91Ó°ÊÓ

Centripetal acceleration is an essential concept when dealing with circular motion, like that experienced on a Ferris wheel. It refers to the acceleration needed to keep an object moving in a circular path.
In our example, to find the centripetal acceleration (\(a_c\)), we use the formula:

• \(a_c = \omega^2 \cdot r\)

Here, \(\omega\) is the angular velocity, and \(r\) is the radius of the Ferris wheel.
Since the Ferris wheel rotates four times per minute, this information helps us calculate the angular velocity:
\(\omega = \frac{2 \pi n}{60}\)
Once you substitute \(\omega\) and the radius (half of the diameter, i.e., \(\frac{18.0 \, \text{m}}{2}\)) into the formula, you can find the centripetal acceleration to be \(2.96 \, \text{m/s}^2\).
This acceleration acts towards the center of the circle, allowing the child to remain seated on the Ferris wheel as it turns.

Angular velocity represents how fast an object rotates around a central point. It's an important factor when calculating centripetal acceleration. For a Ferris wheel, the angular velocity tells us how many times the wheel turns in a given time period.
We determine angular velocity using the formula:
\(\omega = \frac{2 \pi n}{60}\)
where \(n\) is the number of rotations per minute.
In this exercise, the Ferris wheel completes four rotations per minute. By substituting this into our formula, we calculate the angular velocity.

• \(\omega = \frac{2 \pi \times 4}{60} = \frac{8 \pi}{60} \approx 0.418 \, \text{rad/s}\)

Angular velocity is essential for determining the centripetal acceleration, which in turn affects the forces at different points on the Ferris wheel.

Gravitational force is the force exerted by the Earth pulling objects toward its center. It is always directed downward and plays a crucial role in Ferris wheel dynamics.
For the child on the Ferris wheel, the gravitational force \(F_g\) can be calculated using:
\(F_g = mg\)
where \(m\) is the mass of the child and \(g\) is the gravitational acceleration, approximately \(9.81 \, \text{m/s}^2\). In our case, with a \(40.0\)-kg child, the gravitational force is:

• \(F_g = 40.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 392.4 \, \text{N}\)

This force is essential in analyzing the net forces that act on the child at various points along the Ferris wheel ride, such as the top, bottom, and halfway points.

Understanding Ferris wheel dynamics involves looking at forces at different points as the wheel rotates. These dynamics highlight how both gravitational and centripetal forces interact.
At the lowest point, the seat exerts an upward force balancing both gravity and providing the necessary centripetal force. Hence, the formula becomes:

• \(F = mg + ma_c\)
• This results in an upward force of \(586.4 \, \text{N}\).

At the highest point, the seat's upward force is less because it's overcoming gravitational pull, measuring:

• \(F = mg - ma_c\), resulting in \(306.4 \, \text{N}\).

Halfway, the direction of the forces means only gravity plays a role, causing the seat's force to be:

• \(392.4 \, \text{N}\) directed along the radius, inward.

These dynamics make riding a Ferris wheel a steady, exhilarating experience.