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Chapter 3: Problem 29

A Ferris wheel with radius 14.0 \(\mathrm{m}\) is turning about a horizontal axis through its center (Fig. E3.29). The linear speed of a passenger on the rim is constant and equal to 7.00 \(\mathrm{m} / \mathrm{s} .\) What are the magnitude and direction of the passenger's acceleration as she passes through (a) the lowest point in her circular motion? (b) The highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution?

Short Answer

Expert verified
(a) Acceleration is 3.50 m/s虏 upwards. (b) Acceleration is 3.50 m/s虏 downwards. (c) Time for one revolution is 4蟺 seconds.

Step by step solution

01

Understand Circular Motion

Since the passenger is moving in circular motion, their acceleration will be centripetal, meaning it points towards the center of the Ferris wheel. The formula for centripetal acceleration \(a_c\) is \( a_c = \frac{v^2}{r} \), where \(v\) is the linear speed and \(r\) is the radius of the wheel.
02

Calculate Centripetal Acceleration

Given the linear speed \(v = 7.00 \ m/s\) and radius \(r = 14.0 \ m\), substitute these values into the centripetal acceleration formula: \[a_c = \frac{(7.00 \, m/s)^2}{14.0 \, m} = \frac{49 \, m^2/s^2}{14.0 \, m} = 3.50 \, m/s^2\]This acceleration is always directed towards the center of the Ferris wheel.
03

Determine Acceleration at the Lowest Point

At the lowest point of the circle, the direction of the centripetal acceleration is upwards, towards the center of the Ferris wheel.
04

Determine Acceleration at the Highest Point

At the highest point, the centripetal acceleration is directed downwards, again towards the center of the Ferris wheel.
05

Calculate Time for One Revolution

The time for one complete revolution \( T \) can be found using the circumference formula of the circle and the constant speed: \[T = \frac{2 \pi r}{v} = \frac{2 \pi (14.0 \, m)}{7.00 \, m/s} = \frac{28 \pi}{7} \, s = 4 \pi \, s \]

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is a key concept in understanding circular motion. This type of acceleration is always directed towards the center of the path on which an object is moving. This keeps the object in a circular trajectory. The formula for centripetal acceleration is given by: \[ a_c = \frac{v^2}{r} \]where:
  • \( a_c \) is the centripetal acceleration,
  • \( v \) is the linear speed, and
  • \( r \) is the radius of the circle.
In the context of a Ferris wheel, this acceleration is crucial as it determines how fast the wheel can spin without passengers being flung outwards. For example, if a passenger sits at the edge of a Ferris wheel turning with a linear speed of 7.00 m/s, and the radius is 14.0 m, we can calculate the centripetal acceleration using the above formula. By plugging the values in, we find that \\( a_c = 3.50 \, m/s^2 \), which means this is the acceleration needed to keep the passenger moving in a circle along the rim.
Ferris Wheel Physics
Ferris wheels are a fascinating example of circular motion in real life. As you ride on a Ferris wheel, you experience changes in speed and direction due to centripetal acceleration. Whenever you are at the lowest point of a Ferris wheel, your direction of acceleration is upwards, towards the center. Conversely, at the highest point of your trip, your acceleration is downwards, still pointing towards the center. This is because the force keeping you in circular motion is always directed towards the center of the circular path. Essentially, at both the top and bottom of the wheel, the centripetal acceleration remains the same in magnitude (3.50 m/s虏 in our example) but changes direction based on your location around the wheel. Understanding these dynamics helps engineers design Ferris wheels that are both safe and exhilarating.
Time for One Revolution
The time it takes for a Ferris wheel to complete one full spin is known as the time for one revolution. This is an important measure as it gives us insight into the motion's speed and rhythm. To determine this, we use the formula: \\[ T = \frac{2\pi r}{v} \] \where:
  • \( T \) is the time for one revolution,
  • \( r \) is the radius of the Ferris wheel, and
  • \( v \) is the linear speed.
For a Ferris wheel with a radius of 14.0 m and a constant speed of 7.00 m/s, we insert these values into the formula. This results in a time of \(4\pi \, s\) for one complete revolution. Converting \(4\pi\) into seconds gives approximately 12.57 seconds. This calculation confirms that each spin of the Ferris wheel takes about 12.57 seconds, allowing passengers to gauge how long the ride will last.

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Most popular questions from this chapter

A rocket designed to place small payloads into orbit is carried to an altitude of 12.0 \(\mathrm{km}\) above sea level by a converted airliner. When the airliner is flying in a straight line at a constant speed of 850 \(\mathrm{km} / \mathrm{h}\) , the rocket is dropped. After the drop, the airliner maintains the same altitude and speed and continues to fly in a straight line. The rocket falls for a brief time, after which its rocket motor turns on. Once its rocket motor is on, the combined effects of thrust and gravity give the rocket a constant acceleration of magnitude 3.00\(g\) directed at an angle of \(30.0^{\circ}\) above the horizontal. For reasons of safety, the rocket should be at least 1.00 \(\mathrm{km}\) in front of the airliner when it climbs the airliner's altitude. Your job is to determine the minimum time that the rocket must fall before its engine starts. You can ignore air resistance. Your answer should include (i) a diagram showing the flight paths of both the rocket and the airliner, labeled at several points with vectors for their velocities and accelerations; (ii) an \(x-t\) graph showing the motions of both the rocket and the airliner; and (iii) a \(y-t\) graph showing the motions of both the rocket and the airliner. In the diagram and the graphs, indicate when the rocket is dropped, when the rocket motor turns on, and when the rocket climbs through the altitude of the airliner.

The earth has a radius of 6380 \(\mathrm{km}\) and turns around once on its axis in 24 \(\mathrm{h}\) . (a) What is the radial acceleration of an object at the earth's equator? Give your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and as a fraction of \(g .\) (b) If \(a_{\text { rad }}\) at the equator is greater than \(g,\) objects will fly off the earth's surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

Martian Athletics. In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the greatest horizontal distance. Suppose that on earth she is in the air for time \(T\) , reaches a maximum height \(h,\) and achieves a horizontal distance \(D .\) If she jumped in exactly the same way during a competition on Mars, where \(g_{\text {Mars}}\) is 0.379 of its earth value, find her time in the air, maximum height, and horizontal distance. Express each of these three quantities in terms of its earth value. Air resistance can be neglected on both planets.

Our balance is maintained, at least in part, by the endolymph fluid in the inner ear. Spinning displaces this fluid, causing dizziness. Suppose a dancer (or skater) is spinning at a very fast 3.0 revolutions per second about a vertical axis through the center of his head. Although the distance varies from person to person, the inner ear is approximately 7.0 \(\mathrm{cm}\) from the axis of spin. What is the radial acceleration ( in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g^{\prime} \mathrm{s}\) ) of the endolymph fluid?

Two piers, \(A\) and \(B,\) are located on a river: \(B\) is 1500 \(\mathrm{m}\) downstream from \(A\) (Fig. E3.34). Two friends must make round trips from pier \(A\) to pier \(B\) and return. One rows a boat at a constant speed of 4.00 \(\mathrm{km} / \mathrm{h}\) relative to the water; the other walks on the shore at a constant speed of 4.00 \(\mathrm{km} / \mathrm{h}\) . The velocity of the river is 2.80 \(\mathrm{km} / \mathrm{h}\) in the direction from \(A\) to \(B .\) How much time does it take each person to make the round trip?
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