Answer:
Angular acceleration, [tex]\alpha =-2.168\ rad/s^2[/tex]
Explanation:
Given that,
Number of revolution, [tex]\theta=63\ rev=395.84\ rad/s[/tex]
Diameter of the car, d = 0.88 m
Radius, r = 0.44 m
Initial speed of the car, u = 91 km/h = 25.27 m/s
Initial angular speed,
[tex]\omega_i=\dfrac{u}{r}[/tex]
[tex]\omega_i=\dfrac{25.27}{0.44}[/tex]
[tex]\omega_i=57.43\ rad/s[/tex]
Final speed of the car, v = 63 km/h = 17.5 m/s
Final angular speed,
[tex]\omega_f=\dfrac{v}{r}[/tex]
[tex]\omega_f=\dfrac{17.5}{0.44}[/tex]
[tex]\omega_f=39.77\ rad/s[/tex]
To find,
The angular acceleration of the tires.
Solution,
The angular acceleration of the wire can be calculated using third equation of kinematics as :
[tex]\alpha =\dfrac{\omega_f^2-\omega_i^2}{2\theta}[/tex]
[tex]\alpha =\dfrac{39.77^2-57.43^2}{2(395.84)}[/tex]
[tex]\alpha =-2.168\ rad/s^2[/tex]
So, the angular acceleration of the tires is [tex]-2.168\ rad/s^2[/tex].
