A wheel rotates about a fixed axis so that its angle changes…

A wheel rotates about a fixed axis so that its angle changes as $\theta(t) = kt^2$, with $k = 1\ \mathrm{rad/s^2}$. Find the total acceleration of a point on its rim at $t = 5\ \mathrm{s}$, if at that moment the point's speed is $v = 1\ \mathrm{m/s}$.

Explanation of Method

Total acceleration of a rim point from θ = kt².

Step 1: Angular velocity and angular acceleration

\[ \omega = \frac{d\theta}{dt} = 2kt = 10\ \mathrm{rad/s}\ \ (t = 5\ \mathrm{s}), \]

\[ \varepsilon = \frac{d\omega}{dt} = 2k = 2\ \mathrm{rad/s^2}. \]

Step 2: Radius of the wheel

The radius is not given, but the speed is: \[R = \frac{v}{\omega} = \frac{1}{10} = 0.1\ \mathrm{m}.\]

Step 3: Accelerations

\[ a_t = \varepsilon R = 0.2\ \mathrm{m/s^2}, \]

\[ a_n = \omega^2 R = 100 \cdot 0.1 = 10\ \mathrm{m/s^2}, \]

\[ a = \sqrt{0.2^2 + 10^2} = \sqrt{100.04}. \]

\[a \approx 10.0\ \mathrm{m/s^2}\]

Shortcut without $R$

$ a_n = \omega v = 10 \cdot 1 = 10\ \mathrm{m/s^2} $, $ a_t = \tfrac{\varepsilon}{\omega}\,v = 0.2\ \mathrm{m/s^2}\ \checkmark $

$a_n = 10$ $a_t = 0.2$ $a \approx 10.0$ $O$ $\omega = 10\ \mathrm{rad/s},\ \varepsilon = 2\ \mathrm{rad/s^2}$ $R = v/\omega = 0.1\ \mathrm{m}$ $a = \sqrt{0.2^2 + 10^2} \approx 10.002$

tangential (drawn 5× longer)
normal
total

At $t = 5\ \mathrm{s}$ the rim point's acceleration is almost entirely normal (toward the centre). Even drawn five times too long, the tangential part barely tilts the total.

Result

\[\boxed{\,a \approx 10.0\ \mathrm{m/s^2}\,}\]