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Stocke's law and terminal velocity

  1. Jul 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Figure shows the motion of a rubber ball,of mass m,released (from rest),from the bottom of a pond.Explain the motion of the ball,with reference to its change in velocity and any other forces present,at points A,B,C,D and E
    Also comment on the path and velocity of the ball after it reaches point F.
    State any assumptions you make at arriving at the answer



    2. Relevant equations

    Stoke's law
    Archimedes' principle

    3. The attempt at a solution
    At the beginning(at A) ,u=0,so a=0,so,
    I guess,
    [tex]\uparrow[/tex]U - mg = ma
    and U (upthrust)=mg

    As the rubberball moves upwards, the velocity increases(a increases),therefore according to Stockes law,the drag force(F )also increases,thus the ball decelerates & at 1 point(maybe at B?), the acceleration (a) becomes 0,and,
    U = mg + F ?
    so when a = 0, the ball attains terminal velocity (V)

    With this velocity(V ),I guess the ball would move upwards to position C?
    At D,there would only be 2 foces acting on the ball? U & mg,(I think?),but how does this bring about the projectile-like motion of the ball from D to F?
    I'm really confused. :(

    Thank you
     

    Attached Files:

  2. jcsd
  3. Jul 12, 2009 #2
    Nevermind. My mistake.
     
    Last edited: Jul 12, 2009
  4. Jul 15, 2009 #3
    Could someone take a look at this problem for me?please
    I'm still not sure how I'm supposed to explain the motion of the rubberball from D onwards
     
  5. Jul 15, 2009 #4
    I suppose we can say that it will be at terminal velocity at point C. At point D the only force is the weight since it has left the water, and since it has some initial velocity, it follows a projectile-like trajectory (it's just like any other projectile problem)

    When it falls back down to the water (F) I guess it just repeats the entire cycle, sinking and floating again, but at a diminishing magnitude...
     
  6. Jul 15, 2009 #5
    Thanks for replying,queenofbabes.
    For the part about assumptions,would neglecting air resistance be it?

    And if I were to sketch a velocity-time graph to explain the motion of the rubberball,would it be like this?
     

    Attached Files:

    • v-t.bmp
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  7. Jul 15, 2009 #6

    turin

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    By a, do you mean acceleration? If so, then you are probably not understanding the concept of acceleration.

    What do you suppose causes upthrust when the ball is out of the water?
     
  8. Jul 16, 2009 #7
    OK, I'm going to start again.from A.
    The instant the ball is released,it accelerates upwards ,therefore the velocity increases,
    U - (F + mg) = ma
    U =upthrust
    F=drag force
    a= acceleratin
    As the velocity increases,the drag force acting down on the moving ball increases until,
    U-( F + mg) = 0
    i.e, the acceleration is now 0 and the ball attains terminal velocity?(at B?)
    It continues with this terminal velocity to C?
    At D ,as queenofbabes pointed out,only the weight,mg,of the ball acts downwards,but I'm not quitesure how this helps it to project out of water ?
     
  9. Jul 16, 2009 #8
    It propels itself out of the water because it hadn't lost its momentum. It should emerge from the water at the terminal velocity.
     
  10. Jul 16, 2009 #9

    turin

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    Excellent!

    I don't know. This cannot be determined from the information given. For example, it may reach the surface before it achieves terminal velocity. Anyway, for the sake of argument I will assume that point B is the point where the ball achieves terminal velocity.

    Excellent!

    Excellent!

    Does motion require force? There is a Newton's law about this. E.g. how did you decide that it continues with the terminal velocity from B to C?
     
  11. Jul 17, 2009 #10
    I don't know for sure,but this is what I thought,
    At B,there are only 3 forces acting on the body,
    the weight and the viscous force downwards,and upthrust,giving
    U - (F+mg) = 0
    therfore U = F+mg
    and since the forces balance and there are no other external forces(or resultant force) acting on the body,according to Newton's first law,it would continue to move upwards with this terminal velocity?
    So I think the answer to your question,
    "does motion require a force?" would be no?
    I think a force is only needed to cause an acceleration or for a change in momentum,as according to Newton's 2nd law?

    But,here's my problem
    at D,only the weight of the body is acting down on the ball,so there's a resultant force downwards and so a change in momentum downwards(which would mean the ball decelerates upwards),and so starts moving with a reduced velocity upwards which just doesn't seem right to me?
     
  12. Jul 17, 2009 #11
    Yes, what you have said is correct. Why doesn't it seem right? It's the same thing as picking up a ball and throwing it upwards into the air. It's your typical projectile motion =)
     
  13. Jul 17, 2009 #12
    Thanks for the reassurance,queenofbabes.
    but now I have my doubts on the velocity-time graph I've attached on post#5.
    I'm not sure how I'm supposed to sketch the change in velocity at and beyond F.

    I hope someone can help,
    Thank you
     
  14. Jul 17, 2009 #13
    i think at F and beyond the graph will move upwards because the speed decrease
     
  15. Jul 17, 2009 #14
    so would the graph look like this ?

    http://img148.imageshack.us/img148/7853/34853828.png" [Broken]
     
    Last edited by a moderator: May 4, 2017
  16. Jul 17, 2009 #15

    turin

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    I presume that the primes are only labels, not derivatives. I agree with the shape of your velocity vs time plot between A and F. After F, I disagree. Describe what happens after F in words.

    Hint: compared to the terminal velocity of the ball in the water (between B and C), at what velocity does the ball enter the water at F?
     
  17. Jul 18, 2009 #16
    At F,the ball would enter the water at the same speed it left the water at B,which is V' .
    Now,beyond F
    once again there are 3 forces acting on the rubber ball,
    the upthrust and drag force upwards,and the weight down,resulting in
    mg - ( U +F) = ma"
    so the ball would initially accelerate(not sure if it's with the same magnitude as in the 1st case?)and again as velocity increases the drag force(F) increases ,the balll decelerates (that is with decreasing negative velocity?)until
    mg - (U+F) = 0
    which is when it starts moving with a new terminal velocity(- V" ),the magnitude (or speed) of which I think should be less than that between C and B (V')
    or is it more than V' ?
    cause at F(the instant it touches thewater surface), (unlike the initial zero velocity at A),the ball moves with a speed V' ,so it decelerates faster and quickly attains a terminal velocity ,the magnitude of which is more than that between B and C (i.e . V" > V' ? )
     
  18. Jul 18, 2009 #17
    You're forgetting that upthrust > weight (that's why it floated in the first place) so it will decelerate and come to a rest at some point not as deep as the original starting point, and then it will float/fly out of water/sink again, repeat, etc.
     
  19. Jul 18, 2009 #18
    Yes.I get it now.
    Thank you very much!
     
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