Question

a ferris wheel with a 9 - m radius makes 9 complete clockwise revolutions in 4 minutes about its horizontal axis. what is the magnitude of a passengers centripetal acceleration at the highest point during the ride? your answer: answer units

Explanation:

Step1: Calculate the angular speed $\omega$

The Ferris - wheel makes $n = 9$ revolutions in $t=4$ minutes. One revolution is $2\pi$ radians. So the total angle $\theta=9\times2\pi$ radians. The time $t = 4\times60s=240s$. The angular speed $\omega=\frac{\theta}{t}=\frac{9\times2\pi}{240}=\frac{3\pi}{40}\text{ rad/s}$.

Step2: Calculate the centripetal acceleration $a_c$

The formula for centripetal acceleration is $a_c = r\omega^{2}$, where $r = 9m$ is the radius of the Ferris - wheel. Substitute $\omega=\frac{3\pi}{40}\text{ rad/s}$ and $r = 9m$ into the formula: $a_c=9\times(\frac{3\pi}{40})^{2}=9\times\frac{9\pi^{2}}{1600}\approx0.50m/s^{2}$.

Answer:

$0.50m/s^{2}$