Question

A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

Answer 1

The answer is below

Step-by-step explanation:

a) Using the formula:

`\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ angular\ velocity,\omega_o=initial\ anglular\ velocity,\alpha= angular\ acceleration,\\\theta=angular\ distance\\\\Given\ that:\\\\initial\ velocity(u)=28.78m/s,distance(s)=50\ m,radius(r)=0.25\ m,\\final/ velocity(v)=0(stop)\\\\\omega=v/r=(28.78m/s)/(0.25m) =115.12\ rad/s,\omega_o=0,\theta=s/r=(50\ m)/(0.25\ m)=200\ rad\\ \\\omega^2=\omega_o^2+2\alpha \theta\\\\115.12^2=0^2+2\alpha(200)\\\\2\alpha(200)=13252.6144\\\\\alpha=33.13\ rad/s^2`

b)

`\theta=200\ rad=200\ rad*(1\ rev)/(2\pi\ rad)=31.83\ rev`