Question

A bicycle wheel with a radius of 0.42 m accelerates uniformly for 6.8 s from an initial angular velocity of 5.5 rad/s to a final angular velocity of 6.7 rad/s. What was its angular acceleration

Answer 1

Answer:

The angular acceleration required  is 0.1765 rad/ $s^2$

Explanation:

The radius of the bicycle wheel has a radius of 0.42 m.

The acceleration is for time, t =  6.8 seconds.

Initial angular velocity is given as  $\omega_{0}$  = 5.5 rad/s

Final angular velocity is given as $\omega_{f}$ = 6.7 rad/s

Therefore from the formula for angular speed we get

$\omega_{f}$ = $\omega_{0}$ + ($\frac{d\omega}{dt}$ $\times$ t),   where t is the time in seconds.

Therefore we get

6.7 =  5.5 + (6.8 × $\frac{d\omega}{dt}$ )

Therefore we get the angular acceleration, $\frac{d\omega}{dt}$ = $\frac{(6.7 - 5.5 }{6.8} = 0.1765 rad/ s^2$

The angular acceleration required  is 0.1765 rad/ $s^2$