Question

An automobile travelling at 80 km hr-1 has tyres of radius 80 cm. On applying brakes, the car is brought to a stop in 30 complete turns of the tyres. What is the magnitude of the angular acceleration of the wheels? How far does the car move while the brakes are applied?

Answer 1

Answer on Question #65494-Mechanics - Relativity

An automobile travelling at $v_0 = 80 \, \text{km hr}^{-1}$ has tyres of radius $r = 80 \, \text{cm}$. On applying brakes, the car is brought to a stop in $N = 30$ complete turns of the tyres. What is the magnitude of the angular acceleration of the wheels? How far does the car move while the brakes are applied?

Solution

The car will travel distance

$$L = 2 \pi N r = 2 \times 3.14 \times 30 \times 0.8 = 150.72 \, \text{m}.$$

Since the final speed of car is equal to zero

$$v^2 = v_0^2 + 2 a L = v_0^2 + 4 \pi a N r = 0.$$

So the linear acceleration (deceleration) is

$$a = - \frac {v_0^2}{4 \pi N r}.$$

The angular acceleration

$$\varepsilon = \frac {a}{r} = - \frac {v_0^2}{4 \pi N r^2} = - \frac {(80/3.6)^2}{4 \times 3.14 \times 30 \times 0.8^2} = -2.04 \, \frac {\mathrm{rad}}{\mathrm{s}^2}.$$

Answer $\varepsilon = -2.04 \, \frac{\mathrm{rad}}{\mathrm{s}^2}$, $L = 150.72 \, \mathrm{m}$.

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