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Homework Help: Pebble falling into water; drag force

  1. Jan 1, 2013 #1
    1. The problem statement, all variables and given/known data

    The drag force experienced by the spherical pebble in water is given by: (0.5)CpwAv2

    mass of pebble = 83.8g
    radius of pebble = 2 cm
    C = 0.7


    (a) Write down the differential equation governing pebble's descent. (done)

    (b) Derive an expression for terminal velocity and evaluate it. (done)

    (c) The pebble is dropped 5m above surface of lake, which is also 5m deep. Show that the pebble travels about 10cm in water before reaching it's terminal velocity.


    3. The attempt at a solution

    Part (a)
    n3b3br.png

    Part (b)
    16h7v6c.png

    Part (c)
    2mfyqhc.png
     
  2. jcsd
  3. Jan 1, 2013 #2

    mfb

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    In your exact analysis, the ball will never reach its terminal velocity - but it will approach it with a typical timescale, and you can determine this and show what it does not travel more than 10cm before it reaches a velocity very close to the terminal velocity.
     
  4. Jan 1, 2013 #3
    Is there anything wrong with my working in part (b)?? I substituted for v2 using the differential equation..
     
    Last edited: Jan 1, 2013
  5. Jan 1, 2013 #4

    mfb

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    I don't see any result in c, and I don't understand what you are doing there.
     
  6. Jan 1, 2013 #5
    sorry i meant part (b)!
     
  7. Jan 1, 2013 #6

    SteamKing

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    1. Are you sure that C is the drag coefficient which uses the entire surface area of the pebble or just the cross-sectional area of the pebble?

    2. You recognize that the acceleration of the pebble is x double dot. What about the velocity of the pebble? Can't the velocity also be represented in terms of x?
     
  8. Jan 1, 2013 #7

    haruspex

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  9. Jan 1, 2013 #8

    haruspex

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    Sorry, misunderstood the point you were making (and for some reason it's not letting me edit the post).
     
  10. Jan 1, 2013 #9
    You need to substitute x double dot = dv/dt, and solve for the velocity vs time first. Also, what happened to the buoyant force?
     
  11. Jan 2, 2013 #10

    SteamKing

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    The point about drag coefficients is that they are usually based on the cross sectional or frontal area of the body, not the total surface area.

    The OP has used (4/3)pi*r^2, which is the total area of the pebble.
     
  12. Jan 2, 2013 #11

    haruspex

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    That's not how I read it. Seems to have written the mass as (4/3)pi*r^3*density, then cancelled one r and π with the expression for cross-sectional area. Not that there was much point in doing that: the mass is given, the density is not.
     
  13. Jan 2, 2013 #12
    sorry, density of pebble is 2.5g/cm3!
     
  14. Jan 3, 2013 #13
    I'm more concerned about the integration part; to find the work done against resistance. Is there anything wrong with that?
     
  15. Jan 3, 2013 #14

    haruspex

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    When you lost k from your equations, you might have realised you were off course. K must feature in the answer. There are errors further along, but they're not relevant to finding a correct solution.
    i don't think you can hope to integrate v2dx straight off. Work with the forces and get the differential equation for the acceleration.
     
  16. Jan 5, 2013 #15
    There are 2 ways to approach this problem:

    Method 1

    Substitute v2 using the differential equation and integrate; which was what I did but I'm not sure what's wrong with my working...


    Method 2

    a) Solve the differential equation to obtain v in terms of x. show that when x = 0.1 m, v = vterm
     
  17. Jan 5, 2013 #16

    haruspex

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    As I said, that substitution lost k, and as soon as that happened you were going nowhere. At best you'd end up with tautology. As it happens, you must have made some other error, but I don't think it's interesting to figure out what.

    You have [itex]\ddot{x}=\dot{v}=g-kv^2[/itex]. You could solve that to find v in terms of t, but there is a trick to eliminate t first and solve for v in terms of x. Is that what you mean?
     
  18. Jan 5, 2013 #17
    Guys,

    You've left out the buoyant force on the pebble.

    Chet
     
  19. Jan 5, 2013 #18

    haruspex

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    Good catch!
     
  20. Jan 6, 2013 #19
    yeah. replace dv/dt by v(dv/dx) then solve the diferential equation, putting initial condition as v = √2gh.

    then substitute v = vterm to find that x = 0.1 m...
     
  21. Jan 6, 2013 #20
    Ignore the buoyant force on the pebble.
     
  22. Jan 6, 2013 #21
    I did some working here using method 2, to express v in terms of distance travelled x:

    vraiko.png

    35n0aol.png


    k = b/m = 0.4398 and 1 < α ≤ 10k

    Next, i let α = 1.1 ≈ 4% above vterm

    I get x = 4m = 40cm....
     
  23. Jan 6, 2013 #22

    haruspex

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    There's something screwy in your second line, but after that it looks right.
    It's a bit hard to guess what they consider to be effectively terminal velocity. How about Plugging in x=0.1 and see what you get for alpha?
    And how do you know to ignore buoyancy? Have you checked how much difference it makes?
     
  24. Jan 6, 2013 #23
    This is a prelim question in maths paper, and we're told to ignore buoyance force..

    I got a crazy a = 4.. which is slightly below the maximum a = 10k = 4.398
     
  25. Jan 6, 2013 #24
    Are you considering that the first five meters are in the air above the lake?
     
  26. Jan 6, 2013 #25

    haruspex

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    I get slightly different numbers... 2k = Cρw A / ρp = 3 Cρw / (ρp 4r) = 3*0.7*1000/(2500*0.02) , k = 5.25. (With buoyancy, k = 8.75.)
    (v/vt)2-1 = ((v0/vt)2-1)e-2kx = (10k-1)e-2kx
    So with x =.1, (v/vt)2-1 = 51.5e-1.05 = 18. As you say, much too large. With buoyancy, 15, hardly better (about twice terminal velocity.).

    FWIW, imagining it as a real experiment, I don't believe a 10cm wide pebble dropped 5m would slow to around terminal velocity in 10cm.

    Just had a thought... I was always bothered by the question asking for when it slowed to 'about terminal velocity' - much too vague. It would make a lot more sense as 'slowed to about twice terminal velocity' (or some other multiple). Is it possible the question is quoted wrongly?

    EDIT: Where I wrote "about twice terminal velocity" above I should have written "about four times terminal velocity". Must have square rooted twice.
     
    Last edited: Jan 6, 2013
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