An autmobile travelling at `80 km//h` has tyres of `75.0 cm` diameter. What is the angualr speed of the tyres about their axles ? If the car is brought to stop uniformly in `30` complete turns of the tyres without skidding, what is the magnitude of angular acceleration of the wheels ? How far does the car move during the braking ?

Explanation

Here, `v = 80 km//h = (80xx1000)/(60 xx 60) = 22.2 m//s`
This is also the speed of centre of mass of the car
As `r = (75)/(2)cm = 37.5 cm = 0.375 m`
`:. omega_(0) = (v_(cm))/(r ) = (22.2)/(0.375) = 59.2 rad//s`
For `omega =0, theta = 30 xx 2pi radian`
As `omega^(2) = omega_(0)^(2) + 2 alpha theta`
`:. alpha = (omega^(2) - omega_(0)^(2))/(2 theta)`
`alpha = ( 0 - (59.2)^(2))/(2 xx 30(2pi)) = 9.33 rad//s^(2)`
The distance moved by the car during braking,
`s = 30 xx 2pi r = 30 xx 2 xx 3.14 xx 0.375 = 70.65 m`