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Spherical stone launched

  1. Jan 11, 2014 #1
    1. The problem statement, all variables and given/known data

    A spherical stone of mass 0.500 kg and radius 10 cm is launched vertically from ground level with an initial speed of 20.0 m/2. As it moves upwards, it experiences drag from the air as approximated by Stokes drag, F=6η∏Rv, where the viscosity of air is 1.002 mPa*s.
    (a)Which forces are acting on the stone while it moves upward?
    (b) Using Newton's second law of motion, write down an equation of motion for the stone this is a differential equation)
    (c) Solve the differential equation to get an expression for v(t) for the stone.
    (d) From (c) find the time at which the stone reaches its maximum height.
    (e) From v(t), find h(t) for the stone (height as a function of time).
    (f) Using (d) and (e), find the maximum height the stone reaches.
    (h) Find the speed of the stone just before it hits the ground

    2. Relevant equations
    F=ma
    a=dv/dt


    3. The attempt at a solution
    While the stone is moving upward, there is a gravitational force which is directed downwards. At the same time, there is a Stokes drag force directed downwards also(?).
    (b). Newtons second law of motion: F=ma=m*(dv/dt). How should I continue.. Can anybody please help me with working out all these resting questions.. It would really help me with my understanding for such motions, and to develop some feeling for these problems.
    Thank you in advance!!!
     
  2. jcsd
  3. Jan 11, 2014 #2
    ma is the left side of the equation ma = F. You need to specify F? You already noted that F is the sum of the gravitational force and the drag force. What is your equation for the gravitational force? What is your equation for the drag force? What is your equation for F?

    Chet
     
  4. Jan 11, 2014 #3

    tiny-tim

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    hi hvthvt! :smile:
    yes, a drag force means opposite to the direction of motion (relative to the air)
    erm :redface: … you should continue by starting! :wink:

    i think if you write out F = ma in terms of the information given, you will see what they mean by a differential equation

    show us what you get :smile:
     
  5. Jan 11, 2014 #4
    Hey tiny-tim and Chestermiller, thanks for motivating.
    I think that:
    F=mg + 6η∏Rv
    F=ma , so

    mg + 6η∏Rv = m*(dv/dt)
    I know that there is a v in the drag force,
    m*(dv/dt) - 6η∏Rv = mg
    Is this the differential equation?
    Further calculus gives (dv/dt) - ((6η∏R)/m)*v = g
    Now I should determine the integration factor, is this right? Doesn't that get too complicated then?

    e-6η∏R .. ?
     
  6. Jan 11, 2014 #5
    You can do it by using an integrating factor (don't forget to include the m and the t in the integrating factor), of by recognizing that the equation is variable-separable. Either way, it's not too complicated.
     
  7. Jan 11, 2014 #6
    The standard form of this diff equation should be
    (dv/dt) + ((-6η∏R)/m)*v = g
    p(t) = -6η∏R/m so P(t)=-6η∏Rt/m
    So the integration factor is e-6η∏Rt/m

    This gives (e-6η∏Rt/mv)' = g*e-6η∏Rt/m
    e-6η∏Rt/m*v = (mg/-6η∏Rt)*e-6η∏Rt/m

    Is this correct..
     
  8. Jan 11, 2014 #7

    haruspex

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    According to that, the faster it is going the more its speed will increase.
    To make the working easier, I suggest substituting k = 6η∏R/m.
     
  9. Jan 11, 2014 #8
    No. You left out the initial condition on the velocity at t = 0.
     
  10. Jan 12, 2014 #9
    Ok, at this point i'm pretty confused and stuck, especially because this is just part (b)..
    (dv/dt) + (k/m)*v = g ??
     
  11. Jan 12, 2014 #10
    With the initial condition included, your equation for the velocity should be:

    e-6η∏Rt/m*v-v0 = (mg/-6η∏R)*(e-6η∏Rt/m-1)
    Also, v0 is negative because the stone is thrown upward, and you're been taking downward velocity as positive.
    Please try to be more careful about your algebra, and try to include integration constants.

    Chet
     
  12. Jan 12, 2014 #11

    haruspex

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    Except that the sign is wrong on the exponent. (See my post #7.)

    Yes, that gets the sign right (k being positive).
    The way I defined k (= 6η∏R/m) it also removes the m, so it's just (dv/dt) + k*v = g.
    Applying the method you used before, you should now get (d/dt)(ektv) = gekt.
    As Chet says, all you have to do is remember to include the constant of integration and deduce its value from v(0) = v0.
     
  13. Jan 12, 2014 #12
    Yiii. You're right. I want to get the equation formulation right for the record. If v is the upward velocity, then

    [tex]m\frac{dv}{dt}=-mg-kv[/tex]

    If v is the downward velocity (a silly choice in my opinion),

    [tex]m\frac{dv}{dt}=mg-kv[/tex]

    Chet
     
  14. Jan 19, 2014 #13
    Oooooh, I understand it, actually it looks pretty simple?
    So that's the answer to b. Now for (c) I have to solve it.. meaning
    m*(dv/dt) + kv = - mg
    dv/dt + kv:m = -g

    I do not really understand what to do with the initial velocity. This is zero.

    ektv = - g*k*ekt + c
    so v= -k*g + c/(ekt)

    Is this it??
     
    Last edited: Jan 19, 2014
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