0
$\begingroup$

I'm trying to build a tennis ball machine and I'm using a couple resources to help me out. One of them has the following formula for calculating drag force which I am having trouble interpreting.

Formula and some Information

The main question I have is what the V with a bar means and the difference between it and the regular V. I am also wondering if there is a special formula for calculating translational velocity. Any help would be much appreciated, and I'm not sure if it helps but cd is given as anywhere between .55 and .85. The air density could be 1.1376kg/m^3 and the radius of a tennis ball is around 3.35cm.

Thanks for your time and help!

Improve this question
$\endgroup$
2
  • $\begingroup$ "translational velocity" Did you mean terminal velocity? $\endgroup$
    – pho
    Commented Sep 29, 2020 at 0:29
  • $\begingroup$ No, I did mean to say translational velocity but I'm not quite sure what it is. $\endgroup$
    – ShinyWhaleFood
    Commented Sep 30, 2020 at 21:18

1 Answer 1

Reset to default
3
$\begingroup$

The main question I have is what the V with a bar means and the difference between it and the regular V.

$\bar{V}$ is the velocity vector. $V$ is the speed, which is the magnitude of the velocity vector. $\bar{F}$ is the force vector. (A more typical notation for vectors is $\vec{V}$ and $\vec{F}$, with arrows instead of bars, or $\mathbf{V}$ and $\mathbf{F}$, using boldface.)

The formula is saying that the direction of the drag force is opposite to the direction of motion, and the magnitude of the drag force is proportional to the square of the speed.

I am also wondering if there is a special formula for calculating translational velocity.

For motion in a single direction, like falling vertically, yes. For two dimensions, I don’t think so; you have to solve the differential equation numerically.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Doesn't the $\bar V$ notation usually represent a complex conjugate? $\endgroup$
    – JMac
    Commented Sep 28, 2020 at 23:03
  • 1
    $\begingroup$ @JMac Sometimes, but not here. In QFT it’s more than just a complex conjugate. $\endgroup$
    – G. Smith
    Commented Sep 28, 2020 at 23:04
  • $\begingroup$ Ahh I see they are using $\bar F$ too. That seems pretty weird. $\endgroup$
    – JMac
    Commented Sep 28, 2020 at 23:05
  • 1
    $\begingroup$ It just indicates vectors. I’m used to arrows rather than bars, but it’s not that weird. $\endgroup$
    – G. Smith
    Commented Sep 28, 2020 at 23:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.