Question

A Ferris wheel rotating at 20 rad/s slows down with a constant angular acceleration of magnitude 5.0 rad/s2. How many revolutions does it make while slowing down before coming to rest

Answer 1

Answer:

It takes 6.37 revolutions to stop.

Explanation:

The constant angular acceleration is negative if we choose the wheel direction of rotation as positive direction, so radial acceleration is α=5.0$\frac{rad}{s^{2}}$.Because the wheel is changing its velocity and we already know radial acceleration we should use the Galileo's kinematic rotational equation:

$\omega^{2}=\omega_{i}^{2}+2\alpha\varDelta\theta$

with $\omega$ the final angular velocity (is zero because the wheel comes to rest), $\omega_{i}$ the initial angular velocity and Δθ the angular displacement. Solving (1) for Δθ :

$\varDelta\theta=\frac{\omega^{2}-\omega_{i}^{2}}{2\alpha}$

$\varDelta\theta=\frac{0-20^{2}}{(2)(-5.0)}=40rad$

The angular displacement can be converted to revolution knowing that 1 revolution is 2π rad:

$40rad=\frac{40rad}{2\pi\frac{rad}{rev}}$

$40rad=6.37 rev$