A train moves along a circular track of radius R = 0.5 km…

A train moves along a circular track of radius $R = 0.5\ \mathrm{km}$ with angular acceleration $\varepsilon = 0.0049\ \mathrm{rad/s^2}$. What is the train's acceleration at the moment its speed is $v = 60\ \mathrm{km/h}$? What is the angular velocity of its wheels at that moment if their radius is $r = 0.5\ \mathrm{m}$?

Explanation of Method

Acceleration of a train on a curve and the spin of its wheels.

Step 1: Convert the speed

\[ v = \frac{60}{3.6}\ \mathrm{m/s} \approx 16.67\ \mathrm{m/s}. \]

Step 2: Tangential acceleration

The angular acceleration about the centre of the track changes the speed along the track: \[a_t = \varepsilon R = 0.0049 \cdot 500 = 2.45\ \mathrm{m/s^2}.\]

Step 3: Normal (centripetal) acceleration

Because the track curves, there is also an acceleration toward the centre: $a_n = \frac{v^2}{R} = \frac{16.67^2}{500} \approx 0.556\ \mathrm{m/s^2}$.

Step 4: Total acceleration

The two parts are perpendicular: \[a = \sqrt{a_t^2 + a_n^2} = \sqrt{2.45^2 + 0.556^2} = \sqrt{6.003 + 0.309}.\]

\[a \approx 2.51\ \mathrm{m/s^2}\]

It points mostly forward, tilted toward the centre by $\arctan\tfrac{0.556}{2.45} \approx 12.8^\circ$.

$a_t = 2.45\ \mathrm{m/s^2}$ $a_n \approx 0.56\ \mathrm{m/s^2}$ $a \approx 2.51\ \mathrm{m/s^2}$ $\text{train}$ $\text{to the centre } (R = 500\ \mathrm{m})$ $v$ $r$ $\omega_w$ $\omega_w = \tfrac{v}{r} \approx 33.3\ \mathrm{rad/s}$ $\text{a wheel of the train}$

tangential
normal (to centre)
total

Left, seen from above and drawn to scale: the train speeds up along the curve (tangential part) and turns (normal part, toward the centre). The total acceleration is the diagonal. Right: a wheel rolling without slipping turns with $\omega_w = v/r$.

Step 5: Angular velocity of the wheels

The wheels roll without slipping, so a wheel's rim speed relative to its axle equals the train's speed: $\omega_w = \frac{v}{r} = \frac{16.67\ \mathrm{m/s}}{0.5\ \mathrm{m}}$.

\[\omega_w \approx 33.3\ \mathrm{rad/s} \ \ (\approx 318\ \mathrm{rpm})\]

Result

\[\boxed{\,a \approx 2.51\ \mathrm{m/s^2}\,}\]

\[\boxed{\,\omega_w \approx 33.3\ \mathrm{rad/s} \ \ (\approx 318\ \mathrm{rpm})\,}\]

Remark

Note. Two different angular quantities appear here: the train's angular acceleration about the centre of the curve ($\varepsilon$, with radius 500 m) and the spin of the wheels about their axles ($\omega_w$, with radius 0.5 m). The wheel spin is the same as on a straight track: it depends only on the speed $v$. Compare with the rolling wheel in Rotational Kinematics 004.