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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$m^2$. If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its brakes to wheel is:
A
2.86 kg $m^2s^{-2}$
B
6.66 kg $m^2s^{-2}$
C
8.58 kg $m^2s^{-2}$
D
10.86 kg $m^2s^{-2}$
Detailed Solution
$v = 54\times\frac{5}{18} = 15$ m/s
$\omega_0 = \frac{v}{R} = \frac{15}{0.45} = \frac{100}{3}$ rad/s
$\alpha = \frac{\omega_f - \omega_0}{t} = -\frac{100}{45}$ rad/$s^2$
$|\tau| = I\alpha = 3\times\frac{100}{45} = 6.66$ kg $m^2s^{-2}$
$\omega_0 = \frac{v}{R} = \frac{15}{0.45} = \frac{100}{3}$ rad/s
$\alpha = \frac{\omega_f - \omega_0}{t} = -\frac{100}{45}$ rad/$s^2$
$|\tau| = I\alpha = 3\times\frac{100}{45} = 6.66$ kg $m^2s^{-2}$