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Circular Motion, artificial gravity

  1. Aug 21, 2009 #1
    My brain hangs when i try to draw a free body diagram of a person stand on Earth and then a person standing inside a "space-colony"(a large wheel that spins,at the correct angular velocity and radius, and so the person standing inside feels the same amount of force pulling him "downwards")

    Free body diagram on Earth: Upward --- Normal reaction force
    Downward --- Weight = mg
    Therefore acceleration = 0
    Free body diagram in the wheel: Upward --- Normal reaction force
    Therefore "upward" acceleration = mg

    Ahh, someone help! I need to ctrl-alt-del already.
    My phy teacher explained to me about the concept of Pseudo forces to help me better understand but says that we can't use that for A-level exams, so I hope someone can explain in great details and in different approaches to enhance the visualisation. Thanks! =)
  2. jcsd
  3. Aug 21, 2009 #2
    The person in the space ship is always accelerating at g if v^2/r =~10m/s^2 simply because he is always changing direction (I am also assuming the speed is uniform). There has to be something on the spaceship that holds the person in that circular path, some force. We call that centripetal force and it is supplied by the "walls" of the spaceship. Or a seatbelt or something if he is on an "inside" wall.

    On the surface of the earth you are not accelerating with respect to the surface of the earth... (your mg=normal force), but you actually are accelerating with respect to the axis of the earth the largest acceleration being at the equator. I think the equatorial speed is about 465 m/s, but the mean radius of the earth is 6.36? X 10^6 m so v^2/r or centripetal acceleration is a miniscule 0.033 m/s^2 which points basically in the same direction as the acceleration due to gravity at the equator.

    I hope this helps...
    And I hope I have not made too many errors.

    The next question would be would one weigh more or less if the earth stopped spinning on its axis and one was on the equator. We could use a long night in Texas as it is hot as hell this summer.

    The psuedo force he is probably referring to is probably inertia. on the spaceship your mass "wants" to keep moving in a straight path but the spaceship walls make you accelerate towards the center, they keep you from "flying off" along a straight path. I am certain someone can draw a picture for you if needed. im not into graphics this late...night. This can be a bit strange however if we pick a different frame of reference which I dont think your teacher was referring to... at least I hope not.
    Last edited: Aug 21, 2009
  4. Aug 21, 2009 #3
    Now think about the bolded again. There really is only one force and its always pointed towards the center of rotation of the spaceship/wheel. Up and down are really not good terms for this.
  5. Aug 22, 2009 #4
    Thanks for the help but i think you've missed the point.

    My confusion/question is that i find that the 2 free body diagrams "looks very different", why is it that the person feels the same way---a seeming gravitational strength of 9.81N/kg.

    Lets say the person is on standing on the North Pole of the Earth.
    And there's a reason why i put quotation marks on the "upward" coz im aware its not entirely meaningful, but that's not the point of my question either. No offence! =)
  6. Aug 22, 2009 #5


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    The normal force is not the reaction force from newton's 3rd law of motion.

    What a person would feel in a rotating wheel is a "centrifugal force" because the person is in a rotating (and noninertial) frame of reference. In actuality, the person just has a tendency to travel in a straight line (inertia). What the person feels as the "centrifugal force" is the wheel exerting a centripetal (center pointing) force that pushes the person towards the center (and away from the straight line the person wants to follow).
  7. Aug 22, 2009 #6

    Doc Al

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    Staff: Mentor

    What makes you "feel" that you have weight is the presence of the normal force (and the stresses it produces in your body), which appears in both diagrams. You don't feel gravity directly. If you step off of a roof into the air, gravity still pulls you down but there's no longer a support force--thus you feel "weightless".
  8. Aug 22, 2009 #7
    As stated above on the wheel/spaceship the wall of the spaceship pushes on the person and the person pushes back with an equal force (third law). But the net force on the person on the spaceship is in towards the center of rotation and is supplied by the spaceship.

    When you draw a free body diagram you have to choose which body you are going to assess.

    1. So lets say you are looking at all the forces on a person on the North pole. There is a force straight towards the center of the earth acting on the person, mg. There is a force acting opposite of the force due to gravity. This force is supplied by the surface of the earth and is called the normal force. So we have two forces acting on the person. The are equal in magnitude, but opposite in direction and give a net force of zero. The person is not accelerating, no net force.

    2. The spaceship... There is one force acting on the person. It is centripetal force mv^2/r. It is supplied by something on the spaceship like a wall. So there is a net force on this person. There must be because the person is constantly changing direction (accelerating) unlike the person on the earth who has two forces acting on him that "cancel" each other out. From the third law point of view however, the spaceship is pushing on the person and the person pushes back on the spaceship. The person "pushing back" on the spaceship is NOT a part of the free body diagram. It is not acting on the person.

    In both cases the person is going to feel like gravity is present. If he were to drop a ball in each case it would appear to accelerate and strike the "ground"... Except in the case on the spaceship, the ball is really not accelerating, its moving in a straight path with respect to the center of rotation. On the earth, the ball is being accelerated towards the center of the earth (by gravity) and will continue to accelerate until it hits the ground and the Normal force supplied by the ground is equal to mg for the ball.

    And I did completely go off on a tangent, sorry. And maybe I did again.
    And I did not notice your quotes, sorry about that also.
    It was late I guess. Maybe this helps in understanding the Free body diagrams, if not, I apologize. So, Many apologies.
    Last edited: Aug 22, 2009
  9. Aug 22, 2009 #8
    Does it help if i were to take note that the force mg and the normal force are 2 different forces, gravitational and electromagnetic respectively?
    And that electromagnetic force does not simultaneously accelerate every single atom in the person's body, thus having a bit of lag (which is essentially inertia), which creates the effect of "pulling" everything in the person's body down.
    Whereas if the man were to be in mid-air on Earth, every part of his body is accelerating at the same rate (well as long as he isnt 999999km tall to make a head-toe difference) due to gravity, thus not feeling anything.
    (I just realise my question could be clearer if the comparison is between the man in free-fall and man in the wheel) So refering to this concept and answering my new question, it can be explain why the man under the only force, gravity, accelerating at 9.81, feels weightless. Whereas the man in the wheel, under the only force, normal force from the wheel, accelerating at 9.81 feels "gravity".

    And then the fact that one, the net acceleration is 0 and the other is 9.81 would then explain my original question right?

    Thanks zcd, i visualise the scenario the same way you do, just less vivid as yours haha.
    And pgardn for writing so much XD (you always only elaborate what i wrote) but nah, no need for apologies.
  10. Aug 23, 2009 #9

    Doc Al

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    Staff: Mentor

    Sounds good to me.
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