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Drag on a marble shot through water

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data

    An object moving in a liquid experiences a linear drag force: D⃗ =(bv, direction opposite the motion), where b is a constant called the drag coefficient. For a sphere of radius R, the drag constant can be computed as b=6πηR, where η is the viscosity of the liquid.

    2. Relevant equations

    Drag equation is given in the problem.

    3. The attempt at a solution

    I am working toward a solution but there is something that troubles me. Drag, D = bv, depends on velocity. But, velocity changes continually from the point the marble first begins to slow down, until it reaches zero. So how can I come up with net force for this to solve the kinnematic equations, when they are dependant on a constantly changing velocity?

    Thanks
     
  2. jcsd
  3. Sep 21, 2013 #2
    And the question being?

    Yes, the velocity and hence the force will change. You will need to make use of differential equations and solve them.
     
  4. Sep 21, 2013 #3
    The question is:

    Water at 20 ∘C has viscosity η=1.0×10−3Ns/m2. Suppose a 1.0-cm-diameter, 1.2g marble is shot horizontally into a tank of 20 ∘C water at 15cm/s . How far will it travel before stopping?

    My confusion is with the general principle of using v in the equation. This class requires only calc 1 so I don't think a differential equation is what they're looking for.
     
  5. Sep 21, 2013 #4
    D=bv
    m(dv/dt)=bv
    integrate (EDIT->) and you get time taken to stop and also if you let the limits be in terms of variables you will get a similar equation for v and t.
     
    Last edited: Sep 21, 2013
  6. Sep 22, 2013 #5
    I will try that out, thank you.
     
  7. Sep 22, 2013 #6
    How does that help? I integrate and still get either the derivative of velocity on the left, or velocity (integral of the derivative of velocity, is velocity). So I still have velocity in the equation...
     
  8. Sep 22, 2013 #7
    $$m ln(v)=bt$$
    $$ln(v)=bt/m$$
    $$v=e^{bt/m}$$
    $$\frac{dx}{dt}=e^{bt/m}$$
    Integrate.
     
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