Mistake Master
Mistake Master
A torque that turns a body through an angle does work on it. For a constant torque, $W_{rot} = \tau\,\Delta\theta$. When $\tau$ varies, $W_{rot}$ is the signed area under the $\tau$-vs-$\theta$ graph. That work is what changes the body's rotational kinetic energy: $W_{net,\,rot} = \Delta K_{rot}$. Negative work isn't "no work"; it's energy taken back out.
Topic 6.1 named the energy of a spinning body: $K_{rot} = \tfrac{1}{2}I\omega^2$. This topic is about what changes it. A torque acting through an angular displacement does work on the body, and that work raises or lowers $K_{rot}$, exactly like $W = Fd$ for a sliding block.
The headline: $W_{rot} = \tau\,\Delta\theta$ when $\tau$ is constant, and the signed area under the $\tau$-vs-$\theta$ graph when it varies. Same shape as the translational case with rotational variables.
For a constant torque $\tau$ that turns a body through angular displacement $\Delta\theta$:
$$W_{rot} = \tau\,\Delta\theta.$$Units: N$\cdot$m for $\tau$, rad for $\Delta\theta$, joules for $W_{rot}$ (radians are dimensionless, so $1\text{ N}\cdot\text{m}\cdot\text{rad} = 1\text{ J}$). Same shape as translational $W = Fd$: torque takes the place of force, angular displacement takes the place of distance.
When $\tau$ varies with $\theta$, $W_{rot} = \tau\,\Delta\theta$ stops working. The general rule: the signed area under the $\tau$-vs-$\theta$ graph is the work done by the torque. Rectangle, triangle, trapezoid, two-piece profile, all the same idea. Area above the $\theta$ axis counts positive; area below counts negative.
A constant torque $\tau = 3$ N$\cdot$m acts on a wheel that rotates through $\Delta\theta = 4$ rad. Find the work done by the torque.
$$W_{rot} = \tau\,\Delta\theta = (3)(4) = 12\text{ J}.$$
Twelve joules, scalar, positive. On a $\tau$-vs-$\theta$ graph this would be a rectangle of height $3$ and width $4$: same area, same answer.
A $\tau$-vs-$\theta$ graph shows a triangle: $\tau$ rises linearly from $0$ at $\theta = 0$ to a peak of $6$ N$\cdot$m at $\theta = 4$ rad, then drops linearly back to $0$ at $\theta = 8$ rad. Find the total work.
The shape is a triangle of base $8$ rad and height $6$ N$\cdot$m:
$$W_{rot} = \dfrac{1}{2}(\text{base})(\text{height}) = \dfrac{1}{2}(8)(6) = 24\text{ J}.$$Using $\tau\,\Delta\theta$ with the peak torque gives $48$ J, twice too big. The torque sits at its peak for only an instant; the area accounts for that.
The sign of $W_{rot}$ is the sign of $\tau$ relative to $\Delta\theta$. Torque along the rotation: same sign on $\tau$ and $\Delta\theta$, positive work, energy going in. Torque opposing the rotation (a friction torque, say): opposite signs, negative work, energy coming out.
The rotational work-energy theorem says it directly. For the net torque on a rigid body:
$$W_{net,\,rot} = \Delta K_{rot} = \tfrac{1}{2}I\omega_f^2 - \tfrac{1}{2}I\omega_i^2.$$Positive net work raises $K_{rot}$. Negative net work lowers it. Zero net work leaves $K_{rot}$ where it was, even if the wheel turned through a real angle, as long as the positive and negative areas of the $\tau$-vs-$\theta$ graph cancelled.
A wheel with $I = 0.5\text{ kg}\cdot\text{m}^2$ starts from rest. A constant net torque does $W_{net} = 9$ J of work on it. Find the wheel's final angular speed.
$$W_{net,\,rot} = \Delta K_{rot} = \tfrac{1}{2}I\omega_f^2 - 0 \;\Longrightarrow\; \omega_f = \sqrt{\dfrac{2 W_{net}}{I}} = \sqrt{\dfrac{2(9)}{0.5}} = \sqrt{36} = 6\text{ rad/s}.$$
Nine joules of work made a half-kilogram-meter-squared wheel spin at six radians per second. Same machinery as $W = \Delta K$ for a sliding block, just with the rotational variables.
One distinction worth knowing now. The area under $\tau$-vs-$\theta$ is work; the area under $\tau$-vs-$t$ is angular impulse, which changes angular momentum, not energy. Same vertical axis, different horizontal axis, different physics. Topic 6.3 picks up the impulse side.
Rotational work goes wrong when students misread a graph's axes, misjudge a negative sign, or hand a scalar a direction. All three show up in your diagnostic.
Two graphs share the same $\tau$ axis. The area under $\tau$-vs-$\theta$ is the work done by the torque; the area under $\tau$-vs-$t$ is the angular impulse, $J_{rot} = \tau\,\Delta t$, which changes angular momentum, not energy. Same vertical axis, different meaning.
Fix. Check the horizontal axis before reading the area. If it says $\theta$ in radians, the area is in joules: work. If it says $t$ in seconds, the area is angular impulse. The horizontal axis tells you which physical quantity the area is.
Negative work is real work. It moves energy out of the rotating body instead of into it. A friction torque doing $-8$ J of work on a wheel removes $8$ J of $K_{rot}$. That is more change in $K_{rot}$ than zero work would have produced, not less.
Fix. Read $W < 0$ as "the torque took $|W|$ joules out of the rotation." The work-energy theorem reads $W_{net} = \Delta K_{rot}$, signs and all. Negative work isn't an edge case; it's how every braking torque shows up.
Torque has a rotation sense (CCW or CW). So does angular displacement. It's natural to want their product to carry a direction forward. The formula doesn't.
Fix. $W_{rot} = \tau\,\Delta\theta$ is a product of two signed scalars, not a cross product. The result is one signed number in joules. The sign tells you whether energy went in or came out; nothing else.