Understanding centripetal force

  1. 1. The problem statement, all variables and given/known data
    a person is standing stationairy on earth (whilst it is rotating) there is a net force towards the centre of the earth.

    What would the free body diagram look like.


    Please could someone explain why

    thanks

    2. Relevant equations


    I know the formula for acceleration (v^2/r) and Force = m(v^2/r).
    3. The attempt at a solution
    I would suggest that the free body digram shows:
    R + mv^2/r = mg
    so mv^2/r = mg-R
    But I get really confused by R and mg as they should cancel out.

    Please could someone explain it
     
  2. jcsd
  3. PeterO

    PeterO 2,319
    Homework Helper

    First let's ignore the centripetal force.

    If you stand in a stationary lift, the reaction force from the floor will be equal in magnitude and opposite in direction to mg.
    If the lift was accelerating up, the Reaction force from the floor will be greater than mg [you feel heavier when the lift starts going up], Net force is up - you accelerate up.
    If the lift was accelerating down, the Reaction force from the floor will be less than mg [you feel lighter when the lift starts going down] Net force is down - you accelerate down.

    Now lets get centripetal Force involved.
    When you are on the spinning Earth you have a centripetal acceleration down [like in the lift accelerating down] but the size of the acceleration will be MUCH smaller than what you achieve in the average lift.
    As such the reaction force will be slightly smaller than mg. The difference possibly only shows up in the decimal places of the Force magnitude.

    So R and mg only cancel out if the acceleration is zero.
    In a lift, the acceleration can be significant. The centripetal acceleration is very small.
     
  4. so we have a free body diagram looking like:

    Up: reaction force
    down: mg, mv^2/r

    so there is no reaction force to mv^2/r?

    therefore
    mv^2/r = mg - R

    what confuses now is what is the Newton pair of the centripetal force?
     
  5. also, if we have a body orbiting a planet what is the reaction force? or does it not exist?
     
  6. Finally,
    Q:If the earth rotated faster on its axis would would happen to your weight (as measured by a scale)
    A: decrease

    Q:As the rotational speed of a space habitat increases what happens to the apparent weight of the people inside
    A: increases

    could someone please explain the answers (i have tried 2 as below but can't do 1)


    Example: Q2:

    The force, mg, of the space station acts towards the earth

    F = ma
    F = mv2/r
    BUT v = omega (w) * r
    SO

    F = m*w2*r

    F = (GMm)/r2

    (GMm)/r2 = m*w2*r

    If w (omega) increases then Force increases
    this seems wrong because GMm/r2 should always be constant and therefore not affected by speed :S
     
  7. Doc Al

    Staff: Mentor

    Never put 'centripetal force' on a free body diagram. There are only 2 forces acting on the body: The weight, mg, and the force of the earth pushing up, the 'reaction force'.

    Don't think of mv^2/r as a separate force; think of it as applying Newton's 2nd law in the case of centripetal acceleration.

    What we call 'centripetal force' is just the net force that produces the centripetal acceleration.

    OK. (For the simple case where the mass is on the equator.) But think of it as:
    ΣF = ma
    mg - R = m(v^2/r)

    Again, the term 'centripetal force' is just another name for the net force. The only forces involved in this are mg and R, and they both have Newton's 3rd law pairs. (What are they?)
     
  8. I would presume there has to be something wrong with my working because it is not possible for it to be correct but it follows the advice i've be given :S
     
  9. Doc Al

    Staff: Mentor

    What working are you referring to? Which post?
     
  10. post 6
     
  11. Doc Al

    Staff: Mentor

    A scale measures 'apparent weight', which is what you called R (the 'reaction' force) above.

    You'll have to redo this one. They are talking about a space station 'habitat' that creates its own artificial 'gravity' by rotation. No real gravity involved. (Hopefully they have some sort of diagram of this thing in action.)
     
  12. Q:If the earth rotated faster on its axis would would happen to your weight (as measured by a scale)


    Answer
    Here, the centripetal force increases due to an increase in speed.
    As we said before, mv2/r = mg - R

    Is this correct


    For the next question I will break it into two questions


    Q1:As the rotational speed of a space habitat increases what happens to the apparent weight of the people inside

    I don't understant how this would be done as I have never come across the idea of simulated gravity

    Q2:As the rotational speed of a satellite increases what happens to the apparent weight of the people inside
    for this question, which is what the working was meant to be for (post 6), I presume the above working is still flawed
     
  13. Last edited: Sep 15, 2011
  14. post #12

    I just to check also, is your weight given by the reaction force or mg???
     
  15. Doc Al

    Staff: Mentor

    But what about your 'weight'?
     
  16. Doc Al

    Staff: Mentor

    The scale measures the force pressing on it, which is the reaction force.
     
  17. Q:If the earth rotated faster on its axis would would happen to your weight (as measured by a scale)


    Answer
    Here, the centripetal force increases due to an increase in speed.
    As we said before, mv2/r = mg - R

    My weight = the reaction force so R = mv^2/r + mg therefore weight increases
    I have said weight = reaction force as you said this in the last post
     
  18. Doc Al

    Staff: Mentor

    This is correct.

    This is not correct.

    Take the correct equation and solve for R.
     
  19. sorry
    R = mg - mv^2/r
     
  20. Doc Al

    Staff: Mentor

    Good. Now how would you answer the question?
     
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