Problem 59 - Fundamentals of Physics

Question

A woman rides a carnival Ferris wheel at radius \(15 \mathrm{~m}\), completing five turns about its horizontal axis every minute. What are (a) the period of the motion, the (b) magnitude and (c) direction of her centripetal acceleration at the highest point, and the (d) magnitude and (e) direction of her centripetal acceleration at the lowest point?

Step-by-Step Solution

Answer

(a) 12s; (b) ~4.11m/s² down; (c) downward; (d) ~4.11m/s² up; (e) upward.

Step 1: Determine the Period of Motion

The period of motion, denoted by \( T \), is the time taken to complete one full rotation. Given that the Ferris wheel completes five rotations every minute (60 seconds), the period can be calculated as follows: \[ T = \frac{60 \text{ seconds}}{5} = 12 \text{ seconds} \]. Therefore, the period of motion is 12 seconds.

Step 2: Calculate Angular Velocity

To find the centripetal acceleration, we first need the angular velocity, \( \omega \). Angular velocity is calculated using the formula \( \omega = \frac{2\pi}{T} \). Given \( T = 12 \text{ s} \), we have: \[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \text{ rad/s} \].

Step 3: Find the Magnitude of Centripetal Acceleration at Highest Point

The magnitude of the centripetal acceleration, \( a_c \), is given by \( a_c = R \omega^2 \), where \( R = 15 \text{ m} \). Substituting the values, we have: \[ a_c = 15 \times \left( \frac{\pi}{6} \right)^2 = 15 \times \frac{\pi^2}{36} = \frac{15\pi^2}{36} \approx 4.11 \text{ m/s}^2 \].

Step 4: Determine the Direction of Centripetal Acceleration at Highest Point

At the highest point of the Ferris wheel, the direction of the centripetal acceleration is directly downward, toward the center of the Ferris wheel.

Step 5: Find the Magnitude of Centripetal Acceleration at Lowest Point

The magnitude of centripetal acceleration at the lowest point is the same as at the highest point, as it depends only on the radius and angular velocity. Thus, the magnitude is still approximately \( 4.11 \text{ m/s}^2 \).

Step 6: Determine the Direction of Centripetal Acceleration at Lowest Point

At the lowest point of the Ferris wheel, the direction of centripetal acceleration is directly upward, toward the center of the Ferris wheel.

Key Concepts

Period of Motion

The period of motion, denoted as \( T \), is a fundamental concept in circular motion. It refers to the amount of time it takes for an object to complete one full cycle of its circular path. In the context of the Ferris wheel problem, understanding the period helps us determine other important aspects like angular velocity and centripetal acceleration.

For the Ferris wheel, which completes five rotations every minute, we calculate the period by dividing the total time by the number of rotations. This gives us \( T = \frac{60 \text{ seconds}}{5} = 12 \text{ seconds} \). This means that it takes 12 seconds for one complete revolution of the Ferris wheel.

Knowing the period of motion is crucial because it serves as the basis for determining other characteristics like how fast the wheel is spinning or how much force is needed to keep someone safe while riding. By utilizing the period, we can simplify calculations for properties like the angular velocity, which directly ties into how quickly the wheel is moving through its rotation.

Angular Velocity

Angular velocity, symbolized by \( \omega \), describes how fast an object rotates or turns around a central point. It essentially tells us the angular displacement per unit time. This concept is especially important in scenarios where understanding the speed of rotation is vital, such as with a Ferris wheel.

To find the angular velocity, we use the relationship \( \omega = \frac{2\pi}{T} \). Here, \( T \) is the period of motion. In the example from the exercise, with a period of 12 seconds, the angular velocity becomes \( \omega = \frac{2\pi}{12} = \frac{\pi}{6} \text{ rad/s} \).

This angular velocity indicates that the wheel completes its rotation at a rate of \( \frac{\pi}{6} \) radians per second. Understanding angular velocity allows us to connect with centripetal forces, as a greater angular velocity would require a stronger centripetal acceleration to keep the motion smooth and steady. This knowledge is pivotal in safety and mechanical design for rides like a Ferris wheel.

Centripetal Acceleration

Centripetal acceleration is a key concept in circular motion that describes the acceleration required to keep an object moving in a curved path. It points towards the center of the circular path, maintaining the object’s trajectory. For the Ferris wheel in our example, understanding the magnitude and direction of centripetal acceleration is essential for both engineering and safety considerations.

The formula for centripetal acceleration \( a_c \) is \( a_c = R \omega^2 \), where \( R \) is the radius of the path, and \( \omega \) is the angular velocity. In our Ferris wheel problem, with \( R = 15 \text{ m} \) and \( \omega = \frac{\pi}{6} \text{ rad/s} \), the acceleration computes to \( a_c = 15 \times \left( \frac{\pi}{6} \right)^2 = \frac{15\pi^2}{36} \approx 4.11 \text{ m/s}^2 \).

It's worth noting the direction of this acceleration changes based on your position on the wheel. At the highest point, the centripetal acceleration points downward toward the center of the wheel, while at the lowest point, it points upwards, again toward the center. This ensures the rider remains securely on the ride as the forces work towards the center, preventing any outward fly-off.