Free fall equations

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Homework Statement



A particle of mass m falls from rest from height h above the ground. The retarding force per unit mass due to air resistance is kv^2 where k is a constant and v is the velocity of the particle

a) write and simplify the equation of motion for the particle
b) find an expression for the velocity with which the particle reaches the ground
c) if h is very large, determine the expression for the velocity of the particle
d) explain what happens to the velocity when h is very large

Homework Equations





The Attempt at a Solution



a) i thought a particle falls with equation F=ma and the air resistance works against this so F=ma-kv^2

b) no idea just yet :(

c) when h is very large there is an equilibrium in net force and so terminal velocity is reached

terminal velocity = √2mg/pA(cd)
where p=density A=area of object cd=drag coefficient m=mass g=gravity

d) pretty much covered in part c) an object stops accelerating because the force of gravity and the air resistance are in equilibrium
 

Answers and Replies

  • #2
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re (a):

First draw a FBD for the paraticle.
What will be the net force downwards?
 
  • #3
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the net force acting on the particle would be mass x gravity?
 
  • #4
993
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How many forces act on the particle?
 
  • #5
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gravity is acting downwards air resistance is working against gravity?
 
  • #6
993
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Exactly. So the net force downwards is ...
 
  • #7
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net force = g-kv^2 ?
 
  • #8
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The retarding force per unit mass due to air resistance is kv^2 where k is a constant and v is the velocity of the particle
hence retarding force is mkv^2.
therefore net force downwards is Fnet = mg - mkv^2.
 
  • #9
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ahh i get it now :)

so part b would use the equation v^2=u^2=2gh where v = final velocity and u = initial velocity so therefore final velocity v=root (2gh)
 
  • #10
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sorry i meant v^2=u^2+2gh
 
  • #11
cepheid
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sorry i meant v^2=u^2+2gh
Nope. This equation is only valid when you have constant acceleration. The acceleration here is not constant, because the force is not constant.
 
  • #12
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Just as i think i am getting somewhere :)
back to my notes it is then
 
  • #13
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sorry i meant v^2=u^2+2gh
acceleration is not g because g is for FREE FALL and here the falling is not freely but there is a dragging force. Instead of g you have to use the acc that you find from Fnet = ma
 
  • #14
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That is the reason why you were asked to do part (a).
 
  • #15
cepheid
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acceleration is not g because g is for FREE FALL and here the falling is not freely but there is a dragging force. Instead of g you have to use the acc that you find from Fnet = ma
No. See my previous post. This entire equation is invalid because the acceleration is not a constant.

To the original poster: I would encourage you to look into the concept of terminal velocity. Hint: what happens when the velocity becomes large enough that kv2 = g ?
 
  • #16
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No. See my previous post. This entire equation is invalid because the acceleration is not a constant.

To the original poster: I would encourage you to look into the concept of terminal velocity. Hint: what happens when the velocity becomes large enough that kv2 = g ?
Yes cepheid is correct.

v^2 = u^2 + 2ah holds only constant acc and here the acc is not constant. Forgive my mistake!
I think that one has to integrate the acc eqn to get v.
 
  • #17
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arghh more integration :(

so acceleration = dv/dt

and then substitute into v-u=ah ?

ohh and thanks a lot for all your help
 
  • #18
cepheid
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arghh more integration :(

so acceleration = dv/dt

and then substitute into v-u=ah ?

ohh and thanks a lot for all your help
No integration is required to solve part b. :smile:

To the original poster: I would encourage you to look into the concept of terminal velocity. Hint: what happens when the velocity becomes large enough that kv2 = g ?
 
  • #19
cepheid
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Actually, I just looked at parts c and d, and it seems like they refer to the terminal velocity case, whereas part b wants a general solution. If that's true, then you have a non-linear differential equation to solve. Uh, good luck with that.
 
  • #20
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no integration needed is the best news i have heard for a while :)

i have looked at terminal velocity and think i am getting somewhere with that

terminal velocity is reached when force due to gravity = air resistance

mg=kv^2

air resistance can be written as 1/2 (drag coefficient x density x frontal area x velocity^2)

mg=1/2 (dc p Af V^2)

then its simply manipulating for velocity (just about the only part I can do :)

as for part b) more time looking at my scribbled notes is required
 
  • #21
cepheid
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no integration needed is the best news i have heard for a while :)

i have looked at terminal velocity and think i am getting somewhere with that

terminal velocity is reached when force due to gravity = air resistance

mg=kv^2

air resistance can be written as 1/2 (drag coefficient x density x frontal area x velocity^2)

mg=1/2 (dc p Af V^2)

then its simply manipulating for velocity (just about the only part I can do :)
None of this stuff in red is really necessary. For this problem, you're given the drag force. It's kv2. So, all these pesky constants like the density and drag coeff. etc are just absorbed into k. You really only need to give your answer in terms of k.

So for part c, terminal velocity, it's literally just: g = kv2. Solve for v.

as for part b) more time looking at my scribbled notes is required
I can help you set up the differential equation, but I can't help you solve it. To set up the differential equation, start with the equation of motion you got for part a, and write everything in terms of velocity and its derivatives. Again, once you have that, good luck solving it. :tongue2:
 
  • #22
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part a) f=mg-mkv^2 = ma =(this is where i may go wrong) m(dv/dx)(dx/dt) =m(dv/dx)v

limited knowledge of differential equations but here goes

x = -∫(v/(kv² + g))dv = -(1/(2k))ln(kv² + g) + C.
final velocity = initial velocity (v=u) when x = 0, C = (1/(2k))ln(ku² + g).
Therefore -2kx = ln(kv² + g) - ln(ku² + g),
and so e^(-2kx) = (kv² + g)/(ku² + g) = (v² + g/k)/(u² + g/k).
manipulating gives v² = (g/k + u²)e^(-2kx) - g/k.

and finally is it too late to take chemistry instead of physics?
 
  • #23
cepheid
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To be honest, I'm having trouble understanding your steps. That doesn't mean they're wrong, it just means I can't follow them.

Here's what I had:

[tex] a = g - kv^2 [/tex]

[tex] \frac{dv}{dt} = g - kv^2 [/tex]

Then you did a smart change of variables using the chain rule, which I approve of:

[tex] \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx} = g - kv^2 [/tex]

Now we can divide both sides through by v:

[tex] \frac{dv}{dx} = \frac{g}{v} - kv [/tex]

I still don't know how to solve this differential equation.:rofl:
 
  • #24
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Calculus i didnt find too bad, just dont like it :)

mechanics is a problem didnt do any at A level

Thanks for all your help tonight
 
  • #25
cepheid
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Calculus i didnt find too bad, just dont like it :)

mechanics is a problem didnt do any at A level

Thanks for all your help tonight
I have figured out how to solve it, but it is complicated:

Start with

[tex] v\frac{dv}{dx} = g - kv^2 [/tex]

Now, notice that the left hand side looks like something that has already been differentiated using the chain rule. In math, they call such an expression an exact differential, since it is the derivative of something. In particular, "reversing" the chain rule in your head, you can see that:

[tex] v\frac{dv}{dx} = \frac{d}{dx}\left[\frac{1}{2}v^2\right] [/tex].

To verify that this is true, just differentiate the expression in square brackets using the chain rule, and you'll get back what is on the left hand side. So we end up with:

[tex]\frac{d}{dx}\left[\frac{1}{2}v^2\right] = g - kv^2 [/tex].

Let's now introduce a change of variables i.e. a substitution. Let u = (1/2)v2. Then the differential equation becomes:

[tex] \frac{du}{dx} = g - 2ku [/tex]

[tex] \frac{du}{dx} + 2ku = g [/tex]

Now this is a linear differential equation that I DO know how to solve. It can be solved by a method known as the method of integrating factors. If/when you take a course in ordinary differential equations, you will learn this method. Basically, it involves multiplying both sides of the equation by an integrating factor. The integrating factor is a factor that turns the left-hand side into an exact differential. In this case, the left hand side will look like something that has been differentiated using the product rule. You can then "reverse" this differentiation by integrating both sides w.r.t. x. After applying the initial condition u(0) = 0, you'll be able to solve for u(x) and hence for v(x).

The integrating factor is always exp(∫ψdx) where ψ is the coefficient on the second term (the term that has only u in it and no derivatives). So, in this case, ψ = 2k, and the integrating factor becomes exp(2kx). If you multiply both sides by this factor, you should be able to solve the equation.

Edit: I just realized that the initial condition is actually u(h) = 0, not u(0) = 0. So that will change the form of the solution somewhat.
 
Last edited:

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