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Chapter 5: Problem 54

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 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 s). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 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?

Short Answer

Expert verified
(a) 5.24 m/s. (b) 832.8 N (high) and 931.2 N (low). (c) 14.2 s. (d) 1764 N.

Step by step solution

01

Find Passenger Speed

The circumference of the Ferris wheel is given by the formula for the circumference of a circle, \( C = \pi \times d \), where \( d = 100 \) m is the diameter. So, \( C = \pi \times 100 = 314.16 \) m. Since the Ferris wheel makes one revolution every 60 seconds, the speed \( v \) of a passenger is \( v = \frac{C}{T} = \frac{314.16}{60} \approx 5.24 \) m/s.
02

Determine Apparent Weight High and Low

At the highest point, the apparent weight \( W_h \) is the actual weight minus the centripetal force \( F_c \). \( F_c = m \times a_c = m \times \frac{v^2}{r} \), where \( m \) is the mass, \( v = 5.24 \) m/s, and \( r = 50 \) m. First, find \( m = \frac{882}{9.8} = 90 \) kg. So, \( F_c = 90 \times \frac{(5.24)^2}{50} \approx 49.2 \) N. Thus, \( W_h = 882 - 49.2 = 832.8 \) N. At the lowest point, apparent weight \( W_l = 882 + 49.2 = 931.2 \) N.
03

Calculate Time for Zero Apparent Weight

To find the time for one revolution where the apparent weight at the highest point is zero, set \( W_h = 0 \). Then, \( F_c = 882 \), so \( \frac{(90)v^2}{50} = 882 \). Solving \( v^2 = 490 \) gives \( v = 22.14 \) m/s. The circumference is still 314.16 m, thus\( T = \frac{314.16}{22.14} \approx 14.2 \) s.
04

Find Apparent Weight at Lowest Point During Zero Weight Condition

Here, the centripetal force \( F_c = 882 \) N and acts upwards, so the apparent weight \( W_l = 882 + 882 = 1764 \) N.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Apparent Weight
The concept of apparent weight is an interesting one, especially in the context of circular motion like that experienced on a Ferris wheel. Apparent weight is essentially the normal force exerted by a surface to support an object, which can differ from the actual gravitational weight of the object.
  • At the highest point of a Ferris wheel, the centripetal force needed to keep you in circular motion acts in the same direction as gravity, making you feel lighter. Hence, your apparent weight is less than your actual weight.
  • Conversely, at the lowest point, the centripetal force and gravity work in opposite directions on your body, causing the apparent weight to increase.
In simpler terms, your body feels these changes as variations in weight, which is why you may feel lighter at the top and heavier at the bottom of the Ferris wheel's rotation.
Centripetal Force
Centripetal force plays a pivotal role in circular motion phenomena, especially for passengers on a Ferris wheel. This force keeps an object moving in a circular path and is always directed toward the center of the circle. On the Ferris wheel:
  • The passengers are constantly changing direction, which requires a force acting toward the center of the wheel, provided by the tension in the supporting structure.
  • Centripetal force can be calculated using the formula \( F_c = m \cdot a_c = m \cdot \frac{v^2}{r} \), where \( m \) is mass, \( v \) is velocity, and \( r \) is the radius of the circle.
This force doesn鈥檛 do any work because it acts perpendicular to the direction of motion, maintaining speed but altering direction. Understanding centripetal force helps explain why you remain seated comfortably as the Ferris wheel spins.
Ferris Wheel Physics
Ferris wheel physics combines elements of circular motion and Newtonian mechanics. The motion of a Ferris wheel is a classic example of uniform circular motion where:
  • Speed remains constant, but velocity changes due to the continual change in direction.
  • At various points on the Ferris wheel, the apparent weight of the passengers changes due to the interplay between gravitational forces and centripetal forces.
The Ferris wheel provides a controlled environment to study how forces operate in circular motion, making it an excellent real-world application of physics principles. As the Ferris wheel rotates, passengers experience variation in apparent weight, often noticed as the feeling of being lifted off the seat or pressed downward, illustrating the direct effect of circular motion forces.
Revolution Time
Revolution time describes the time it takes for one complete turn of the Ferris wheel. It's important for calculating the speed of rotation and understanding how forces change throughout the motion. For instance:
  • If the time for one revolution decreases while the diameter remains the same, the speed increases, affecting both centripetal force and apparent weight of the passengers.
  • By modifying the revolution time, we can control the experience of the passengers, illustrated by alterations in apparent weight.
In the context of the exercise, adjusting the revolution time explains how conditions like zero apparent weight at certain points are achieved, highlighting its significance in designing thrill rides like Ferris wheels.

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Most popular questions from this chapter

A box is sliding with a constant speed of 4.00 m/s in the \(+x\)-direction on a horizontal, frictionless surface. At \(x =\) 0 the box encounters a rough patch of the surface, and then the surface becomes even rougher. Between \(x =\) 0 and \(x =\) 2.00 m, the coefficient of kinetic friction between the box and the surface is 0.200; between x = 2.00 m and x = 4.00 m, it is 0.400. (a) What is the \(x\)-coordinate of the point where the box comes to rest? (b) How much time does it take the box to come to rest after it first encounters the rough patch at \(x =\) 0?

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