Terminal velocity and drag coefficient?

In summary, the author is attempting to solve two homework problems. One deals with air resistance, and the other with drag. The author believes that the equation for case (2) is correct, but does not know what would change for case (1).
  • #1
MaGG
13
0
Ok, I think I'm getting a hold on this problem, but I want to make sure I'm in the direction...

Homework Statement


Let [tex]\inline{v_\infty}[/tex] be the terminal velocity of a falling body in air. If the body is dropped from rest, find the expression for the speed as a function of time (1) if the air resistance is proportional to the speed, and (2) if it is proportional to the square of the speed. In each case, relate the terminal velocity to the drag coefficient (proportionality constant) [tex]\inline\rho[/tex], mass m, and g, the acceleration due to gravity.

Homework Equations


1.[tex]f=ma=m\frac{dv}{dt}[/tex]

2.[tex]f=f_a-f_g=\rho v^2-mg[/tex]

3.[tex]a=\frac{dv}{dt}=\frac{d^2x}{dt^2}[/tex]

The Attempt at a Solution


I believe this is the solution for case (2), but I don't really know what it is for case (1). I know that case (1) involves slow moving bodies, and vice versa for case 2.

If you take eqn. 1 and divide by m, that gives you the acceleration.

[tex]\frac{f}{m}=\frac{dv}{dt}=\frac{\rho}{m}v^2-g\\
=g\left(\frac{\rho}{mg}v^2-1\right)\\
=g\left(\frac{v^2}{v_\infty}-1\right)[/tex]

where [tex]\inline v_\infty[/tex] is the terminal velocity [tex]\inline{\sqrt{\frac{mg}{\rho}}}[/tex]

Using separation of variables, I got to this:

[tex]\frac{v_\infty^2dv}{v^2-v_\infty^2}=gdt[/tex]

Then after taking the integral of each side, and then doing a bunch of algebra which I won't show here [unless you need me to put it to better understand the process], I eventually got this:

[tex]v(t)=(v_\infty)\frac{e^{\frac{-2gt}{v_\infty}}-1}{e^{\frac{-2gt}{v_\infty}}+1}[/tex]

Is this correct for case (2)? If so, what would change for case (1)? Thanks for your help! :smile:
 
Last edited:
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  • #2
The physics here is writing the differential equations. The rest is just math.

For the first, the differential equation is as follows:

[tex]
m\frac{dv}{dt} = \rho v - mg
[/tex]

For the second:

[tex]
m\frac{dv}{dt} = \rho v^2 - mg
[/tex]

The terminal velocity is just the limit as [tex]t \rightarrow \infty[/tex].
 
  • #3
So that final answer is correct? I would just have to re-do the math with a v instead of a v²?
 

Related to Terminal velocity and drag coefficient?

1. What is terminal velocity?

Terminal velocity is the maximum speed that an object can reach when falling through a fluid, such as air or water. This occurs when the upward force of drag equals the downward force of gravity, resulting in a constant velocity.

2. How is terminal velocity affected by the mass of an object?

The mass of an object does not affect its terminal velocity. This is because the force of gravity is directly proportional to an object's mass, while the force of drag is dependent on the object's surface area and shape. As a result, all objects of different masses will reach the same terminal velocity when falling through the same fluid.

3. What is the relationship between drag coefficient and terminal velocity?

The drag coefficient is a measure of an object's ability to overcome air resistance. It is directly related to an object's surface area and shape. As the drag coefficient increases, the force of drag also increases, resulting in a lower terminal velocity.

4. Can terminal velocity be reached in a vacuum?

No, terminal velocity can only occur in a fluid because there is no air resistance or drag force in a vacuum. Without these opposing forces, an object will continue to accelerate due to the force of gravity.

5. How does altitude affect terminal velocity?

At higher altitudes, air density decreases, which means there is less air resistance. This results in a higher terminal velocity for objects falling at high altitudes. In fact, at extremely high altitudes, the terminal velocity can reach speeds close to the speed of sound.

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