A wheel rotates so that the angle of rotation is
A wheel rotates so that the angle of rotation is proportional to the square of the time, $\varphi = ct^{2}$. The first revolution is completed in $8\ \text{s}$. Find the angular acceleration $32\ \text{s}$ after the start.
Explanation of Method
A quadratic rotation law means constant angular acceleration.
Step 1: Fix the constant
One revolution is $2\pi$ radians, reached at $t = 8$: \[c\cdot8^{2} = 2\pi \;\Longrightarrow\; 64c = 2\pi \;\Longrightarrow\; c = \frac{\pi}{32}, \qquad \varphi(t) = \frac{\pi t^{2}}{32}.\]
Step 2: Angular velocity
\[ \omega(t) = \frac{d\varphi}{dt} = \frac{\pi t}{16}. \]
Step 3: Angular acceleration
\[ \varepsilon = \frac{d\omega}{dt} = \frac{d^{2}\varphi}{dt^{2}} = \frac{\pi}{16}. \]
The answer does not contain $t$: the angular acceleration is the same at every instant, so the value at $t = 32\ \text{s}$ is the value at any other moment.
Step 4: Verification
\[\varphi(8) = \frac{\pi\cdot64}{32} = 2\pi \ \text{ (one revolution)} \;\;\checkmark\]
\[\varphi(32) = \frac{\pi\cdot1024}{32} = 32\pi = 16 \text{ revolutions in the first } 32\ \text{s} \;\;\checkmark\]
\[\omega(32) = \frac{32\pi}{16} = 2\pi\ \frac{\text{rad}}{\text{s}} = 1 \text{ revolution per second} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\varepsilon = \dfrac{\pi}{16} \approx 0.1963\ \frac{\text{rad}}{\text{s}^{2}}\,}\]
Remark
Remark. The mention of $32\ \text{s}$ is a red herring in the sense that any time would give the same answer — but it is a useful check that the reader has noticed $\varepsilon$ is constant. The revolution count grows as $t^{2}$: the first revolution takes $8\ \text{s}$, the second only $8(\sqrt2-1) \approx 3.3\ \text{s}$.