The angular velocity of a wheel decreases uniformly from ω_1…

The angular velocity of a wheel decreases uniformly from $\omega_1 = 1200\ \mathrm{rpm}$ to $\omega_2 = 600\ \mathrm{rpm}$ in $\Delta t = 10\ \mathrm{s}$. (a) How long does it take for the wheel to stop? (b) How long does its last revolution take?

Explanation of Method

Time to stop and the duration of the last revolution.

Step 1: Angular acceleration

\[ \omega_1 = 1200 \cdot \frac{2\pi}{60} = 40\pi\ \mathrm{rad/s}, \]

\[ \omega_2 = 20\pi\ \mathrm{rad/s}, \]

\[ \alpha = \frac{\omega_2 - \omega_1}{\Delta t} = \frac{20\pi - 40\pi}{10} = -2\pi\ \mathrm{rad/s^2}. \]

Step 2: (a) Time to stop

The angular velocity halved in 10 s; losing the other half at the same rate takes another 10 s: \[t = \frac{\omega_1}{|\alpha|} = \frac{40\pi}{2\pi} = 20\ \mathrm{s}.\]

\[\begin{gathered} t = 20\ \mathrm{s} \text{ from } 1200\ \mathrm{rpm} \\ (10\ \mathrm{s} \text{ after reaching } 600\ \mathrm{rpm}) \end{gathered}\]

$\omega\ (\mathrm{rev/s})$ $t\ (\mathrm{s})$ $0$ $10$ $20$ $10$ $20$ $18$ $20\ \mathrm{s}$ $1\ \mathrm{rev}$ $\sqrt2\ \mathrm{s}$ $\sqrt2$ $600\ \mathrm{rpm}\ \text{at } 10\ \mathrm{s}$

Angular velocity in rev/s: 1200 rpm = 20 rev/s falls in a straight line, reaching 10 rev/s at 10 s and zero at 20 s. The inset magnifies the end: the last revolution is the small triangle of area 1 rev, which lasts $\sqrt2$ s.

Step 3: (b) The last revolution

Run the film backward: the last revolution looks like a wheel starting from rest and speeding up at $|\alpha| = 2\pi\ \mathrm{rad/s^2}$ through one turn ($2\pi$ rad): \[2\pi = \tfrac12 |\alpha|\, t_{\text{last}}^2 = \tfrac12 \cdot 2\pi \cdot t_{\text{last}}^2 \;\Longrightarrow\; t_{\text{last}}^2 = 2.\]

\[t_{\text{last}} = \sqrt2\ \mathrm{s} \approx 1.41\ \mathrm{s}\]

Check

Total angle to stop: $ \tfrac{\omega_1^2}{2|\alpha|} = \tfrac{1600\pi^2}{4\pi} = 400\pi\ \mathrm{rad} = 200 $ revolutions

At the start of the last one (after $398\pi$ rad): $ \omega^2 = 1600\pi^2 - 4\pi \cdot 398\pi = 8\pi^2 $, so $ \omega = 2\sqrt2\,\pi $; then $ t = \tfrac{\omega}{|\alpha|} = \sqrt2\ \mathrm{s}\ \checkmark $

Result

\[\boxed{\,\begin{gathered} t = 20\ \mathrm{s} \text{ from } 1200\ \mathrm{rpm} \\ (10\ \mathrm{s} \text{ after reaching } 600\ \mathrm{rpm}) \end{gathered}\,}\]

\[\boxed{\,t_{\text{last}} = \sqrt2\ \mathrm{s} \approx 1.41\ \mathrm{s}\,}\]

Remark

Note. The original solution reached the right answer after a detour (an impossible $\sqrt{-4\pi^2}$ along the way). The time-reversal trick above avoids that. Compare with Rotational Kinematics 006, which uses the same graph idea.