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 the wheel is

Options

A. `10 . 86 kgm^(2) s^(-2)`
B. `2.86 kg m^(2) s^(-2)`
C. `6.66 kg m^(2) s^(-2)`
D. `8.58 kg m^(2) s^(-2)`

Answer

C

Explanation

Here , speed of the automobile , `v = 54 km h^(-1) = 54 xx (5)/(18) m s^(-1) = 15 m s^(-1)`
Radius of the wheel of the automobile , `R = 0.45` m
Moment of inertia of the wheel about its axis of rotation , I = 3 kg `m^(2)`
Time in which the vehicle brought to rest , t = 15 s
The initial angular speed of the wheel is
`omega_(1) = (v)/(R) = (15 ms^(-1))/(0.45 m) = (1500)/(45) rad s^(-1) = (100)/(3) rad s^(-1)`
and its final angular speed is `omega_(f) = 0` (as the vehicle comes to rest)
`therefore` The angular retardation of the wheel is `alpha = (omega_(f) - omega_(i))/(t) = (0 - (100)/(3))/(15 s) = - (100)/(45) rad s^(-2)`
The magnitude of required torque is `tau = I|alpha| = (3 kg m^(2)) ((100)/(45) "rad" s^(-2))`
`= (20)/(3) kg m^(2) s^(-2) = 6.66 kg m^(2) s^(-2)`