Differential Equation: Falling Object

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Homework Statement



For small, slowly falling objects, the assumption made in equation (1) (see below) is good. For larger, more rapidly falling objects, it is more accurate to assume that the magnitude of the drag force is proportional to the square of the velocity with the force orientation opposite to that of velocity.

Write a differential equation for the velocity of a falling object of mass [itex]m[/itex] if the magnitude of the drag force is proportional to the square of the velocity

Homework Equations



(1) [itex]m \frac{dv}{dt} = -mg - \gamma v[/itex]
[itex] \gamma > 0[/itex]

The Attempt at a Solution



Since the drag force ([itex]- \gamma v[/itex]) is proportional to the square of the velocity:

[itex] | - \gamma v | = \gamma | v | \propto v^2[/itex]

and since the orientation is opposite to that of velocity

[itex] \gamma | v | = -kv^2 [/itex] where [itex]k[/itex] is some constant.

but this is where I get stuck.

The answer in the back of the book has the drag force as [itex]- \gamma v |v|[/itex].
Where did I go wrong?

Thank you anyone for your help!
 

Answers and Replies

  • #2
Dick
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-kv^2 is a scalar. It doesn't have any direction associated with it. That would work fine if you assume the direction of v (and hence the force) is vertical. In the more general case where you don't assume the direction of v is vertical, then -kv|v| would describe both the direction of the frictional force (since v is a vector) and the magnitude.
 
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  • #3
rude man
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mdv/dt = -mg -kv^2
is your equation.

It's non-linear since the dependent variable (or any of its derivatives) appears as any other than to the 1st power. So it's a ***** to solve, which is why they didn't ask you to.

The drag force IS a vector. Its direction is opposite to the direction of motion.

For example, if y is the vertical axis, the complete force equation is
mdv/dt j = - mg j - kv^2 j where j is the unit vector in the +y direction.

The term -kv^2 j is a vector in this equation, with the coefficient kv^2 a scalar.
 
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  • #4
PAllen
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Well, actually the equation is separable, so it is trivial to solve.
 
  • #5
Dick
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mdv/dt = -mg -kv^2
is your equation.

It's non-linear since the dependent variable (or any of its derivatives) appears as any other than to the 1st power. So it's a ***** to solve, which is why they didn't ask you to.

The drag force IS a vector. Its direction is opposite to the direction of motion.

For example, if y is the vertical axis, the complete force equation is
mdv/dt j = - mg j - kv^2 j where j is the unit vector in the +y direction.

The term -kv^2 j is a vector in this equation, with the coefficient kv^2 a scalar.
It also doesn't work if v isn't a vector proportional to j. Which I think was the whole point of the question.
 
  • #6
rude man
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By definition, the drag force IS in the direction (opposite to) the velocity.

Of course, air friction for example can cause an orthogonal force component as well. An aircraft wing experiences such a force, but it's called "lift."

There is "lift" and then there is "drag". And never the twain shall meet.

PS - most of the lift on an airfoil is due to the difference in air velocities above & below the wing (Bernoulli). But some of the lift is due simply to air resistance generating a lift component.
 
  • #7
Dick
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By definition, the drag force IS in the direction (opposite to) the velocity.

Of course, air friction for example can cause an orthogonal force component as well. An aircraft wing experiences such a force, but it's called "lift."

There is "lift" and then there is "drag". And never the twain shall meet.

PS - most of the lift on an airfoil is due to the difference in air velocities above & below the wing (Bernoulli). But some of the lift is due simply to air resistance generating a lift component.
The question was why the book wrote -kv|v| instead of -kv^2.
 
  • #8
rude man
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The reason they wrote kv|v| is that that way the vector aspect of the term is retained: it's a vector with magnitude |v|^2 and direction in the direction of v.
 
  • #9
rude man
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BTW PAllen says mdv/dt = -mg -kv^2 is solvable by separation of variables. Could he/she or someone else show me please?
 
  • #10
PAllen
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BTW PAllen says mdv/dt = -mg -kv^2 is solvable by separation of variables. Could he/she or someone else show me please?
First, note that I was assuming the co-linear case. If the body starts with a non-vertical component, so -kv|v| is not equivalent to -k v^2, my observation is false. So, assuming pure vertical motion, you have:

v' = -g - k v^2

v' / (g + k v^2) = -1

integral (1/(g + k v^2)) dv = integral (-1) dt
= c -t

etc.
 
  • #11
rude man
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Thank you PAllen! OK, it works only for collinear motion. Too bad since most elementary problems have some kid toss a ball at a certain angle to the horizontal ... blah blah ...still, it's an eye-opener for me - I thought the g term precluded separation of variables, but obviously it does not.

Thanks again!
 

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