/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 A space station, in the form of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A space station, in the form of a wheel 120 m in diameter, rotates to provide an "artificial gravity" of \(3.00 \mathrm{m} / \mathrm{s}^{2}\) for persons who walk around on the inner wall of the outer rim. Find the rate of rotation of the wheel (in revolutions per minute) that will produce this effect.

Short Answer

Expert verified
To solve for the rate of rotation that produces the 'artificial gravity', calculate the radius of the wheel first. Then, use the given 'artificial gravity' and the calculated radius to find the velocity. Finally, use the calculated velocity and radius to find the rate of rotation in revolutions per minute.

Step by step solution

01

Calculating the radius of the wheel

The radius of the wheel is half the diameter. Given that the diameter of the wheel is 120 m, the radius \( r \) would be \(\frac{120m}{2} = 60 m\).
02

Calculating the velocity

The formula for centripetal acceleration is \(a = \frac{{v^2}}{{r}}\), you can rearrange for velocity \( v = \sqrt{{a*r}}\). Substituting the given acceleration 3.00 \(m/s^2\) and the calculated radius 60 m, we find \(v = \sqrt{{3.00 m/s^2 * 60 m}}\). Calculate this to find the velocity.
03

Calculating the rate of rotation

The formula for the rate of rotation is \(v = 2*\pi*r*N\), where \(N\) is the revolution per minute. Rearranging for \(N\) yields \(N = \frac{v}{{2*\pi*r}}\). Substitute the calculated velocity and radius into the formula to obtain the rate of rotation in revolutions per minute.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is crucial for understanding why artificial gravity can exist in a rotating space station. It is the acceleration required to keep an object moving in a circular path and is directed toward the center of the circle. This occurs because when an object travels around a circle, it constantly changes direction, and acceleration is required for this change.
The formula for centripetal acceleration is \[ a = \frac{v^2}{r} \]where \( a \) is the centripetal acceleration, \( v \) is the velocity of the object, and \( r \) is the radius of the circle.
In the context of the space station, centripetal acceleration simulates gravity by creating a force toward the center of the station's rotation. This allows astronauts to feel as though they are being "pulled" down onto the floor of the station's outer rim, mimicking Earth’s gravity to some degree. This concept helps astronauts maintain their physiological functions over long missions.
Rate of Rotation
The rate of rotation refers to how quickly the space station completes a circular path, and it is usually measured in revolutions per minute (RPM). Knowing the rate of rotation is vital in calculating how much artificial gravity you can generate.
To find this, we rearrange the velocity formula to solve for the rate of rotation. Essentially, as the station rotates, people experience the centripetal force as gravity, allowing them to stand and walk as they would on Earth. For this space station:
  • The diameter is 120 m, so the radius \( r \) is 60 m.
  • Given centripetal acceleration \( a = 3.00 \mathrm{m/s}^2 \).
Calculate the velocity \( v \) using:\[ v = \sqrt{a \times r} \]After obtaining the velocity, use the formula:\[ N = \frac{v}{2 \pi r} \]This calculation lets us find \( N \), or the number of revolutions per minute needed for the desired artificial gravity. Understanding this helps in designing rotations that safely replicate gravity for long-term space dwellers.
Space Station Dynamics
Space station dynamics involve understanding the physics that govern the movement and stabilization of these immense structures in outer space. Rotational motion is key here.
For a wheel-shaped space station, achieving artificial gravity involves careful balancing of forces to maintain a stable rotation. The structure must spin at a rate that ensures crew comfort and safety, simulating conditions on Earth as closely as possible.
  • Stable rotation keeps space debris and various unnecessary vibrations from destabilizing the station.
  • Systems need to manage any energy losses due to friction or external forces acting on the rotating system.
In this dynamic environment, engineers must also consider aspects like energy sources, materials used to construct the station, and the effects of radiation. By understanding these dynamics, astronauts can live and work in space with minimized risks, making space exploration more accessible and sustainable. Such designs not only promote crew comfort but might one day facilitate longer missions and even future space tourism.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a cyclotron (one type of particle accelerator), a deuteron (of atomic mass \(2.00 \mathrm{u}\) ) reaches a final speed of \(10.0 \%\) of the speed of light while moving in a circular path of radius \(0.480 \mathrm{m}\). The deuteron is maintained in the circular path by a magnetic force. What magnitude of force is required?\(214.897 N\)

The pilot of an airplane executes a constant-speed loop-theloop maneuver in a vertical circle. The speed of the airplane is \(300 \mathrm{mi} / \mathrm{h},\) and the radius of the circle is \(1200 \mathrm{ft}\). (a) What is the pilot's apparent weight at the lowest point if his true weight is 160 lb? (b) What is his apparent weight at the highest point? (c) What If? Describe how the pilot could experience weightlessness if both the radius and the speed can be varied. (Note: His apparent weight is equal to the magnitude of the force exerted by the seat on his body.)

A \(3.00-\mathrm{g}\) leaf is dropped from a height of \(2.00 \mathrm{m}\) above the ground. Assume the net downward force exerted on the leaf is \(F=m g-b v,\) where the drag factor is \(b=0.0300 \mathrm{kg} / \mathrm{s} .\) (a) Calculate the terminal speed of the leaf. (b) Use Euler's method of numerical analysis to find the speed and position of the leaf, as functions of time, from the instant it is released until \(99 \%\) of terminal speed is reached. (Suggestion: Try \(\Delta t=0.005 \mathrm{s} .\) )

An amusement park ride consists of a rotating circular platform \(8.00 \mathrm{m}\) in diameter from which \(10.0-\mathrm{kg}\) seats are suspended at the end of \(2.50-\mathrm{m}\) massless chains (Fig. \(P(6.63) .\) When the system rotates, the chains make an angle \(\theta=28.0^{\circ}\) with the vertical. (a) What is the speed of each seat? (b) Draw a free-body diagram of a 40.0 -kg child riding in a seat and find the tension in the chain.

A light string can support a stationary hanging load of \(25.0 \mathrm{kg}\) before breaking. A \(3.00-\mathrm{kg}\) object attached to the string rotates on a horizontal, frictionless table in a circle of radius \(0.800 \mathrm{m},\) while the other end of the string is held fixed. What range of speeds can the object have before the string breaks?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.