In a potters' lane in Khurja, a potter sets a heavy stone wheel turning with a stick. He fits the stick in a notch near the rim and swings it through a quarter turn, again and again. Then he puts the stick down and shapes a pot while the wheel turns by itself for a minute or more. The pushes stored something in the wheel, and the wheel spends it slowly against friction.
Energy in a turning body
A flywheel on a fixed axle goes nowhere. Its centre of mass stays at rest. Yet it clearly holds energy: it can keep a machine running, and a brake grows hot when it stops it. The energy is in its particles. Each particle off the axis moves on its circle at \(v = \omega r\) and carries the ordinary kinetic energy \(\tfrac12 mv^2\).
The chapter on Rigid Bodies and Moment of Inertia added these energies and found \(K = \tfrac12 I\omega^2\). That result is the rotational kinetic energy. It is not a new kind of energy. It is \(\tfrac12 mv^2\) summed over the particles, with \(I\) doing the job that mass does for a body in translation.
A flywheel is a uniform disc of mass \(8\,\text{kg}\) and radius \(0.5\,\text{m}\), turning at \(20\,\text{rad/s}\) about its central axis. Find its kinetic energy. How much would it hold at \(40\,\text{rad/s}\)?
\(I = \tfrac12 MR^2 = \tfrac12 \times 8 \times 0.25 = 1\,\text{kg m}^2\), so \(K = \tfrac12 \times 1 \times 20^2 = 200\,\text{J}\).
At \(40\,\text{rad/s}\), \(K = \tfrac12 \times 1 \times 40^2 = 800\,\text{J}\). Twice the speed stores four times the energy, because \(K\) grows as \(\omega^2\).
A flywheel is built to use this. It is a heavy wheel with most of its mass in the rim, so its \(I\) is large. It absorbs energy when the machine pushes harder than the load needs, and gives it back when the push weakens. The flywheel of a diesel engine smooths the jerky push of the pistons in this way.
A flywheel spins on a fixed axle, so its centre of mass is at rest. Which statement about its energy is correct?
A ceiling fan has three blades, each about \(0.4\,\text{kg}\) and \(0.6\,\text{m}\) long, and turns at about 300 turns a minute. Treat each blade as a thin rod turning about one end. Estimate the kinetic energy of the turning blades, in joules.
The work done by a torque
A force does work when its point of application moves along it. Let a force \(F_t\) act along the tangent at a point a distance \(r\) from the axis. When the body turns through a small angle \(d\theta\), that point moves along its circle through an arc \(ds = r\,d\theta\), in the direction of the force.
\[ dW = F_t\,ds \]
Why. the work of a force along the displacement of its point of application.
\[ ds = r\,d\theta \]
Why. the point moves on a circle of radius \(r\) through the angle \(d\theta\), in radians.
\[ dW = F_t\,r\,d\theta = \tau\,d\theta \]
Why. \(F_t r\) is the torque of the force about the axis.
\[ W = \int_{\theta_1}^{\theta_2}\tau\,d\theta \]
Read it. add the work of all the small turns from the first angle to the last.
The work done by a torque is therefore \(\int\tau\,d\theta\). On a graph of torque against angle it is the area under the curve. In the same way, the work of a force was the area under its force-position graph. A steady torque gives a rectangle, and \(W = \tau\,\Delta\theta\). Only the angle turned enters. A torque of \(10\,\text{N m}\) acting through \(20\,\text{rad}\) does \(200\,\text{J}\) of work, whether the turn takes one second or one hour.
A steady torque of \(10\,\text{N m}\) turns a wheel through \(20\,\text{rad}\) in \(4\,\text{s}\). How much work does the torque do?
Power: torque times angular velocity
Power is the rate of doing work. Divide \(dW = \tau\,d\theta\) by \(dt\). Since \(d\theta/dt = \omega\), the power delivered by a torque is \(P = \tau\omega\). It is the rotational twin of \(P = Fv\). A torque along the sense of turning delivers positive power and feeds energy in. A torque against it, like a brake, gives negative power and drains energy out.
Motors and engines quote their speed in revolutions per minute, written rpm. Every formula here needs rad/s. One revolution is \(2\pi\) radians and one minute is \(60\) seconds. So a motor giving \(50\,\text{N m}\) at 300 rpm turns at \(300 \times 2\pi/60 = 10\pi \approx 31.4\,\text{rad/s}\). It delivers \(P = 50 \times 31.4 \approx 1570\,\text{W}\), about \(1.6\,\text{kW}\).
An angle grinder's motor gives a torque of \(2\,\text{N m}\) at 3000 rpm. Find the power it delivers, in watts.
James Watt sold steam engines to mill owners who used horses to turn their machines. He needed to say how many horses one engine could replace. According to the usual account, he watched a mill horse pulling with about \(180\) pounds of force. It walked round a circle about \(24\) feet across, \(2.4\) times a minute. That is a torque of about \(2900\,\text{N m}\) at \(0.25\,\text{rad/s}\), so \(P = \tau\omega \approx 740\,\text{W}\). Watt rounded the figure up to \(33\,000\) foot-pounds per minute, the horsepower, about \(746\,\text{W}\). The unit of power, the watt, now carries his name.
The work-energy theorem for rotation
Start from \(\tau_{net} = I\alpha\). Write \(\alpha\) with the chain rule, so that the angle appears instead of the time. Then integrate over the turn. The result is the work-energy theorem of the chapter on Work, Kinetic Energy and Power, in its rotational form.
\[ \tau_{net} = I\dfrac{d\omega}{dt} = I\dfrac{d\omega}{d\theta}\dfrac{d\theta}{dt} \]
Why. the chain rule; the angular velocity changes as the angle changes.
\[ \tau_{net} = I\omega\dfrac{d\omega}{d\theta} \]
Why. \(d\theta/dt = \omega\).
\[ \int\tau_{net}\,d\theta = \int_{\omega_i}^{\omega_f} I\omega\,d\omega \]
Why. multiply by \(d\theta\) and add over the whole turn.
\[ W_{net} = \tfrac12 I\omega_f^2 - \tfrac12 I\omega_i^2 \]
Read it. the left side is the net work of the torques; integrate the right side.
Take a rigid body turning about a fixed axis. The net work done by all the torques on it equals the change in its rotational kinetic energy: \(W_{net} = \tfrac12 I\omega_f^2 - \tfrac12 I\omega_i^2\).
The derivation never assumed that the torque is constant. So the theorem holds for any torque, steady or varying, and for a braking torque, whose work is negative. It links angles to speeds directly. The equations for constant \(\alpha\) link times to speeds. Choose the tool by what the problem gives: an angle points to work and energy, a time points to the equations of motion.
Play both chips of Fig 4.2 and watch the two readouts. The steady \(10\,\text{N m}\) turns the flywheel through \(62.5\,\text{rad}\) in \(5\,\text{s}\). The area is \(10 \times 62.5 = 625\,\text{J}\), and the flywheel ends at \(25\,\text{rad/s}\), with \(\tfrac12 \times 2 \times 25^2 = 625\,\text{J}\). The second torque falls from \(15\) to \(5\,\text{N m}\). Its area over the same \(62.5\,\text{rad}\) is also \(625\,\text{J}\), so the flywheel again ends at \(25\,\text{rad/s}\). It reaches that speed sooner, in about \(4.35\,\text{s}\), but the energy cares only about the area.
A wheel of moment of inertia \(0.5\,\text{kg m}^2\) turns at \(40\,\text{rad/s}\). A brake applies a steady opposing torque of \(4\,\text{N m}\). Through what angle does the wheel turn before it stops, and how long does it take?
The brake is the only torque, and it opposes the turning. So its work is negative, and it removes all of the kinetic energy.
Work-energy: \(-\tau\theta = 0 - \tfrac12 I\omega_0^2\). For the time: \(\alpha = \tau/I\) and \(t = \omega_0/\alpha\).
\(\tfrac12 \times 0.5 \times 40^2 = 400\,\text{J}\), so \(\theta = 400/4 = 100\,\text{rad}\), about 16 turns. Then \(\alpha = 4/0.5 = 8\,\text{rad/s}^2\) and \(t = 40/8 = 5\,\text{s}\).
The wheel stops after \(100\,\text{rad}\), in \(5\,\text{s}\). The \(400\,\text{J}\) of kinetic energy becomes heat in the brake.
The same brake stops the same wheel, but now the wheel starts at \(80\,\text{rad/s}\) instead of \(40\,\text{rad/s}\). Through what angle does it turn before stopping?
When does \(W_{net} = \tfrac12 I\omega_f^2 - \tfrac12 I\omega_i^2\) hold for a body turning about a fixed axis?
Now return to the potter's wheel. Take the stone wheel as a uniform disc of \(40\,\text{kg}\) and radius \(0.4\,\text{m}\), so \(I = 3.2\,\text{kg m}^2\). At one turn a second, \(\omega = 2\pi\,\text{rad/s}\), and the wheel holds \(\tfrac12 \times 3.2 \times (2\pi)^2 \approx 63\,\text{J}\). A push of \(20\,\text{N}\) at the rim through a quarter turn does \(20 \times 0.4 \times \pi/2 \approx 12.6\,\text{J}\) of work. So about five pushes bring the wheel to that speed, if friction is small.
- Using rpm or revolutions per second in \(P = \tau\omega\) or \(\tfrac12 I\omega^2\).Convert to rad/s first: multiply rpm by \(2\pi/60\).
- Multiplying the torque by the time to get work.Work is the torque times the angle turned, \(\tau\,\Delta\theta\).
- Doubling the energy when the angular velocity doubles.\(K\) grows as \(\omega^2\). Twice the speed, four times the energy.
- Using energy to find a stopping time.Energy gives angles. For times, use \(\alpha = \tau/I\) and the equations of motion.