2

This is what links the work done by a torque to the change in rotational kinetic energy.

Example Problem: A Motor and Friction on a Grinding Wheel

Problem: a grinding wheel with

I = 0.25

kg·m² starts from rest. A motor exerts a constant 2.4 N·m torque on it, and friction in the bearings exerts a constant 0.40 N·m torque the opposite way. (a) Find the work done by each torque, and the net work, during the first 25 rad of rotation. (b) Find the wheel’s angular speed after 25 rad. (c) The motor then shuts off. Through what angle does the wheel turn before friction stops it?

(a) Take the direction of rotation as positive. Both torques are constant, so

W = \tau\Delta\theta

W_{motor} = (2.4\ \text{N·m})(25\ \text{rad}) = 60\ \text{J} \qquad W_{friction} = (-0.40\ \text{N·m})(25\ \text{rad}) = -10\ \text{J}

The net work is

60 + (-10) = 50

J. The motor transfers 60 J into the wheel, and friction transfers 10 J out (it ends up as thermal energy in the bearings).

(b) Work-energy theorem, starting from rest:

50\ \text{J} = \tfrac{1}{2}(0.25\ \text{kg·m}^2)\omega^2 - 0 \quad\Rightarrow\quad \omega = \sqrt{\frac{2(50)}{0.25}} = 20\ \text{rad/s}

(c) Now only friction acts, and it has to remove all 50 J:

-50\ \text{J} = (-0.40\ \text{N·m})\Delta\theta \quad\Rightarrow\quad \Delta\theta = 125\ \text{rad} \approx 1.3 \times 10^2\ \text{rad}

That’s about 20 turns (

125 \div 2\pi \approx 19.9

Getting a Rotational Kinematics Equation from the Work-Energy Theorem

The work-energy theorem can hand you a rotational kinematics equation.

Start with:

W = \tau \Delta\theta

and

W = \Delta K_{\text{rot}} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2

Set them equal:

\tau \Delta\theta = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2

For constant torque (

\tau = I\alpha

from Newton’s second law):

I\alpha \Delta\theta = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2

Divide by

I

\alpha \Delta\theta = \frac{1}{2}\omega_f^2 - \frac{1}{2}\omega_i^2

Multiply by 2:

2\alpha \Delta\theta = \omega_f^2 - \omega_i^2

That’s the rotational twin of

2ad = v_f^2 - v_i^2

Rearranged:

\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta

And that’s one of the CAKE equations from Unit 5.

Example Problem: Torsion Spring

Problem: a torsion spring is attached to the end of a rod with rotational inertia

I_R

. The torsion spring is anchored to a horizontal table with negligible friction, as shown in the top view. When the rod gets displaced by an angle

\theta

from equilibrium, the torsion spring pushes back with a restoring torque of magnitude:

\tau = 2\kappa\theta^2

where

\kappa

is a positive constant with the right units. After being displaced by an angle

\theta_0

, the rod is released and swings through its equilibrium position with angular speed

\omega_0

How much work went into moving the rod out to the displaced position $\theta_0$ ?

Part (a): Finding the Work Done

Step 1: Set up the work integral

This is the work an outside agent does against the restoring torque to twist the rod:

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

We’re going from equilibrium (

\theta = 0

) out to the displaced position (

\theta = \theta_0

W = \int_0^{\theta_0} \tau \, d\theta

Step 2: Plug in the torque function

We’re given

\tau = 2\kappa\theta^2

W = \int_0^{\theta_0} 2\kappa\theta^2 \, d\theta

Step 3: Pull the constant out

W = 2\kappa \int_0^{\theta_0} \theta^2 \, d\theta

Step 4: Integrate with the power rule

\int \theta^2 \, d\theta = \frac{\theta^3}{3}

W = 2\kappa \left[\frac{\theta^3}{3}\right]_0^{\theta_0}

Step 5: Evaluate the definite integral

W = 2\kappa \left(\frac{\theta_0^3}{3} - \frac{0^3}{3}\right)

W = 2\kappa \cdot \frac{\theta_0^3}{3}

W = \frac{2\kappa\theta_0^3}{3}

Answer: the work done is:

\boxed{W = \frac{2\kappa\theta_0^3}{3}}

What it means: that work gets stored as potential energy in the twisted torsion spring, the same way a regular spring stores elastic potential energy.

Extension: Finding the Final Angular Speed

Follow-up: use the work-energy theorem to find the angular speed

\omega_0

as the rod swings back through equilibrium.

Solution:

Step 1: Apply the work-energy theorem

The energy stored in the spring comes back out as rotational kinetic energy:

W = \Delta K_{\text{rot}}

\frac{2\kappa\theta_0^3}{3} = \frac{1}{2}I_R\omega_0^2 - 0

(The initial kinetic energy is zero since it’s released from rest.)

Step 2: Solve for $\omega_0$

\frac{2\kappa\theta_0^3}{3} = \frac{1}{2}I_R\omega_0^2

Multiply both sides by 2:

\frac{4\kappa\theta_0^3}{3} = I_R\omega_0^2

\omega_0^2 = \frac{4\kappa\theta_0^3}{3I_R}

\omega_0 = \sqrt{\frac{4\kappa\theta_0^3}{3I_R}}

Or, written another way:

\omega_0 = \frac{2\theta_0^{3/2}\sqrt{\kappa}}{\sqrt{3I_R}}

What it tells you: - A bigger displacement

\theta_0

stores more energy, so you get a higher angular speed. - A bigger rotational inertia

I_R

is harder to accelerate, so you get a lower angular speed. - A bigger spring constant

\kappa

means a stiffer spring, so you get a higher angular speed.

Try It with Numbers

Say

\kappa = 0.60

N·m/rad², the rod is twisted to

\theta_0 = 1.2

rad, and

I_R = 0.050

kg·m².

W = \frac{2\kappa\theta_0^3}{3} = \frac{2(0.60)(1.2)^3}{3} = 0.69\ \text{J}

\omega_0 = \sqrt{\frac{2W}{I_R}} = \sqrt{\frac{2(0.6912)}{0.050}} = 5.3\ \text{rad/s}

Keep the unrounded 0.6912 J in the second step and round only at the end. Units check:

\tau = 2\kappa\theta^2

is in N·m and

\theta^2

is in rad², so

\kappa

is in N·m/rad², and

\kappa\theta^3

comes out in N·m = J (radians are dimensionless).

Summary of Relationships

Linear Quantity	Rotational Quantity	Relationship
Force: $F$	Torque: $\tau$	$\tau = rF$
Displacement: $d$	Angular displacement: $\theta$	$d = r\theta$
Work: $W = Fd$	Work: $W = \tau\theta$	Same form
Work (variable): $W = \int F \, dx$	Work (variable): $W = \int \tau \, d\theta$	Same form
KE: $K = \frac{1}{2}mv^2$	KE: $K = \frac{1}{2}I\omega^2$	Same form
Work-Energy: $W = \Delta K$	Work-Energy: $W = \Delta K_{\text{rot}}$	Same principle

The pattern: swap the linear quantities for the rotational ones and the math stays exactly parallel.

Area Under the Curve Interpretation

Graphically, the work done is the area under a torque vs. angular displacement graph.

W = \int_{\theta_i}^{\theta_f} \tau(\theta) \, d\theta = \text{Area under the } \tau\text{-}\theta \text{ curve}

Which mirrors: - Work = area under an

F

vs.

x

graph (linear motion). - Impulse = area under an

F

vs.

t

graph.

Example Problem: Work from a Torque vs. Angle Graph

Problem: a flywheel with

I = 0.24

kg·m² starts from rest. The graph shows the net torque on it as a function of its angular position. Find (a) the work done from

\theta = 0

to 5.0 rad, (b) the angular speed at 5.0 rad, and (c) the angular speed at 6.0 rad.

(a) Split the area into shapes. From 0 to 3.0 rad it’s a rectangle. From 3.0 to 5.0 rad it’s a triangle above the axis:

W_{0\to 5} = (12\ \text{N·m})(3.0\ \text{rad}) + \tfrac{1}{2}(2.0\ \text{rad})(12\ \text{N·m}) = 36\ \text{J} + 12\ \text{J} = 48\ \text{J}

(b) It starts from rest, so all 48 J becomes rotational kinetic energy:

\omega_5 = \sqrt{\frac{2W}{I}} = \sqrt{\frac{2(48)}{0.24}} = \sqrt{400} = 20\ \text{rad/s}

(c) From 5.0 to 6.0 rad the torque is negative, so the area below the axis counts as negative work:

\frac{1}{2}(1.0\text{ rad})(-6.0\text{ N·m}) = -3.0

J. The net work from 0 to 6.0 rad is

48 - 3.0 = 45

\omega_6 = \sqrt{\frac{2(45)}{0.24}} = \sqrt{375} \approx 19\ \text{rad/s}

What it tells you: the flywheel spins fastest at

\theta = 5.0

rad, right where the torque crosses zero. Before that, the torque adds energy. After that, it takes energy back out. The negative torque doesn’t reverse the spin here; it just slows the flywheel down.

Units Check

Always make sure the dimensions line up.

Work from torque:

[W] = [\tau][\theta]

[\text{J}] = [\mathrm{N·m}][\text{rad}]

And since radians are dimensionless:

[\text{J}] = [\mathrm{N·m}] = [\text{J}]

✓

So torque times angular displacement really does give you energy, and the units back that up.

Common Mistakes

Using the wrong angle for work. Work uses the angular displacement $\Delta\theta$ in radians, not degrees, and not the absolute angle unless you happen to start at zero.

Not integrating for variable torque. If $\tau$ depends on $\theta$ , you MUST use $W = \int \tau \, d\theta$ . Only use $W = \tau\theta$ when the torque is actually constant.

Confusing work and torque. Work is in joules (J), and torque is in N·m. Numerically 1 J = 1 N·m, but they’re different physical things. Work is torque times angular displacement.

Not squaring the angular velocity in the kinetic energy. It’s $K = \frac{1}{2}I\omega^2$ , not $\frac{1}{2}I\omega$ . Always square it.

Wrong limits of integration. Integrate from the initial angle to the final angle: $\int_{\theta_i}^{\theta_f}$ . In the example that’s 0 (equilibrium) to $\theta_0$ (the displaced spot).

Missing that the lower limit vanishes. When you evaluate $\left[\frac{\theta^3}{3}\right]_0^{\theta_0}$ , the lower limit gives 0, which cleans things up.

Fraction slip-ups. $2\kappa \cdot \frac{\theta_0^3}{3}$ equals $\frac{2\kappa\theta_0^3}{3}$ , not $\frac{2\kappa\theta_0^3}{6}$ or $\frac{\kappa\theta_0^3}{3}$ .

Mixing up positive and negative work. If the torque fights the motion, the work done by that torque is negative. If you’re computing the work done against a restoring torque, that work is positive (it’s getting stored).

Not spotting when the work-energy theorem applies. When a problem says “find the work done” and then “find the angular speed,” that’s your cue to connect them with $W = \Delta K_{\text{rot}}$ .

Botching the power rule. $\int \theta^n \, d\theta = \frac{\theta^{n+1}}{n+1} + C$ . For $\theta^2$ that’s $\frac{\theta^3}{3}$ , not $\frac{\theta^3}{2}$ .

Forgetting the initial kinetic energy. The theorem is $W = K_f - K_i$ . If it’s released from rest, $K_i = 0$ and things simplify, but always check the initial conditions.

Failing dimensional analysis. Work should come out in joules. If it doesn’t, something went sideways.

Not pulling constants out of the integral. In $\int 2\kappa\theta^2 \, d\theta$ , pull the $2\kappa$ out first: $2\kappa\int\theta^2 \, d\theta$ . Much cleaner.

Mistaking the torque function for the work. The given $\tau = 2\kappa\theta^2$ is the torque at angle $\theta$ , not the total work. You have to integrate to get work.

Sign confusion with restoring torques. A restoring torque opposes the displacement (so the torque itself does negative work), but the work an external agent does to create that displacement is positive and gets stored as potential energy.

Summary of Key Formulas

Concept	Formula	When to Use
Work (constant torque)	$W = \tau\Delta\theta$	Torque is constant
Work (variable torque)	$W = \int_{\theta_i}^{\theta_f} \tau(\theta) \, d\theta$	Torque changes with angle
Work-Energy Theorem	$W = \Delta K_{\text{rot}} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$	Linking work to energy
Rotational kinetic energy	$K_{\text{rot}} = \frac{1}{2}I\omega^2$	Energy of rotation
Kinematics (from W-E)	$\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$	The no-time equation

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