Torque and work

A torque acting through an angle does work, just as a force pushing through a distance does. For a constant torque the work is $W=\tau\,\Delta\theta$: torque times the angle turned (in radians), in joules. It is the rotational version of $W=Fd$.

When the torque changes as the wheel turns, that single product no longer works. The work becomes an integral, $W=\int\tau\,d\theta$: the area under the torque-versus-angle graph. The rate of delivering that work is the power, $P=\tau\omega$, in watts.

Four habits carry the topic: integrate when $\tau$ varies, keep the sign when a torque brakes the rotation, tell work apart from power, and pair torque with angular speed (or force with linear speed) in the power formula.

Start with a constant torque. As the wheel turns through an angle $\Delta\theta$, it does work $W=\tau\,\Delta\theta$. On a graph of torque against angle, that is a flat line at height $\tau$, and the work is the rectangle beneath it.

Now let the torque vary. Cut the turn into tiny steps $d\theta$. Over each step the torque is almost constant and does $dW=\tau\,d\theta$. Adding the steps gives

• $W=\displaystyle\int_{\theta_1}^{\theta_2}\tau\,d\theta$, the area under the $\tau$-versus-$\theta$ curve between the start and end angles.

For example, $\tau(\theta)=6\theta$ N$\cdot$m from $\theta=0$ to $\theta=4$ rad does $W=\int_0^4 6\theta\,d\theta=3\theta^2\big|_0^4=48$ J: the triangle under the line, not $\tau\,\Delta\theta$.

Work is the area under the $\tau$-versus-$\theta$ curve. For a constant torque that area is the rectangle $\tau\,\Delta\theta$; for a varying torque, integrate.

The product $\tau\,\Delta\theta$ is the area only when the torque is constant, since only then is the region a rectangle. If $\tau$ grows or shrinks as the wheel turns, grabbing the torque at the start or end and multiplying by $\Delta\theta$ gives the wrong work.

Take $\tau(\theta)=6\theta$ N$\cdot$m over $\theta=0$ to $4$ rad. The final torque $\tau(4)=24$ N$\cdot$m gives $24\times4=96$ J, twice too big. The starting torque $\tau(0)=0$ gives $0$ J, far too small. The right answer is the integral, $W=\int_0^4 6\theta\,d\theta=48$ J.

The test is simple: if $\tau$ depends on $\theta$, integrate. Pull a torque out in front of $\Delta\theta$ only when it never changes.

$\tau\,\Delta\theta$ works only for a constant torque. Whenever $\tau$ depends on the angle, use $W=\int\tau\,d\theta$.

The work integral has a sign, because $\theta$ is the signed angular coordinate about the rotation axis and $\tau$ is the torque component about that axis. A torque that opposes the rotation does negative work, shown as area below the axis, and that area is subtracted, not added. Flipping the limits flips the sign of the whole result, so always integrate from the start angle to the end angle.

When a torque drives the rotation, then brakes it, the net work is the signed area: the part above the axis minus the part below. For $\tau(\theta)=8-2\theta$ N$\cdot$m from $0$ to $6$ rad, it drives up to $\theta=4$ rad and opposes after. The driving area is $16$ J and the braking area is $4$ J, so $W=\int_0^6(8-2\theta)\,d\theta=48-36=12$ J, not the $20$ J from adding magnitudes.

Keep the sign. A braking torque does negative work, and net work is the signed area from start to finish: the positive part minus the negative.

Work and power are different. Work is the energy delivered over a turn, $W=\int\tau\,d\theta$, in joules. Power is how fast it is delivered, $P=\tau\omega$, in watts. Giving one when the question asks for the other is a common slip, and so is using $\tau t$ (torque times time), which is angular impulse, not work.

Rotational power matches linear power: $P=\tau\omega$ is the twin of $P=Fv$. For a belt on a wheel of radius $R$ they agree, since the rim force $F=\tau/R$ and rim speed $v=\omega R$ give $Fv=\tau\omega$. The trap is mixing the partners, like $F\omega$ or $\tau v$. Pair torque with angular speed, or force with linear speed, never force with angular speed.

$P=\tau\omega$ is a rate (watts); $W=\int\tau\,d\theta$ is the total energy (joules). They share the torque but answer different questions.

Almost every rotational-work misstep comes from shortcutting a varying torque, blurring work into power, or losing a sign along the way.

• Constant-torque shortcutUsing $\tau\,\Delta\theta$ when $\tau$ varies. Fix: integrate, $W=\int\tau\,d\theta$.
• Work vs powerGiving the power $\tau\omega$ (or $\tau t$) as work, or work as power. Fix: match the quantity to its formula and units.
• Sign and limitsDropping the sign of braking work, or flipping the limits. Fix: keep the signed area, integrate start to finish.
• Power analogUsing $P=Fv$ with mismatched partners like $F\omega$ or $\tau v$. Fix: pair $\tau$ with $\omega$, or $F$ with $v$.

Integrate when $\tau$ varies, keep the sign, tell work from power, and match the partners in $P=\tau\omega$. The whole topic comes down to those four habits.