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 #65508, Physics / Mechanics | Relativity

An automobile travelling at $80~\mathrm{km}$ hr$^{-1}$ has tyres of radius $80~\mathrm{cm}$. 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:

$$v = 80~\mathrm{km/h} = 22.2~\mathrm{m/s}$$ $$r = 80~\mathrm{cm} = 0.8~\mathrm{m}$$

Initial angular velocity of wheels

$$\omega_0 = \frac{v}{r} = 22.2~\mathrm{m/s} / 0.8~\mathrm{m} = 27.75~\mathrm{rad/s}$$

The angular displacement of wheels

$$\theta = 30 \times 2\pi = 60\pi~\mathrm{radian}$$

Final angular velocity of wheel

$$\omega = 0$$

From equation of rotational motion

$$\omega^2 = \omega_0^2 + 2\alpha \times \theta$$

Finally we get

$$0^2 = (27.75)^2 + 2 \times \alpha \times 60\pi$$ $$\alpha = - (27.75)^2 / 120~\pi = -2.03~\mathrm{rad/s^2}$$

Since they turn 30 rev, the distance will be 30 times the circumference of the tires:

$$d = (30~\mathrm{rev})(\pi \times 0.80~\mathrm{m}) = 75.4~\mathrm{m}.$$

Answer: 27.75 rad/s; -2.03 rad/s²; 75.4 m

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