A wheel of radius R = 20 cm starts rotating from rest with…

A wheel of radius $R = 20\ \mathrm{cm}$ starts rotating from rest with constant angular acceleration $\varepsilon = 6.28\ \mathrm{rad/s^2}$. Find the speed and the acceleration of a point on its rim $t = 5\ \mathrm{s}$ after the start.

Explanation of Method

Speed and acceleration of a rim point after 5 s.

Step 1: Angular velocity and speed

\[ \omega = \varepsilon t = 6.28 \cdot 5 = 31.4\ \mathrm{rad/s}, \]

\[ v = \omega R = 31.4 \cdot 0.2. \]

\[v \approx 6.28\ \mathrm{m/s}\]

Step 2: Tangential and normal acceleration

\[ a_t = \varepsilon R = 6.28 \cdot 0.2 = 1.256\ \mathrm{m/s^2}, \]

\[ a_n = \omega^2 R = 31.4^2 \cdot 0.2 \approx 197.2\ \mathrm{m/s^2}. \]

Step 3: Total acceleration

\[ a = \sqrt{1.256^2 + 197.2^2} \approx 197.2\ \mathrm{m/s^2}. \]

\[a \approx 197\ \mathrm{m/s^2}\]

$a\ (\mathrm{m/s^2})$ $t\ (\mathrm{s})$ $0$ $1$ $2$ $3$ $4$ $5$ $50$ $100$ $150$ $197$ $a \approx a_n \approx 197\ \mathrm{m/s^2}$ $a_t = 1.256\ \mathrm{m/s^2}$ $a_n = \varepsilon^2 R\, t^2$

normal acceleration
tangential acceleration

The tangential acceleration stays constant, while the normal acceleration grows with the square of time. They are equal only at $t = 1/\sqrt{\varepsilon} \approx 0.4\ \mathrm{s}$; after 5 s the normal part is about 157 times larger.

Result

\[\boxed{\,v \approx 6.28\ \mathrm{m/s}\,}\]

\[\boxed{\,a \approx 197\ \mathrm{m/s^2}\,}\]

Remark

Note. $\varepsilon = 6.28 \approx 2\pi\ \mathrm{rad/s^2}$ means the wheel gains one revolution per second every second: after 5 s it turns 5 times per second. The tangential part is so small that it changes the total by less than 0.01 %.