A disk of radius R = 12 cm starts rotating from rest with…

A disk of radius $R = 12\ \mathrm{cm}$ starts rotating from rest with angular acceleration $\varepsilon = 2\ \mathrm{rad/s^2}$. Find the acceleration of a point on its edge $t = 2\ \mathrm{s}$ after the start.

Explanation of Method

Acceleration of a point on the edge of a spinning-up disk.

Step 1: Angular velocity at 2 s

\[ \omega = \varepsilon t = 2 \cdot 2 = 4\ \mathrm{rad/s}. \]

Step 2: The two parts of the acceleration

Tangential (from the change of speed) and normal (from the change of direction): \[a_t = \varepsilon R = 2 \cdot 0.12 = 0.24\ \mathrm{m/s^2},\]

\[ a_n = \omega^2 R = 16 \cdot 0.12 = 1.92\ \mathrm{m/s^2}. \]

Step 3: Total acceleration

\[ a = \sqrt{a_t^2 + a_n^2} = \sqrt{0.0576 + 3.6864} = \sqrt{3.744}. \]

\[a \approx 1.93\ \mathrm{m/s^2}\]

$a_n = 1.92$ $a_t = 0.24$ $a \approx 1.94\ \mathrm{m/s^2}$ $O$ $\omega = \varepsilon t = 4\ \mathrm{rad/s}$ $R = 0.12\ \mathrm{m}$ $\text{angle between } \vec a \text{ and radius: } 7.1^\circ$

tangential
normal
total

A point on the rim at $t = 2\ \mathrm{s}$, drawn to scale. By then the normal acceleration is 8 times the tangential one, so the total acceleration points almost to the centre.

Result

\[\boxed{\,a \approx 1.93\ \mathrm{m/s^2}\,}\]

Remark

Note. A shortcut: $a = R\sqrt{\varepsilon^2 + \omega^4} = 0.12\sqrt{4 + 256} \approx 1.93\ \mathrm{m/s^2}$. At the very start ($\omega = 0$) only the tangential part exists; it is quickly overtaken because $a_n$ grows with $t^2$.