To solve for the radius $r$ of the Ferris wheel's motion, we can use the formula for centripetal acceleration:

$a_c = \frac{v^2}{r}$

where:

• $a_c$ is the centripetal acceleration,
• $v$ is the speed,
• $r$ is the radius of the circular motion.

Given:

• $a_c = 8.11 \, \text{m/s}^2$
• $v = 4.62 \, \text{m/s}$

Rearrange the formula to solve for $r$:

$r = \frac{v^2}{a_c}$

Now substitute the values:

$r = \frac{(4.62)^2}{8.11}$

$r = \frac{21.3444}{8.11}$

$r \approx 2.63 \, \text{m}$

So, the radius of the Ferris wheel's motion is approximately 2.63 meters.

Would you like further details on this solution or have any other questions?

Here are five related questions to explore further:

1. What happens to the centripetal acceleration if the speed of the Ferris wheel increases?
2. How would the radius change if the centripetal acceleration were reduced to half its current value?
3. What would the required centripetal force be for a passenger with a specific mass at the top of the Ferris wheel?
4. How would you calculate the period of rotation for this Ferris wheel?
5. What other factors could affect the centripetal acceleration experienced by passengers?

Tip: When solving problems involving circular motion, make sure to identify if centripetal force, velocity, or acceleration values are given, as they can lead you directly to the radius or other needed quantities.