A car with wheels of diameter D = 0.60 m moves along a…

A car with wheels of diameter $D = 0.60\ \mathrm{m}$ moves along a straight road at $v = 60\ \mathrm{km/h}$. When braking, it stops after $s = 20\ \mathrm{m}$. Assuming uniform deceleration, find the angular deceleration of its wheels during braking.

Explanation of Method

Angular deceleration of the wheels of a braking car.

Step 1: Deceleration of the car

\[ v = \frac{60}{3.6}\ \mathrm{m/s} = \frac{50}{3}\ \mathrm{m/s} \approx 16.67\ \mathrm{m/s}. \]

From $0 = v^2 - 2|a|s$: $|a| = \frac{v^2}{2s} = \frac{277.8}{40}\ \mathrm{m/s^2} \approx 6.94\ \mathrm{m/s^2}$.

Step 2: Angular deceleration of the wheels

The wheels roll without slipping, so the car's speed and a wheel's spin are tied together by $v = \omega R$ at every moment, with $R = D/2 = 0.3\ \mathrm{m}$. Their rates of change are tied the same way: $a = \alpha R$.

\[ |\alpha| = \frac{|a|}{R} = \frac{v^2}{2sR} = \frac{277.8}{2 \cdot 20 \cdot 0.3}\ \mathrm{rad/s^2}. \]

\[\alpha \approx -23.1\ \mathrm{rad/s^2}\]

Check

Braking time: $ t = \tfrac{2s}{v} = \tfrac{40}{16.67} = 2.4\ \mathrm{s} $

Initial wheel spin $ \omega_0 = \tfrac{v}{R} \approx 55.6\ \mathrm{rad/s} $; $ \tfrac{55.6}{2.4} \approx 23.1\ \mathrm{rad/s^2}\ \checkmark $

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $1$ $2$ $2.4$ $16.7$ $s = 20\ \mathrm{m}$ $a \approx -6.94\ \mathrm{m/s^2}$ $v$ $a$ $\omega$ $R = 0.3\ \mathrm{m}$ $\alpha = a/R \approx -23.1\ \mathrm{rad/s^2}$

Left: the car's speed falls uniformly to zero in 2.4 s; the area is the 20 m braking distance. Right: a wheel rolling without slipping slows its spin in step with the car, so $\alpha = a/R$.

Result

\[\boxed{\,\alpha \approx -23.1\ \mathrm{rad/s^2}\,}\]

Remark

About the original solution. It squared 16.67 as 278.89 instead of 277.89, which gave 6.97 m/s² and 23.23 rad/s². The correct values are 6.94 m/s² and 23.1 rad/s².