An autmobile moves on road with a speed of `54 km//h`. The radius of its wheel 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 tansmitted by its brakes to the wheel is :

Options

A. `2.86` kg `m^2 s^-2`
B. `6.66` kg `m^2 s^-2`
C. `8.58` kg `m^2 s^-2`
D. `10.86` kg `m^(2) s^-2`

Answer

B

Explanation

(b) Velocity of the automobile
`v = 54 xx (5)/(18) = 15 m//s`
`omega_0 = (v)/(R) =(15)/(0.45) = (100)/(3) rad//s`
So angular acceleration
`prop = (Delta omega)/(f) =(omega_f - omega_0)/(t) = -(100)/(45) rad//s^2`
So, torque `I prop = 3 xx (100)/(45) = 6.66 kg - m^2s^-2`.