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Drag on a marble shot through water

  • Thread starter oneamp
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  • #1
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Homework Statement



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.

Homework Equations



Drag equation is given in the problem.

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
 

Answers and Replies

  • #2
662
307

Homework Statement



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.
And the question being?

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?
Yes, the velocity and hence the force will change. You will need to make use of differential equations and solve them.
 
  • #3
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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.
 
  • #4
662
307
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:
  • #5
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I will try that out, thank you.
 
  • #6
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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...
 
  • #7
662
307
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.
$$m ln(v)=bt$$
$$ln(v)=bt/m$$
$$v=e^{bt/m}$$
$$\frac{dx}{dt}=e^{bt/m}$$
Integrate.
 

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