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Artificial gravity in space stations. One problem for humans living in outer space is that they are apparently weight- less. One way around this problem is to design a cylindrical space station that spins about an axis through its center at a constant rate. (See Figure $6.32 . )$ This spin creates "artificial gravity" at the outside rim of the station. (a) If the diameter of the space station is $800.0 \mathrm{m},$ how fast must the rim be moving in order for the "artificial gravity" acceleration to be $g$ at the outer rim? (b) If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity on the Martian surface. How fast must the rim move in this case? (c) Make a free-body diagram of an astronaut at the outer rim.

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Physics 101 Mechanics

Chapter 6

Circular Motion and Gravitatio

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

McMaster University

Lectures

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In physics, kinematics is the description of the motion of objects and systems in the frame of reference defined by the observer. An observer has to be specified, otherwise the term is meaningless.

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In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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Okay, So in the space station, if we want to tell Billy the artificial gravity acceleration, we should be using this a question. Where is the real explanation? Because there's a cylinder that's moving on itself because the center is along the center of the cylinder. So a or the acceleration off the radial acceleration is V squared over R, where are is given its radius, which is 400 meters now. Before solving for the V or the velocity, we should know a few things. First off, all we know that the gravity on earth is 9.8 meters per second squared. And if you wanna figure out the gravity on Mars, we can use the following question where gravity in Mars is equal to the gravitational, constant times the mass ofthe the planet Mars and, ah, that we divide that by the radius off mar squared. So if we put all the numbers, we see that the acceleration is three point 71 meters per second squared. Now, if we think if you want to figure out the velocity as we mentioned that we'll be using this equation so we can solve velocity in terms of the other two terms. So there's gonna be V, which is equal to a relax elation times are. Then there'll be a square root because we have a square square there. So that's what we're using here for E. If we substitute that by the earth's gravity, we see that backs Alicia our the velocity 62 0.6 meters second. Similarly, if we I wanna make this space station is a za waiting area for Mars, we see that the only thing changing from in this equation is thie acceleration. Instead, off the arts exploration with using the marshes, gravitation, acceleration. And if you plug in the numbers, we see that the velocities 38.5 meters for a second. If we can draw the everybody diagram for R C three also before drawing, let's consider that the dream is on the right. So let's say that dream off the station is around here. Then there will be one force that that will be towards the center, and that's the way it does in normal force. So we have a normal force which is acting on the left, notice that we don't have any force on the wind direction The reason for that is we don't have gravity. That's or we don't have thie Earth's gravity that acting downward anymore because it's in space and in space. There's no gravity outside. So that's why the only force that acting here is the one that's towards the center. And because of that normal force, we have a radial acceleration. Thank you.

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