A Ferris wheel rotating at 20 rad/s slows down with a constant angular acceleration of magnitude $5.0 \mathrm { rad } / \mathrm { s } ^ { 2 }$
How many revolutions does it make while slowing down before coming to rest?
A) 3.2
B) 20
C) 40
D) 6.4

Question

A Ferris wheel rotating at  20 rad/s slows down with a constant angular acceleration of magnitude $5.0 \mathrm { rad } / \mathrm { s } ^ { 2 }$
How many revolutions does it make while slowing down before coming to rest?
A)  3.2 
B) 20
C) 40
D)  6.4

Quizplus · Accepted Answer

To find the number of revolutions the Ferris wheel makes before coming to rest, we can use the equation that relates angular displacement ($\theta$), initial angular velocity ($\omega_0$), final angular velocity ($\omega$), and angular acceleration ($\alpha$). Since the wheel comes to rest, $\omega = 0$, and we are given $\omega_0 = 20 \, \text{rad/s}$ and $\alpha = -5.0 \, \text{rad/s}^2$ (negative because it's decelerating). The equation we use is:$$\omega^2 = \omega_0^2 + 2\alpha\theta$$Solving for $\theta$ when $\omega = 0$:$$0 = (20)^2 + 2(-5.0)\theta$$$$400 = 10\theta$$$$\theta = 40 \, \text{rad}$$To convert radians to revolutions, we divide by $2\pi$ (since there are $2\pi$ radians in one revolution):$$\text{Revolutions} = \frac{40}{2\pi} \approx 6.4$$So, the Ferris wheel makes approximately 6.4 revolutions before coming to rest.

A Ferris Wheel Rotating at 20 Rad/s Slows Down 5.0rad/s25.0 \mathrm { rad } / \mathrm { s } ^ { 2 }5.0rad/s2

A Ferris Wheel Rotating at 20 Rad/s Slows Down $5.0 \mathrm { rad } / \mathrm { s } ^ { 2 }$