Question

A rotating wheel requires 2.90-s to rotate through 37.0 revolutions. Its angular speed at the end of the 2.90-s interval is 97.2 rad/s. What is the constant angular acceleration of the wheel?

Answer 1

Answer:

Angular acceleration, $\alpha =20.32\ rad/s^2$

Explanation:

It is given that,

Displacement of the rotating wheel, $\theta=37\ rev=232.47\ radian$

Time taken, t = 2.9 s

Initial speed of the wheel, $\omega_i=0$

Final speed of the wheel, $\omega_f=97.2\ rad/s$

Let $\alpha$ is the angular acceleration of the wheel. Using the third equation of kinematics to find it as :

$\alpha=\dfrac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha=\dfrac{(97.2)^2}{2\times 232.47}$

$\alpha =20.32\ rad/s^2$

So, the angular acceleration of the wheel is $20.32\ rad/s^2$. Hence, this is the required solution.