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(II) How many revolutions per minute would a \(22-\mathrm{m}-\) diameter Ferris wheel need to make for the passengers to feel "weightless" at the topmost point?

Short Answer

Expert verified
The Ferris wheel must rotate at approximately 2.83 revolutions per minute.

Step by step solution

01

Understanding the Problem

At the topmost point of the Ferris wheel, passengers feel 'weightless' when the only force acting on them is the centripetal force required to keep them moving in a circle. This occurs when the gravitational force equals the centripetal force.
02

Expressing Forces Mathematically

The force due to gravity is given by \( F_g = m \cdot g \), where \( m \) is the mass of the passenger and \( g \) is the acceleration due to gravity. The centripetal force required is \( F_c = m \cdot a_c \), with \( a_c \) being the centripetal acceleration, \( a_c = \frac{v^2}{r} \).
03

Establishing the Weightless Condition

For weightlessness, gravitational force equals centripetal force, thus: \( m \cdot g = m \cdot \frac{v^2}{r} \). Simplifying gives \( g = \frac{v^2}{r} \).
04

Solving for Velocity

To find the velocity \( v \), rearrange \( g = \frac{v^2}{r} \) to get \( v = \sqrt{g \cdot r} \). Here, \( r \) is the radius of the Ferris wheel. Since the diameter is 22 m, the radius \( r = \frac{22}{2} = 11 \) m.
05

Substitute Known Values

Substituting \( g = 9.8 \ \text{m/s}^2 \) and \( r = 11 \ \text{m} \) into \( v = \sqrt{g \cdot r} \) gives: \( v = \sqrt{9.8 \cdot 11} = \sqrt{107.8} \approx 10.38 \ \text{m/s} \).
06

Convert Velocity to Revolutions Per Minute

The circumference of the Ferris wheel is \( C = 2 \pi r = 2 \pi \cdot 11 = 22\pi \) meters. The time for one revolution is \( \frac{C}{v} = \frac{22\pi}{10.38} \) seconds. To find revolutions per minute, calculate \( \frac{60}{\frac{22\pi}{10.38}} = \frac{60 \cdot 10.38}{22\pi} \approx 2.83 \).
07

Conclusion

Thus, the Ferris wheel must rotate at approximately 2.83 revolutions per minute for passengers to feel 'weightless' at the top.

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Key Concepts

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

Weightlessness
When we talk about feeling "weightless," it doesn't mean there is no gravity acting on you. Instead, it refers to a sensation experienced when the only force acting on you is the one keeping you in circular motion, which is the centripetal force.

In this scenario, as you reach the top of the Ferris wheel, your body isn't pressing against the seat like it does at the bottom. This gives you the feeling of floating or being "weightless."

The feeling is caused because, at this point, the gravitational force which pulls you downward matches exactly with the centripetal force needed to keep you moving along the wheel. The result is that there seems to be no force exerted by the seat on you, leading to this sensation.

This also happens in other situations in physics, like astronauts orbiting Earth. Even though gravity is present, they feel "weightless" because both they and the spaceship are in free fall, always missing Earth."

  • Feeling of floating or "weightlessness" occurs at the topmost point.
  • Gravitational and centripetal forces are equal.
  • No apparent force exerted by the seat, causing the sensation of weightlessness.
Gravitational Force
Gravitational force is what we commonly refer to as "weight." It's the force exerted by the Earth that pulls objects downward. This force depends on the mass of the object and the gravitational acceleration (\( g \)), which on Earth is approximately 9.8 m/s².

Mathematically, it is expressed as:
\[ F_g = m \cdot g \]
where \( F_g \) is the gravitational force, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity. At the top of the Ferris wheel, the passengers experience gravitational force pulling them downward.

Understanding gravitational force is key in knowing why objects fall toward the Earth and why we feel our "weight." It's a crucial part of achieving the balance needed to create the sensation of weightlessness as the centripetal force must be equal to this gravitational pull.
  • Gravitational force is Earth pulling objects down.
  • Calculated with \( F_g = m \cdot g \) formula.
  • Equal to centripetal force when "weightless" at the top of the Ferris wheel.
Centripetal Acceleration
Centripetal acceleration is essential for any object moving in a circular path. It's the acceleration required to change the direction of an object so that it keeps circulating along the path.

This acceleration points toward the center of the circular path and is responsible for making sure the object stays on its curved trajectory.

Mathematically, centripetal acceleration \( a_c \) is expressed as:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the velocity of the object and \( r \) is the radius of the circle.

In the case of the Ferris wheel, centripetal acceleration needs to match the gravitational force for the passengers to experience weightlessness. At this moment, both forces are perfectly balanced, ensuring the sensation of floating at the topmost position.

  • Centripetal acceleration keeps objects in circular motion.
  • Directed toward the path's center, vital for maintaining circular travel.
  • Must balance with gravitational force for weightlessness.

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Most popular questions from this chapter

(II) Two blocks, with masses \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}},\) are connected to each other and to a central post by cords as shown in Fig. \(5-46 .\) They rotate about the post at frequency \(f\) (revolutions per second) on a frictionless horizontal surface at distances \(r_{\mathrm{A}}\) and \(r_{\mathrm{B}}\) from the post. Derive an algebraic expression for the tension in each segment of the cord (assumed massless).

(II) The terminal velocity of a \(3 \times 10^{-5} \mathrm{~kg}\) raindrop is about \(9 \mathrm{~m} / \mathrm{s}\). Assuming a drag force \(F_{\mathrm{D}}=-b v\), determine \((a)\) the value of the constant \(b\) and \((b)\) the time required for such a drop, starting from rest, to reach \(63 \%\) of terminal velocity.

(II) A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at \(28^{\circ}\) above the horizontal. The coefficients of static and kinetic friction between the box and wall are 0.40 and 0.30 , respectively. The box slides down unless the applied force has magnitude \(23 \mathrm{~N}\). What is the mass of the box?

A coffee cup on the horizontal dashboard of a car slides forward when the driver decelerates from \(45 \mathrm{~km} / \mathrm{h}\) to rest in \(3.5 \mathrm{~s}\) or less, but not if she decelerates in a longer time. What is the coefficient of static friction between the cup and the dash? Assume the road and the dashboard are level (horizontal).

A motorcyclist is coasting with the engine off at a steady speed of \(20.0 \mathrm{~m} / \mathrm{s}\) but enters a sandy stretch where the coefficient of kinetic friction is \(0.70 .\) Will the cyclist emerge from the sandy stretch without having to start the engine if the sand lasts for \(15 \mathrm{~m} ?\) If so, what will be the speed upon emerging?

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