Question

question 13 (probelm reference 5 - 2) the ferris wheel at a county fair has a radius of 32.0 m. each of the gondolas on the wheel takes 1.00 minute to complete one revolution when it is running at full speed. ignore gravitational effects. what is the centripetal acceleration of a gondola when the ferris wheel is running at full speed? 11.2 m/s² 0.351 m/s² 0.105 m/s² 12.6 m/s²

Explanation:

Step1: Find the angular velocity $\omega$.

The gondola takes $t = 1.00\ min=60\ s$ to complete one - revolution. One - revolution is $\theta = 2\pi$ radians. The angular velocity $\omega=\frac{\theta}{t}$. So, $\omega=\frac{2\pi}{60}\ rad/s$.

Step2: Use the centripetal - acceleration formula.

The formula for centripetal acceleration is $a_c = r\omega^{2}$, where $r = 32.0\ m$ is the radius of the Ferris wheel. Substitute $\omega=\frac{2\pi}{60}\ rad/s$ and $r = 32.0\ m$ into the formula:
\[

$$\begin{align*} a_c&=r\omega^{2}\\ &=32\times(\frac{2\pi}{60})^{2}\\ &=32\times\frac{4\pi^{2}}{3600}\\ &\approx32\times\frac{4\times9.87}{3600}\\ &=32\times\frac{39.48}{3600}\\ &=\frac{1263.36}{3600}\\ &\approx 0.351\ m/s^{2} \end{align*}$$

\]

Answer:

$0.351\ m/s^{2}$