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…

Question

asked Sep 7, 2020 164k views

1 Answer

Jackytse · Answer 1 · 2020-09-11T18:21:06+0000

Answer:

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

Step-by-step 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 :

$\alpha=(\omega_f^2-\omega_i^2)/(2\theta)$

$\alpha=((97.2)^2)/(2* 232.47)$

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

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