In the potters' colony at Uttam Nagar in Delhi, many potters still use a heavy wheel with no motor. The potter fits the end of a stick into a notch near the rim. He gives a few quick pushes, each lasting a fraction of a second. After half a dozen pushes the wheel is humming round, and it keeps turning while he shapes a whole pot. A push held a little longer, at the same force, sets the wheel spinning noticeably faster.
The torque law, added over time
In the chapter on Accelerated Objects, the second law \(F = dp/dt\) was added over time to give the impulse, \(\int F\,dt = \Delta p\). The torque law can be treated the same way. Multiply \(\tau = dL/dt\) by \(dt\) and add over the interval from \(t_1\) to \(t_2\). The left side becomes the angular impulse \(J = \int\tau\,dt\). The right side becomes the change in angular momentum, \(L_2 - L_1\).
The angular impulse of the net external torque about a fixed axis equals the change in angular momentum about that axis. In symbols, \(\int\tau\,dt = \Delta L\) over any time interval.
On a graph of torque against time, \(\int\tau\,dt\) is the area under the curve. So the change in angular momentum depends only on that area, not on how the torque was shaped. A large torque for a short time and a small torque for a long time give the same change if their areas are equal. The unit N m s is the same as kg m²/s, the unit of angular momentum.
The three shaded shapes in Fig 3.1 look nothing alike. The rectangle is long and low, the triangle rises steadily, and the blow is a tall thin strip. Yet each has an area of \(12\,\text{N m s}\): \(4 \times 3\), \(\tfrac12 \times 8 \times 3\) and \(24 \times 0.5\). The flywheel ends every history with \(L = 12\,\text{kg m}^2\text{/s}\) and \(\omega = 6\,\text{rad/s}\). Only the time taken to get there differs.
Four torques act one at a time on the same flywheel, which starts from rest each time. Rank them by the angular velocity they give it, largest first.
Driving and braking a flywheel
For a flywheel on a fixed axle, \(L = I\omega\) with \(I\) constant, so an angular impulse \(J\) changes the angular velocity by \(\Delta\omega = J/I\). The same idea works in reverse for braking. A friction torque \(\tau_f\) removes angular momentum at the rate \(\tau_f\), so a wheel with angular momentum \(L\) stops in the time \(L/\tau_f\). Neither calculation needs the angle turned.
A flywheel with \(I = 2.5\,\text{kg m}^2\) starts from rest. A motor applies a steady \(5\,\text{N m}\) for \(4\,\text{s}\) and is then switched off. A brake then applies a friction torque of \(2\,\text{N m}\). Find the angular velocity when the motor stops, and how long the brake takes to stop the wheel.
Axis: the axle. The motor's torque and the brake's torque are the only external torques about it.
Driving: \(\Delta L = \tau\,\Delta t\). Braking: \(\Delta t = L/\tau_f\).
\(\Delta L = 5 \times 4 = 20\,\text{kg m}^2\text{/s}\), so \(\omega = 20/2.5 = 8\,\text{rad/s}\). The brake removes \(2\,\text{kg m}^2\text{/s}\) every second, so it takes \(20/2 = 10\,\text{s}\).
The flywheel reaches \(8\,\text{rad/s}\), and the brake stops it in \(10\,\text{s}\).
A table fan's blades have \(I = 0.4\,\text{kg m}^2\) and turn at \(30\,\text{rad/s}\). After the fan is switched off, they stop in \(20\,\text{s}\). Find the average friction torque on them, in N m.
A wheel with \(I = 0.5\,\text{kg m}^2\) starts from rest. A torque \(\tau = 3t^2\) acts on it, with \(\tau\) in N m and \(t\) in seconds, from \(t = 0\) to \(t = 2\,\text{s}\). Find its angular velocity at \(t = 2\,\text{s}\), in rad/s.
A short blow
Some torques come from blows too quick to follow: a hammer on a flywheel, a kick on a door, a ball on a bat. The force of such an impulsive force is large, lasts a very short time and is hard to measure. Its linear impulse \(P = \int F\,dt\) is much easier to know. During the blow the body barely moves, so the force keeps the same moment arm \(d\) about the axis. Then \(\int\tau\,dt = \int Fd\,dt = d\int F\,dt\), and the angular impulse is simply \(Pd\).
The blow in Fig 3.2 has the same impulse in both chips. Only the moment arm changes, and the angular impulse changes with it. Struck at the end, \(J = 3 \times 1.5 = 4.5\,\text{kg m}^2\text{/s}\), and \(\omega = 4.5/1.5 = 3\,\text{rad/s}\). Struck at the middle, \(J = 2.25\) and \(\omega = 1.5\,\text{rad/s}\). The pin also pushes on the rod during the blow, but its push acts at the axis and has no torque about it.
A door has a moment of inertia of \(4\,\text{kg m}^2\) about its hinges and is at rest. A football hits it at right angles, \(0.8\,\text{m}\) from the hinges, and delivers an impulse of \(5\,\text{N s}\). Find the angular velocity of the door just after, in rad/s.
A hammer blow gives a flywheel a large force \(F\) and a large torque \(\tau\) for a very short time \(\Delta t\). The change in the flywheel's angular momentum equals
A push on a swing works the same way. Pushing harder helps, and so does pushing for longer, and so does pushing farther from the pivot. The swing responds to the product of all three: force, time and moment arm. That product is the angular impulse.
Now return to the potter's wheel. Suppose the wheel has \(I = 6\,\text{kg m}^2\), about that of a \(50\,\text{kg}\) stone disc of radius \(0.5\,\text{m}\). The potter pushes the rim with \(60\,\text{N}\), a moment arm of \(0.5\,\text{m}\), for \(0.4\,\text{s}\). Each jab gives \(J = 60 \times 0.5 \times 0.4 = 12\,\text{kg m}^2\text{/s}\), and so \(\Delta\omega = 2\,\text{rad/s}\). Five jabs bring the wheel to \(10\,\text{rad/s}\), about one and a half turns every second. A jab held twice as long gives twice the angular impulse.
- Multiplying a varying torque by the whole time.The angular impulse is the area under the torque-time graph. Integrate, or find the area of its shape.
- Mixing the linear impulse with the angular impulse.\(\int F\,dt\) changes \(p\); \(\int\tau\,dt\) changes \(L\). For a short blow, \(J = Pd\).
- Forgetting the pivot during a blow.The pivot pushes on the body, but at the axis. Its torque about the axis is zero, so it does not enter \(J\).