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A Ferris wheel with radius \(14.0 \mathrm{~m}\) is turning about a horizontal axis through its center (Fig. E3.31). The linear speed of a passenger on the rim is constant and equal to \(6.00 \mathrm{~m} / \mathrm{s}\) What are the magnitude and direction of the passenger's acceleration as she passes through (a) the lowest point in her circular motion and (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
The passenger's acceleration at the lowest and highest points is \(2.57 \, m/s^2\) directed towards the center of the Ferris wheel. It takes 14.7 seconds for the Ferris wheel to make one revolution.

Step by step solution

01

Determine the acceleration at the lowest and highest points

Use the formula \(a = \frac{v^2}{r}\) where \(a\) is the centripetal acceleration, \(v\) is the speed, and \(r\) is the radius. Given that \(v = 6.00 \, m/s\) and \(r = 14.0 \, m\), substituting these values will give \(a = \frac{(6.00 \, m/s)^2}{14.0 \, m} = 2.57 \, m/s^2\). The direction of acceleration at the highest point is downwards towards the center of the Ferris wheel. At the lowest point, the direction of acceleration is upwards towards the center of the Ferris wheel.
02

Determine the time to make one revolution

Use the formula \(T = \frac{2\pi r}{v}\) where \(T\) is the period, \(r\) is the radius, and \(v\) is the speed. Substituting the given values \(r = 14.0 \, m\) and \(v = 6.00 \, m/s\) will give \(T = \frac{2(3.14)(14.0 \, m)}{6.00 \, m/s} = 14.7 \, s\) . So, the time it takes for one full revolution of the Ferris wheel is 14.7 seconds.

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Key Concepts

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

Circular Motion
When an object moves in a circular path at a constant speed, it is undergoing what is known as circular motion. Despite the speed being constant, there is a change in the direction of motion, which means the object is accelerating. This acceleration is always directed towards the center of the circle and is aptly called centripetal acceleration. The presence of centripetal acceleration is critical, as it continuously changes the velocity's direction, keeping the object in circular movement.

Imagine swinging a ball tied to a string in a circular path; the string exerts an inward force on the ball, causing it to move in a circle. If the string breaks, the ball flies off tangentially, showing that without the force (and thus without acceleration), the object will not remain in circular motion.
Uniform Circular Motion
Uniform circular motion occurs when an object travels around a circle at a steady speed. Even though the speed is constant, because the direction of the object's velocity is continually changing, the object experiences a constant change in velocity. We describe this constant change by centripetal acceleration, which is calculated using the formula \( a = \frac{v^2}{r} \), where \( v \) is the linear speed and \( r \) is the radius of the circle. For a Ferris wheel with a radius of 14.0 meters and a passenger traveling at a linear speed of 6.00 meters per second, the centripetal acceleration is \( 2.57 \, m/s^2 \).

This acceleration is crucial in understanding why passengers feel pressed against the seat or pulled into the seat depending on their position on the Ferris wheel.
Ferris Wheel Physics
Let's bring our exploration of circular motion to a practical example — the Ferris wheel. In Ferris wheel physics, various principles come into play. Riders experience the sensation of forces acting on them throughout one revolution of the ride. At the highest point of the wheel, passengers tend to feel lighter, because the force of gravity and the centripetal force required to keep them in circular motion both act downwards.

Conversely, at the lowest point, passengers feel heavier, as the centripetal force required to change their direction of motion acts upwards, adding to the gravitational force. Despite these sensations, the actual acceleration towards the center of the wheel is constant, which is why the motion is uniform. It's the direction of this constant centripetal acceleration in relation to gravity that changes the riders' experiences.
Period of Revolution
The period of a revolution, commonly symbolized as \( T \), is the time it takes for an object to complete one full circle in its path. In uniform circular motion, you can find this period by dividing the circumference of the circle by the linear speed of the object using the formula \( T = \frac{2\pi r}{v} \). If we apply this to our Ferris wheel example with a radius of 14.0 meters and a linear speed of 6.00 meters per second, we can calculate that it takes approximately 14.7 seconds to complete one revolution.

This piece of information is not just a fun fact; it can be used to understand the wheel's performance, determine the ride's duration, and is a critical aspect of the design and operation of amusement rides.

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Most popular questions from this chapter

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of \(58.0^{\circ}\) above the horizontal, some of the tiny critters have reached a maximum height of \(58.7 \mathrm{~cm}\) above the level ground. (See Nature, Vol. 424, July \(31,2003,\) p. 509.) Neglect air resistance in answering the following. (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world- record leap?

An object is projected with initial speed \(v_{0}\) from the edge of the roof of a building that has height \(H .\) The initial velocity of the object makes an angle \(\alpha_{0}\) with the horizontal. Neglect air resistance. (a) If \(\alpha_{0}\) is \(90^{\circ}\), so that the object is thrown straight up (but misses the roof on the way down), what is the speed \(v\) of the object just before it strikes the ground? (b) If \(\alpha_{0}=-90^{\circ}\), so that the object is thrown straight down, what is its speed just before it strikes the ground? (c) Derive an expression for the speed \(v\) of the object just before it strikes the ground for general \(\alpha_{0}\). (d) The final speed \(v\) equals \(v_{1}\) when \(\alpha_{0}\) equals \(\alpha_{1}\). If \(\alpha_{0}\) is increased, does \(v\) increase, decrease, or stay the same?

A "moving sidewalk" in an airport terminal moves at \(1.0 \mathrm{~m} / \mathrm{s}\) and is \(35.0 \mathrm{~m}\) long. If a woman steps on at one end and walks at \(1.5 \mathrm{~m} / \mathrm{s}\) relative to the moving sidewalk, how much time does it take her to reach the opposite end if she walks (a) in the same direction the sidewalk is moving? (b) In the opposite direction?

The radius of the earth's orbit around the sun (assumed to be circular) is \(1.50 \times 10^{8} \mathrm{~km},\) and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in \(\mathrm{m} / \mathrm{s} ?(\mathrm{~b})\) What is the magnitude of the radial acceleration of the earth toward the sun, in \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius \(=5.79 \times 10^{7} \mathrm{~km},\) orbital period \(=88.0\) days).

A baseball thrown at an angle of \(60.0^{\circ}\) above the horizontal strikes a building \(18.0 \mathrm{~m}\) away at a point \(8.00 \mathrm{~m}\) above the point from which it is thrown. Ignore air resistance. (a) Find the magnitude of the ball's initial velocity (the velocity with which the ball is thrown). (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building.

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