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Chapter 6: Problem 32

(II) A Ferris wheel \(22.0 \mathrm{~m}\) in diameter rotates once every \(12.5 \mathrm{~s}\) (see Fig. \(5-19)\). What is the ratio of a person's apparent weight to her real weight \((a)\) at the top, and \((b)\) at the bottom?

Short Answer

Expert verified
At the top, the ratio is less than 1; at the bottom, the ratio is greater than 1.

Step by step solution

01

Identifying problem parameters

The diameter of the Ferris wheel is given as \(22.0 \, \text{m}\). Hence, the radius \(r\) is half of that, \(r = 11.0 \, \text{m}\). The Ferris wheel completes one full rotation every \(12.5 \, \text{s}\).
02

Calculating angular velocity

Angular velocity \(\omega\) is calculated using the formula \(\omega = \frac{2\pi}{T}\), where \(T = 12.5 \, \text{s}\). Thus, \(\omega = \frac{2\pi}{12.5} \, \text{rad/s}\).
03

Finding centripetal acceleration

Centripetal acceleration \(a_c\) is calculated using \(a_c = \omega^2 r\). Substitute the values of \(\omega\) and \(r\) to find \(a_c\).
04

Calculating apparent weight at the top

At the top of the Ferris wheel, the apparent weight \(W_t\) is reduced by the centripetal force: \(W_t = m(g - a_c)\). The ratio of apparent to real weight \(\frac{W_t}{mg}\) is \(1 - \frac{a_c}{g}\).
05

Calculating apparent weight at the bottom

At the bottom of the Ferris wheel, the apparent weight \(W_b\) is increased by the centripetal force: \(W_b = m(g + a_c)\). The ratio of apparent to real weight \(\frac{W_b}{mg}\) is \(1 + \frac{a_c}{g}\).
06

Substituting and simplifying ratios

Use the values calculated and the formulae derived for \(W_t\) and \(W_b\) to determine \(\frac{W_t}{mg}\) and \(\frac{W_b}{mg}\), the ratios at the top and bottom.

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration plays a crucial role in understanding circular motion. When an object moves in a circle, it experiences an inward force that keeps it on that path. This force is the centripetal force, and the resulting acceleration due to this force is called centripetal acceleration. The formula to calculate centripetal acceleration, \(a_c\), is given by \(a_c = \omega^2 \cdot r\), where \(\omega\) is the angular velocity and \(r\) is the radius of the circle.
  • In our example, the Ferris wheel has a radius of 11 meters.
  • Once we determine the angular velocity with \(\omega = \frac{2\pi}{T}\), where \(T\) is the time for one complete rotation, we can find \(a_c\).
  • Centripetal acceleration helps us understand how the apparent weight of a person changes while riding a Ferris wheel.
Knowing the centripetal acceleration is essential for determining how much lighter or heavier a person feels at the top and bottom of the Ferris wheel's rotation.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves around a circle. It is often represented by the symbol \(\omega\) and is calculated as the change in angular displacement over time.To find angular velocity, you can use the formula: \(\omega = \frac{2\pi}{T}\), where \(T\) is the period of one cycle, expressed in seconds.
  • In the context of our Ferris wheel problem, the wheel takes 12.5 seconds to make one complete rotation, leading to \(\omega = \frac{2\pi}{12.5} \, \text{rad/s}\).
  • This value helps us figure out how rapidly the wheel is spinning, which is crucial for calculating centripetal acceleration.
  • Understanding angular velocity gives us insight into the dynamics of rotating systems such as Ferris wheels.
Angular velocity connects to other concepts of motion, making it a foundational element in analyzing circular motions.
Apparent Weight
Apparent weight is an interesting concept that describes how heavy or light a person feels while in motion, compared to their true weight. This is not their actual weight, but the sensation of weight experienced due to acceleration.In the case of the Ferris wheel:
  • At the top of the Ferris wheel, the apparent weight is reduced because the centripetal force required is subtracted from the gravitational force: \(W_t = m(g - a_c)\).
  • At the bottom, the apparent weight is increased because the centripetal force adds to the gravitational force: \(W_b = m(g + a_c)\).
  • Hence, the ratios \(\frac{W_t}{mg}\) and \(\frac{W_b}{mg}\) give us the sense of weight compared to actual weight depending on the wheel's position.
By understanding apparent weight, you gain insight into why you feel heavier or lighter at different points on rides like Ferris wheels. It's a direct result of how the centripetal and gravitational forces interact as they work on your body.

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Most popular questions from this chapter

(II) Suppose the mass of the Earth were doubled, but it kept the same density and spherical shape. How would the weight of objects at the Earth's surface change?

(II) Determine the time it takes for a satellite to orbit the Earth in a circular "near-Earth" orbit. A "near-Earth" orbit is at a height above the surface of the Earth that is very small compared to the radius of the Earth. [Hint: You may take the acceleration due to gravity as essentially the same as that on the surface.] Does your result depend on the mass of the satellite?

(II) A mass \(M\) is ring shaped with radius \(r .\) A small mass \(m\) is placed at a distance \(x\) along the ring's axis as shown in Fig. \(27 .\) Show that the gravitational force on the mass \(m\) due to the ring is directed inward along the axis and has magnitude $$F=\frac{G M m x}{\left(x^{2}+r^{2}\right)^{\frac{3}{2}}}$$ [Hint: Think of the ring as made up of many small point masses \(d M ;\) sum over the forces due to each \(d M,\) and use symmetry.]

Between the orbits of Mars and Jupiter, several thousand small objects called asteroids move in nearly circular orbits around the Sun. Consider an asteroid that is spherically shaped with radius \(r\) and density \(2700 \mathrm{~kg} / \mathrm{m}^{3} .\) ( \(a\) ) You find yourself on the surface of this asteroid and throw a baseball at a speed of \(22 \mathrm{~m} / \mathrm{s}\) (about \(50 \mathrm{mi} / \mathrm{h}\) ). If the baseball is to travel around the asteroid in a circular orbit, what is the largest radius asteroid on which you are capable of accomplishing this feat? (b) After you throw the baseball, you turn around and face the opposite direction and catch the baseball. How much time \(T\) elapses between your throw and your catch?

(I) If you doubled the mass and tripled the radius of a planet, by what factor would \(g\) at its surface change?
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