Equation of motion with proportional drag

In summary: Thank you for your help!In summary, the conversation discusses finding the velocity and place equations for a given force equation and initial conditions. The provided equation for velocity appears to be correct, but there were some input errors in the software. The correct place equation has been found and can be simplified for easier use.
  • #1
Hannibal123
20
0

Homework Statement



the forces will be like this

[tex]m⋅dv/dt=-m⋅g-k⋅v[/tex]

I need to find the velocity equation and the place equation, thereby meaning the integral of the velocity equation. for the condition t,v=0,v_0



The Attempt at a Solution


i have found this
[tex]v(t)=mg/k*(-1+e^(-k/m⋅t))+v_0⋅e^(-k/m⋅t)[/tex] or in picture form http://imgur.com/WjksG

But I am not sure it's correct
 
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  • #2
Hannibal123 said:

Homework Statement



the forces will be like this

[tex]m⋅dv/dt=-m⋅g-k⋅v[/tex]

I need to find the velocity equation and the place equation, thereby meaning the integral of the velocity equation. for the condition t,v=0,v_0

The Attempt at a Solution


i have found this
[tex]v(t)=\frac{mg}{k}(-1+e^{-\frac{k}{m}t})+v_0⋅e^{-\frac{k}{m}t}[/tex] or in picture form http://imgur.com/WjksG

But I am not sure it's correct

It looks correct, but v0=0 .

Edit: I misread vo, so it is correct.

ehild
 
Last edited:
  • #3
Your answer looks right to me. Why do you doubt it?
 
  • #4
ehild said:
It looks correct, but v0=0 .

ehild

I read the OP condition as initially (t,v) = (0, v0)
 
  • #5
haruspex said:
I read the OP condition as initially (t,v) = (0, v0)
Sorry, I misread it. Edit my post.

ehild
 
  • #6
haruspex said:
I read the OP condition as initially (t,v) = (0, v0)

that is what i meant. I doubt the results because when i draw them in a graph they dosen't seem to be correct. I have drawn them i geogebra if you are familiar with that software.
My values are:
m: 0,145 kg
k: 0,0032
g: (gravity acceleration) 9,82
v_0: 9,93 m/s

this is what i have entered in geogebra: g(x) =(0.145*9.82/0.0032)*(-1 + ℯ^(0x / 0.15)) + 9.93ℯ^(0x / 0.15)
if you want to try for yourself (should save you some time). These values are not meant for a proportional drag, but still it seems weird that the graph looks like one of a constant function
http://imgur.com/FFElN (the red one)
 
  • #7
Furthermore when i integrate the velocity equation i get this http://imgur.com/IO6G6
Again not sure if I am correct
 
  • #8
Hannibal123 said:
m: 0,145 kg
k: 0,0032
g: (gravity acceleration) 9,82
v_0: 9,93 m/s

this is what i have entered in geogebra: g(x) =(0.145*9.82/0.0032)*(-1 + ℯ^(0x / 0.15)) + 9.93ℯ^(0x / 0.15)
How did -k/m end up as 0 / 0.15? Looks like rounding error, and the red line graph seems to be a consequence.
 
  • #10
this is the equation of place (if that is the correct term in english) as far as I am concerned the height is not suposed to be negative to positive time values?
http://imgur.com/CxAYC
 
  • #11
Your velocity time graph shows velocity +ve at time 0, so the distance time graph should show distance increasing at time 0.
 
  • #12
Indeed it should, however it dosen't. Is this a wrong integration of the velocity equation?
http://imgur.com/A6BKT
Based on the gaph it does not seem to be correct.
 
  • #13
Just found the correct place equation i think
http://imgur.com/quiLu
can you see a good way to shoten it?
 
  • #14
Looks right. I'd write it as
[itex]\frac{mg}{k}\left(-t+\left(\frac{m}{k}+\frac{v_0}{g}\right)\left(1-e^{-\frac{kt}{m}}\right)\right)[/itex]
 

1. What is the equation of motion with proportional drag?

The equation of motion with proportional drag is a mathematical model used to describe the motion of an object under the influence of both a force and a drag force that is proportional to the object's velocity. It takes into account the object's mass, acceleration, initial velocity, and the drag coefficient.

2. How is the equation of motion with proportional drag derived?

The equation of motion with proportional drag is derived by applying Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration, along with the drag force equation, which is proportional to the object's velocity.

3. What is the significance of the drag coefficient in the equation of motion with proportional drag?

The drag coefficient in the equation of motion with proportional drag represents the amount of resistance an object experiences due to the surrounding medium, such as air or water. It is a dimensionless quantity that depends on the object's shape, size, and the properties of the medium.

4. Can the equation of motion with proportional drag be used for all types of motion?

No, the equation of motion with proportional drag is only applicable to objects moving in a fluid medium, such as air or water. It does not take into account other factors such as friction or non-proportional drag forces that may affect the motion of an object.

5. How does the equation of motion with proportional drag change if the medium's properties change?

The equation of motion with proportional drag remains the same, but the value of the drag coefficient may change if the properties of the medium, such as density or viscosity, change. This can affect the amount of drag force experienced by the object and therefore, its motion.

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