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The period of motion is a fundamental concept when analyzing circular movement, such as the motion experienced on a Ferris wheel. It refers to the time it takes an object to make one complete cycle or rotation.
To find the period of a Ferris wheel that completes five rotations in one minute, we follow simple calculations:

• First, convert one minute into seconds, since time periods are typically measured in seconds. One minute equals 60 seconds.
• Divide the total time by the number of rotations to find the period per rotation. So, 60 seconds divided by 5 rotations equals 12 seconds per rotation.

Thus, each complete rotation of the Ferris wheel takes 12 seconds, making this the period of motion.
Understanding the period is vital for further calculations, such as determining angular velocity and centripetal acceleration.

Angular velocity describes how fast an object rotates or revolves around a central point. It is crucial to understand how quickly a Ferris wheel, or any circular object, is moving in terms of its rotational speed.
Angular velocity is denoted by the symbol \( \omega \) and it can be calculated using the formula: \[ \omega = \frac{2\pi}{T} \]Where \( T \) is the period of motion. In our scenario:

• We have already determined that the period \( T \) is 12 seconds.
• Substitute \( T \) into the equation to find \( \omega = \frac{2\pi}{12} \), which simplifies to \( \frac{\pi}{6} \) rad/s.

This result tells us the Ferris wheel's angular velocity is \( \frac{\pi}{6} \) rad/s.
By understanding angular velocity, we can derive other important properties of rotational motion, such as centripetal acceleration, which helps determine how forces act on moving bodies in circular paths.

To fully appreciate Ferris wheel dynamics, we must focus on the concept of centripetal acceleration, which is the acceleration experienced by an object moving in a circular path. Specifically, it points towards the center of the rotation and keeps the object in its circular trajectory.
On a Ferris wheel, the centripetal force required to keep a rider moving in a circle results in centripetal acceleration.

• We find this using the formula \( a_c = \omega^2 r \), where \( r \) is the radius of the Ferris wheel, here 15 m, and \( \omega \) is the angular velocity \( \frac{\pi}{6} \) rad/s.
• Substitute these values into the formula: \( a_c = \left(\frac{\pi}{6}\right)^2 \times 15 \), which simplifies to approximately 1.10 m/s².

Additionally, the direction of centripetal acceleration changes depending on your position on the Ferris wheel.
At the highest point, it points downward towards the center, and at the lowest point, it points upward, again towards the center.
This constant redirection is what enables the ongoing circular motion of the Ferris wheel, allowing it to function smoothly and safely.