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An automobile moves on a road with a speed of 54 km h⁻¹. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m². If the vehicle is brought to rest in 15s, the magnitude of average torque transmitted by its brakes to the wheel is:

1. 8.58 kg m² s⁻²
2. 10.86 kg m² s⁻²
3. 2.86 kg m² s⁻²
4. 6.66 kg m² s⁻²

Correct answer: 10.86 kg m² s⁻²

Solution

The initial angular velocity of the wheel is calculated as \( \omega = \frac{v}{r} = \frac{54 \times 1000}{3600 \times 0.45} = 33.33 \ \text{rad/s} \). The angular deceleration is \( \alpha = \frac{\Delta \omega}{\Delta t} = \frac{0 - 33.33}{15} = -2.22 \ \text{rad/s}^2 \). Using \( \tau = I \alpha \), the torque is \( \tau = 3 \times 2.22 = 6.66 \ \text{kg m}^2 \text{s}^{-2} \).

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