The electricity goes off on a summer night. The ceiling fan does not stop at once. It slows down gradually and takes half a minute to come to rest, turning all the while. You could count its turns, but they blur at first. Two numbers, the speed at the start and the time to stop, are enough to find the count.
A dictionary between two kinds of motion
Put the angular quantities beside the linear ones. The angular position \(\theta\) plays the part of the position \(x\). The angular velocity \(\omega = d\theta/dt\) plays the part of \(v = dx/dt\). The angular acceleration \(\alpha = d\omega/dt\) plays the part of \(a = dv/dt\). The definitions have exactly the same form, only with new letters.
Motion in One Dimension derived the equations of motion from these definitions and one assumption, a constant \(a\). Every step used only the definitions. So the same steps, with the new letters, give the equations for a constant \(\alpha\). Nothing new has to be proved.
The twin of the distance in the \(n\)th second also carries over. The angle turned during the \(n\)th second, from \(t = n - 1\) to \(t = n\), is \(\theta_n = \omega_0 + \tfrac12\alpha(2n - 1)\). As on a line, the angles turned in successive seconds grow by the same step, \(\alpha\), every second.
\[ \alpha = \dfrac{d\omega}{dt} = \dfrac{d\omega}{d\theta}\,\dfrac{d\theta}{dt} \]
Why. the chain rule. The angular velocity is a function of the angle, and the angle is a function of time.
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Your turn. which line is it? Replace the rate of change of angle with its name.
\[ \int_{\omega_0}^{\omega} \omega\,d\omega = \int_0^{\theta} \alpha\,d\theta \]
Why. separate the variables and add up both sides from the start, where \(\omega = \omega_0\) at \(\theta = 0\).
\[ \tfrac12\left(\omega^2 - \omega_0^2\right) = \alpha\theta \]
Why. \(\alpha\) is constant, so it comes out of the integral on the right.
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
Read it. multiply by two. The time has disappeared from the equation.
A potter's wheel turning at one revolution per second slows at a steady rate and stops in \(8\,\text{s}\). How do the turns in the first \(4\,\text{s}\) compare with the turns in the last \(4\,\text{s}\)?
The numbers behind the figure are worth checking. The wheel starts at \(\omega_0 = 2\pi\,\text{rad/s}\) and stops in \(8\,\text{s}\), so \(\alpha = -2\pi/8 = -\pi/4 \approx -0.785\,\text{rad/s}^2\). The whole angle is \(\theta = \tfrac12(\omega_0 + 0)\,t = \tfrac12 \times 2\pi \times 8 = 8\pi\,\text{rad}\), which is \(4\) turns. In the first \(4\,\text{s}\), \(\theta = 2\pi \times 4 - \tfrac12 \times \tfrac{\pi}{4} \times 16 = 6\pi\,\text{rad}\), which is \(3\) turns.
A ceiling fan turning at \(240\,\text{rpm}\) is switched off and stops in \(30\,\text{s}\), slowing at a steady rate. Find its angular acceleration and the number of turns it makes before it stops.
\(\omega_0 = 240 \times 2\pi/60 = 8\pi \approx 25.1\,\text{rad/s}\). \(\omega = 0\) at \(t = 30\,\text{s}\).
\(\omega = \omega_0 + \alpha t\) gives \(\alpha\). Then \(\theta = \tfrac12(\omega_0 + \omega)\,t\) gives the angle without needing \(\alpha\).
\(\alpha = (0 - 8\pi)/30 \approx -0.84\,\text{rad/s}^2\). \(\theta = \tfrac12 \times 8\pi \times 30 = 120\pi\,\text{rad}\).
Turns \(= 120\pi/2\pi = 60\). The fan makes \(60\) turns before it stops, which answers the question of the power cut.
A wheel starts from rest and turns with a constant angular acceleration of \(2\,\text{rad/s}^2\). Through what angle, in radians, does it turn during the third second?
A motor slows steadily from \(1200\,\text{rpm}\) to \(600\,\text{rpm}\) in \(5\,\text{s}\). How many revolutions does it make in these \(5\,\text{s}\)?
Slopes and areas
The graphs of Graphical Understanding of Motion carry over too. The slope of the \(\theta\)-\(t\) graph at any instant is \(\omega\). The slope of the \(\omega\)-\(t\) graph is \(\alpha\). Going the other way, the area under the \(\omega\)-\(t\) graph between two instants is the angle turned. The area under the \(\alpha\)-\(t\) graph is the change in \(\omega\).
The potter's wheel shows the area rule at work. Its \(\omega\)-\(t\) graph is a straight line from \(2\pi\,\text{rad/s}\) down to zero at \(8\,\text{s}\). The shaded area grows as the wheel turns, and at every instant it equals the angle turned so far. The area rule does not need a constant \(\alpha\). It holds for any shape of the graph, which is what makes it useful.
The \(\omega\)-\(t\) graph of a wheel is a straight line from \(10\,\text{rad/s}\) at \(t = 0\) to \(4\,\text{rad/s}\) at \(t = 3\,\text{s}\). Through what angle does the wheel turn in these \(3\,\text{s}\)?
When the angular acceleration changes
The four equations hold only for a constant \(\alpha\). If \(\alpha\) changes, go back to the definitions and use calculus. Which form to use depends on what \(\alpha\) is given as a function of. If \(\alpha\) is given in terms of \(t\), integrate \(d\omega = \alpha\,dt\) to find \(\omega\), then \(d\theta = \omega\,dt\) to find \(\theta\).
A disc starts at \(t = 0\) with \(\omega_0 = 1\,\text{rad/s}\) and \(\theta_0 = 0\). Its angular acceleration is \(\alpha = 4t\), in rad/s² with \(t\) in s. Find \(\omega\) and \(\theta\) at \(t = 3\,\text{s}\).
Integrate \(d\omega = 4t\,dt\) from the start: \(\omega - 1 = 2t^2\), so \(\omega = 1 + 2t^2\).
Integrate \(d\theta = (1 + 2t^2)\,dt\): \(\theta = t + \tfrac23 t^3\).
At \(t = 3\,\text{s}\): \(\omega = 1 + 18 = 19\,\text{rad/s}\) and \(\theta = 3 + 18 = 21\,\text{rad}\). Using \(\omega = \omega_0 + \alpha t\) with \(\alpha = 12\) would give \(37\,\text{rad/s}\), which is wrong, because \(\alpha\) was smaller earlier.
If \(\alpha\) is given in terms of the angle \(\theta\), time is not the natural variable. Use \(\alpha = \omega\,d\omega/d\theta\) from the gap-derivation above. Separate and integrate: \(\int \omega\,d\omega = \int \alpha(\theta)\,d\theta\). The left side gives \(\tfrac12(\omega^2 - \omega_0^2)\), and the right side is a plain integral in \(\theta\).
A body turning about an axis has \(\omega_0 = 1\,\text{rad/s}\) at \(\theta = 0\), and its angular acceleration is \(\alpha = 2\theta\), in rad/s² with \(\theta\) in rad. Find \(\omega\) when \(\theta = 2\,\text{rad}\).
\(\omega\,d\omega = 2\theta\,d\theta\). Integrate from the start: \(\tfrac12(\omega^2 - 1) = \theta^2\).
So \(\omega^2 = 1 + 2\theta^2\). At \(\theta = 2\): \(\omega^2 = 1 + 8 = 9\), and \(\omega = 3\,\text{rad/s}\).
A third case arises when \(\alpha\) depends on \(\omega\) itself, as for a wheel slowed by air drag. Then write \(dt = d\omega/\alpha(\omega)\) and integrate to find the time. For \(\alpha = -k\omega\), the result is \(\omega = \omega_0\,e^{-kt}\). The wheel's angular velocity halves every \((\ln 2)/k\) seconds and, in this model, never quite reaches zero.
A wheel's angular acceleration is given as \(\alpha = 6\cos\theta\). You need its angular velocity at a certain angle. Which is the right start?
A bicycle wheel has a diameter of about \(0.7\,\text{m}\). Estimate how many turns it makes when you ride to a school \(5\,\text{km}\) away and back home.
- Using the four equations when \(\alpha\) changes with time or with angle.They hold only for a constant \(\alpha\). Otherwise integrate, using \(\alpha = d\omega/dt\) or \(\alpha = \omega\,d\omega/d\theta\).
- Leaving \(\omega\) in rpm inside the equations.Convert to rad/s, or keep every quantity in revolutions and seconds consistently.
- Giving an angle in radians when the question asks for turns.Divide by \(2\pi\): one turn is \(2\pi\,\text{rad}\).
- Reading the angle turned from the height of the \(\omega\)-\(t\) graph.The height is \(\omega\); the angle is the area under the graph.