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A circular-motion addict of mass \(80 \mathrm{~kg}\) rides a Ferris wheel around in a vertical circle of radius \(10 \mathrm{~m}\) at a constant speed of \(6.1 \mathrm{~m} / \mathrm{s}\). (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?

Short Answer

Expert verified
The period is 10.3 s. Normal forces: 486.4 N at the top, 1081.6 N at the bottom.

Step by step solution

01

Calculate the Period of the Motion

The period of motion \( T \) is the time taken to complete one full circle. The speed of the motion \( v \) is given as \( 6.1 \mathrm{~m/s} \). The circumference of the circle \( C \) can be calculated using the formula \( C = 2\pi r \), where \( r \) is the radius of the circle (10 meters).Calculating the circumference: \[ C = 2 \pi \times 10 = 20\pi \approx 62.83 \text{ meters} \]Using the relationship \( v = \frac{C}{T} \), solve for the period \( T \):\[ T = \frac{C}{v} = \frac{62.83}{6.1} \approx 10.3 \mathrm{~s} \]
02

Determine the Normal Force at the Highest Point

At the highest point of the circular path, the normal force \( N_\text{high} \) and the gravitational force \( mg \) act downwards. The centripetal force required for the motion is provided by the difference between gravitational force and the normal force. The equation is given by:\[ m a_c = mg - N_\text{high} \]Where:- \( a_c = \frac{v^2}{r} \) is the centripetal acceleration\[ a_c = \frac{6.1^2}{10} = 3.72 \mathrm{~m/s^2} \]Substituting values in the centripetal force equation:\[ 80 \times 3.72 = 80 \times 9.8 - N_\text{high} \]\[ 297.6 = 784 - N_\text{high} \]\[ N_\text{high} = 784 - 297.6 = 486.4 \mathrm{~N} \]
03

Calculate the Normal Force at the Lowest Point

At the lowest point of the circular path, the gravitational force \( mg \) acts downwards and the normal force \( N_\text{low} \) acts upwards. The centripetal force required for the motion is provided by the sum of gravitational force and the normal force:\[ m a_c = N_\text{low} - mg \]Using the centripetal acceleration calculated earlier:\[ 80 \times 3.72 = N_\text{low} - 80 \times 9.8 \]\[ 297.6 = N_\text{low} - 784 \]\[ N_\text{low} = 297.6 + 784 = 1081.6 \mathrm{~N} \]

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Key Concepts

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

Centripetal Force
Centripetal force is a force that keeps a body moving in a circular path and is always directed towards the center of the circle. This means whenever you're on a circular ride, like a Ferris wheel, the centripetal force is what keeps you in the circular motion by constantly changing your direction.

Mathematically, centripetal force (\( F_c \) ) is calculated using the formula:
  • \[F_c = m a_c\]
  • where \( m \) is the mass of the object (in kilograms)
  • \( a_c \) is the centripetal acceleration
The centripetal acceleration \( a_c \) can be calculated with the formula: \(\[ a_c = \frac{v^2}{r} \]\)
Here, \( v \) is the constant speed of the motion and \( r \) is the radius of the circular path. In our exercise, the radius is 10 meters and the speed is 6.1 meters per second. After plugging these values into the formula, we find \( a_c = 3.72 \text{ meters per second squared} \). Ultimately, using the centripetal force equation, it's possible to find how much force is needed to keep the object moving in a circle.
Normal Force
The normal force is the support force exerted upon an object in contact with another stable object. In the case of the Ferris wheel, the normal force is what the seat exerts on the rider, and it varies depending on the position of the Ferris wheel.

At different points on a Ferris wheel, the normal force has to account for changes in gravitational pull. When you're at the top of the Ferris wheel, both gravity and the normal force act in the same direction towards the center of the circle:
  • At the highest point: \( N_\text{high} = mg - m a_c \)
  • This means the normal force is less than your weight because both the normal force and the gravitational force combine to provide the necessary centripetal force.
  • When calculated, the normal force at the highest point is 486.4 N.
Conversely, at the bottom of the ride, the gravitational force acts downwards, and the seat's normal force acts upwards:
  • At the lowest point: \( N_\text{low} = m a_c + mg \)
  • Here, the normal force is greater than your weight because part of it must counteract gravity while still providing centripetal force.
  • The calculated normal force at the lowest point is 1081.6 N.
Ferris Wheel Calculations
Ferris wheel calculations often involve understanding how forces interact as you move in a vertical circle with a constant speed. Here, you calculate important variables such as the period, the normal forces at different points, and understand the effect of these forces.

**Determining the Period**
Period is the time it takes to complete one full rotation around the Ferris wheel. It can be derived from the speed and circumference of the circle using the formula:
  • \(\[ T = \frac{C}{v} \]\)
  • \( C \) being the circumference: \( C = 2\pi r = 62.83 \text{ meters} \)
  • The period is roughly 10.3 seconds.
**Forces at Play**
While riding a Ferris wheel, several forces are always at play, especially depending on your location on the wheel. At the top, you feel lighter due to the additive nature of gravitational and centripetal forces acting in tandem. However, at the bottom, the forces offset, making the normal force from the seat feel stronger as it has to counteract gravity.
Performing these calculations helps not only in understanding the dynamics of motion on a Ferris wheel but also on other rides and vehicles that operate on similar principles like roller coasters.

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Most popular questions from this chapter

A student of weight \(667 \mathrm{~N}\) rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force \(\vec{F}_{N}\) on the student from the seat is \(556 \mathrm{~N}\). (a) Does the student feel "light" or "heavy" there? (b) What is the magnitude of \(\vec{F}_{N}\) at the lowest point? If the wheel's speed is doubled, what is the magnitude \(F_{N}\) at the (c) highest and (d) lowest point?

A loaded penguin sled weighing \(80 \mathrm{~N}\) rests on plane inclined at angle \(\theta=20^{\circ}\) to the horizontal (Fig. \(6-23\) ). Between the sled and the plane, the coefficient of static friction is \(0.25,\) and the coefficient of kinetic friction is \(0.15 .\) (a) What is the least magnitude of the force \(\vec{F},\) parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude \(F\) that will start the sled moving up the plane? (c) What value of \(F\) is required to move the sled up the plane at constant velocity?

A roller-coaster car at an amusement park has a mass of \(1200 \mathrm{~kg}\) when fully loaded with passengers. As the car passes over the top of a circular hill of radius \(18 \mathrm{~m},\) assume that its speed is not changing. At the top of the hill, what are the (a) magnitude \(F_{N}\) and \((\mathrm{b})\) direction (up or down) of the normal force on the car from the track if the car's speed is \(v=11 \mathrm{~m} / \mathrm{s} ?\) What are \((\mathrm{c}) F_{N}\) and (d) the direction if \(v=14 \mathrm{~m} / \mathrm{s} ?\)

A person pushes horizontally with a force of \(220 \mathrm{~N}\) on a \(55 \mathrm{~kg}\) crate to move it across a level floor. The coefficient of kinetic friction between the crate and the floor is \(0.35 .\) What is the magnitude of (a) the frictional force and (b) the acceleration of the crate?

What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is \(29 \mathrm{~km} / \mathrm{h}\) and the \(\mu_{s}\) between tires and track is \(0.32 ?\)

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