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A person walks on a rotating platform at constant speed

  1. Dec 11, 2016 #1
    1. The problem statement, all variables and given/known data
    A large flat horizontal platform rotates at a constant angular speed ω. A person on the platform walks in a circular path of radius R0 centered on the axis of the platform with a constant linear speed v relative to the platform’s surface. The coefficient of friction between the person and the platform’s surface is µ and the mass of the person is m. How fast can the person walk if: (a) they move in the direction of rotation? (b) if they move opposite the direction of rotation?

    2. Relevant equations
    Equation of motion for rotating body. ma'=Fext+Fcor+Fcentif
    Fcor = -2mωXv'
    Fcentrif = -m(ωX(ωXr'))
    Fext= µmg
    ?a' = v2/r

    ω = ω(khat) ( I chose counter clockwise rotation)
    v = v(jhat)
    r = R0(ihat)
    3. The attempt at a solution
    Part a) I think my problem is with the a', I'm not sure if that is a valid substitution. Honestly pretty confused about how to set up the equation.

    Subbing in the values, all of the m's cancel out and the unit vectors are all (ihat). I'm not entirely sure if the sign on my friction o a' is correct either:

    a' = µg + 2ωv+ω2R0
    a' =v2/r
    Subbing in the a, I get a quadratic for v, which factors to:

    R0µg=(v+ωR0)2, and that lets me solve for v.


    But is it right? This feels very shakey to me, the a' sub, the signs of the forces. I keep trying to wrap my head around it but it doesn't make proper sense.

    I also treating this as a one dimensional problem, and get something entirely different. What is the best way to approach this problem?

    Thanks in advance
     
  2. jcsd
  3. Dec 11, 2016 #2

    TSny

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    Are you required to solve this in the rotating frame? If not, you might try analyzing it in the lab frame.

    For your rotating frame analysis, I don't think you handled the signs properly. Suppose you take radially outward as the positive direction, so radially inward is negative.
    What is the direction of the acceleration a'? Therefore, what is its sign?
    What is the direction of the friction force? What is its sign?
    Check the Coriolis and centrifugal forces to make sure you have the right signs for these.

    Can you elaborate on treating the problem as one dimensional?
     
  4. Dec 11, 2016 #3
    So if outwards is positive.

    Centrifugal would be positive, friction negative, Coriolis I do not know, positive?

    I don't know what the lefthand a is supposed to represent to be honest, and cannot guess it's sign. We only just went over this is in class, and now the semester is over. Textbook gives very little explanation.

    Is the a substitution correct at least? I can try redoing it and paying better attention to the signs
     
  5. Dec 11, 2016 #4
    The only tip given by the professor is to use polar coodinates.
     
  6. Dec 11, 2016 #5

    TSny

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    Yes.
    The direction of the Coriolis force is determined by the direction of the cross product ##-\vec{\omega} \times \vec{v}'##. So, make sure you can visualize the directions of ##\vec{\omega} ## and ##\vec{v}' ##.

    The basic equation in the rotating frame is the vector equation: ##m \vec{a}' = \vec{F}_{\rm ext} +\vec{F}_{\rm cor}+\vec{F}_{\rm cen}##.
    ##\vec{a}'## is the acceleration vector of the object as seen in the rotating frame of reference. The magnitude and direction of ##\vec{a}'## for this problem can be determined by identifying the type of motion of the object relative to the rotating frame.
     
  7. Dec 11, 2016 #6
    So a is zero, as he is moving at a constant velocity? Or do I use the v2/r as he is changing direction
    I really appreciate the help, but still feel lost.

    My new equation is
    a'= -µg + 2(ωezXvetheta + ωezX(ωezXrer)

    Using polar coordinates
    Which solves (if I am computing the crosses correctly) to

    a'= -µg + 2ωv + ω2r all in the er direction

    Rotating counterclockwise, person moving in counter clockwise direction

    If a' is 0 I can solve for v, if it is v2/r I get a quadratic and can solve via factoring.
     
    Last edited: Dec 11, 2016
  8. Dec 11, 2016 #7

    TSny

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    Definitely the latter. The person is walking along a circle on the platform. So, if you were sitting on the platform, you would see the person moving with uniform circular motion. The acceleration ##\vec{a}'## is the acceleration associated with uniform circular motion. Thus, you should be able to state the magnitude and direction of the acceleration.

    OK, but the directions for the term on the left and the first term on the right are not clearly indicated here.

    What can you substutute for a' (including the sign)?
    Did you get the sign of µg right?
     
  9. Dec 11, 2016 #8
    It should be negative as it is pulling the person in towards the center?


    OK, but the directions for the term on the left and the first term on the right are not clearly indicated here.
    I don't understand sorry. Omega is pointing out of the page, as it is rotating CC, so ez. eTheta is moving CC, inline with V

    What can you substutute for a' (including the sign)?
    Did you get the sign of µg right?

    I missed it on the second one, fixed now.
    -v2/r ?
     
  10. Dec 11, 2016 #9

    TSny

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    Yes.
    In the equation a'= -µg + 2(ωezXvetheta + ωezX(ωezXrer), the term on the left side is written as a' without a unit vector to show direction. Or maybe you meant a' to be the acceleration vector? It's hard to tell. Symbols for vectors are usually written with an arrow over them or else they are written in bold face type. Likewise, the first term on the right side is written as -µg without a unit vector to show direction. Maybe you meant -µg er. I guess I'm being picky, but clarity is important.

    OK. The minus sign indicates radially inward. So, when you solve your equation for μ, do all of the signs work out as they should?
     
  11. Dec 11, 2016 #10
    I understand about the unit vectors, I just couldn't figure out how to express them in the forum, sorry about that.

    a' was meant to be acceleration vector from ma'

    If I go with

    -v2/R0 = -µg + 2ωv+ω^2R0 (*All in er direction)

    I end up with R0µg = (v+ωR0)2

    Which gives a vmax of sqrt(R0µg)-2ωR0
     
    Last edited: Dec 11, 2016
  12. Dec 11, 2016 #11
    My bad sorry, it becomes (v+ωR0)^2, have corrected it.
     
  13. Dec 11, 2016 #12

    TSny

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    Looks good.

    OK. Check the last term. Not sure the factor of 2 should be there.
     
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