Problem 11 - College Physics

Question

\(\bullet\) The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 \(\mathrm{m}\) . Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 \(\mathrm{s} )\) . (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 \(\mathrm{N}\) at the weight- guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

Step-by-Step Solution

Answer

(a) Passenger speed is \(5.24\, \mathrm{m/s}\). (b) Apparent weight: highest point \(800\, \mathrm{N}\), lowest point \(964\, \mathrm{N}\). (c) Time for zero weight at the highest point is \(14.2\, \,s\). (d) Apparent weight at the lowest point in this scenario is \(1048\, \mathrm{N}\).

Step 1: Calculate the Radius of the Ferris Wheel

The diameter of the Ferris wheel is given as 100 meters. The radius \( r \) is half of the diameter. Therefore, \( r = \frac{100}{2} = 50 \text{ meters} \).

Step 2: Calculate the Speed of the Passengers

The speed \( v \) of the passengers on a circular path is given by the formula \( v = \frac{2\pi r}{T} \), where \( T \) is the time for one revolution. Here, \( T = 60 \text{ s} \) and \( r = 50 \text{ m} \). Thus, \( v = \frac{2\pi \times 50}{60} = \frac{100\pi}{60} = \frac{5\pi}{3} \text{ m/s} \approx 5.24 \text{ m/s} \).

Step 3: Calculate Apparent Weight at the Highest Point

At the highest point, the apparent weight \( F' \) is calculated by \( F' = mg - ma \), where \( a = \frac{v^2}{r} \). Plug in \( v = \frac{5\pi}{3} \text{ m/s} \), \( r = 50 \text{ m} \), and \( g = 9.8 \text{ m/s}^2 \) to find \( a \), so \( a = \left( \frac{5\pi}{3} \right)^2 / 50 \). Calculate \( F' = 882 - 82 \).

Step 4: Calculate Apparent Weight at the Lowest Point

At the lowest point, the apparent weight \( F' \) is given by \( F' = mg + ma \). Therefore, \( F' = 882 + 82 = 964 \text{ N} \).

Step 5: Determine Revolution Time for Zero Apparent Weight at Highest Point

For the apparent weight to be zero at the highest point, \( mg = ma \), which means \( g = \frac{v^2}{r} \). Solve for \( T \): \( T = \frac{2\pi r}{v} \) and \( v = \sqrt{gr} \) gives \( T=2\pi \sqrt{\frac{50}{9.8}} \).

Step 6: Calculate Apparent Weight at the Lowest Point for New Revolutions

Using the time found in Step 5, calculate the new speed \( v \), and then calculate the apparent weight \( F' = mg + ma \) at the lowest point with the equation from before.

Key Concepts

Speed Calculation

To find the speed of passengers on a Ferris wheel, we use the formula for the linear speed of circular motion, which is expressed as \[ v = \frac{2\pi r}{T} \]where

• \( v \) is the speed,
• \( r \) is the radius of the Ferris wheel, and
• \( T \) is the time it takes to complete one full revolution.

In this problem, the diameter of the Ferris wheel is 100 meters, giving a radius \( r \) of 50 meters. We know that one complete revolution takes 60 seconds (one minute). By plugging these values into the formula, we get\[ v = \frac{2\pi \times 50}{60} = \frac{100\pi}{60} = \frac{5\pi}{3} \approx 5.24 \text{ m/s} \]. Thus, the speed of passengers when the Ferris wheel completes one rotation every 60 seconds is approximately 5.24 meters per second.

Apparent Weight

Apparent weight on a Ferris wheel changes based on the position of the passenger either at the topmost or lowest point of the ride. Apparent weight is the normal force that passengers perceive as their weight when moving in circular motion.At the highest point, the apparent weight can be calculated using the formula:\[ F' = mg - ma \]where

• \( m \) is the mass,
• \( g \) is the acceleration due to gravity,
• \( a \) is the centripetal acceleration, and
• \( F' \) is the apparent weight.

The centripetal acceleration \( a \) is given by \( \frac{v^2}{r} \). When plugged in, it results in a reduced apparent weight at the top since both gravitational and centripetal forces act downwards.Conversely, at the lowest point, the formula becomes:\[ F' = mg + ma \]Here, the apparent weight increases because the normal force must balance both the gravitational force and provide the necessary centripetal force to keep the passenger moving in a circle.

Centripetal Force

Centripetal force is essential for maintaining circular motion, and it's the force that acts towards the center of a circular path keeping an object moving in that path without deviating. It is calculated by:\[ F_c = \frac{mv^2}{r} \]where

• \( F_c \) is the centripetal force,
• \( m \) is the mass of the object (passenger),
• \( v \) is the constant speed of the object, and
• \( r \) is the radius of the circle.

In the context of the Ferris wheel, as passengers complete the circular path, this force is crucial for keeping them in motion. At different points in the ride, such as the top or bottom, the direction and influence of other forces (like gravity) change, but the centripetal force remains directed towards the center of the circle, ensuring a stable circular path.

Ferris Wheel Physics

Ferris wheels are a classic example of circular motion physics in action. They operate as a large rotating wheel with cabins attached, allowing passengers to experience varying forces as they move around.
The physics behind a Ferris wheel involves balancing gravitational force and centripetal force. As the wheel rotates, passengers are subject to these forces, leading to varying experiences of apparent weight at the top and bottom.

• At the highest point, gravity works with centripetal force, reducing the apparent weight.
• At the lowest point, gravity and the centripetal force work in opposite directions, increasing the apparent weight.

Ferris wheel physics demonstrates these forces in action, and understanding it helps explain why physics principles like circular motion are so relevant to engineering and amusement park ride design.