Question

5. A motorcycle accelerates uniformly from rest and reaches a linear speed of 23.8 m/s in a time of 8.10 s. The radius of each tire is 0.270 m. What is the magnitude of the angular acceleration of each tire?

Answer 1

here,

v = 23.8 m/s

u = 0 m/s

Using First equation of motion ,

v = u + at

23.8 = 8.10 * a

a = 2.94 m/s^2

Now,

as angular acceleration = a/r

angular acceleration = 2.94 / 0.270

angular acceleration = 10.89 rad/s^2

angular acceleration of each tire is 10.89 rad/s^2

Answer 2

acceleration =change in speed/time =23.8/8.10 = 2.938 m/s

angular acceleration = acceleration/radius = 2.938/0.27 = 10.88 rad/sec2

Answer 3

angular acceleration = $\alpha = \omega /t = v/(r*t)$ = 23.8 / (0.27*8.1) = 10.883 rad/s^2