InterviewSolution
Saved Bookmarks
| 1. |
A car moves on a road with a speed of `54 km h^(-1)`. The radius of its wheels is `0.35 m`. What is the average negative torque transmitted by its brakes to the wheels if the car is brought to rest in `15 s`? Moment of inertia of the wheels about the axis of rotation is `3 kg m^(2)`. |
|
Answer» Here, `u = 54 km h^(-1) = 15 ms^(-1)`, `r = 0.35 m` `tau = ?, t = 15 s, omega_(2) = 0 , omega_(1) = (u)/(r ) = (15)/(0.35) rad s^(-1)` From `omega_(2) - omega_(1) = alpha t` `alpha = (omega_(2) - omega_(1))/(t) = (0 - 15//0.35)/(15) = - (100)/(35) rad//s^(2)` `tau = I alpha = 3 (-(100)/(3)) = - 8.57 kg m^(2) s^(-2)` |
|