Average Torque by Brakes – Rankers Physics

Average Torque by Brakes

An automobile moves on a road with a speed of $54 \text{ km/h}$. The radius of its wheels is $0.45 \text{ m}$ and the moment of inertia of the wheel about its axis of rotation is $3 \text{ kgm}^2$. If the vehicle is brought to rest in $15 \text{ s}$, the magnitude of average torque transmitted by its brakes to wheel is: (2015 Re)

$2.86 \text{ kg m}^2/\text{s}^2$
$6.66 \text{ kg m}^2/\text{s}^2$
$8.58 \text{ kg m}^2/\text{s}^2$
$10.86 \text{ kg m}^2/\text{s}^2$

Solution:

Initial angular velocity $\omega_0 = \frac{v}{r} = \frac{15}{0.45} = \frac{100}{3} \text{ rad/s}$. The angular acceleration is $\alpha = \frac{\omega_0}{t} = \frac{100/3}{15} = \frac{20}{9} \text{ rad/s}^2$. Torque is $\tau = I\alpha = 3 \times (\frac{20}{9}) = 6.66 \text{ kg m}^2/\text{s}^2$.

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