Question

The wind's speed is 12 knots (6 m/s) NW at 320 degrees. I need to find the x and y components in order to find the air resistance acting on an object. can we discuss?

Answer 1

Maybe this will help:

Answer 2

yeah. I figured that out too @Shane_B .. however, the x has to be negative cause it's west.

Answer 3

Yep

Answer 4

ok. So I'm working with projectile and air resistance.

Answer 5

I need to show how much the wind speed reduces the speed of the ball in flight.
Do you have an idea on what equation I can use?

Answer 6

I think I'd have to see the whole question first...then try to figure it out

Answer 7

a golf ball is projected at 56 degrees with an initial velocity of 8 m/s. The wind's speed is 6 m/s NW 320 degrees. situation one is asking to ignore air resistance and find the components of the displacement. There was no problem with that. Situation 2, I need to take into account the air resistance: So far, I've figured out the x and y of the wind's speed. I need to figure out by how much the displacement of the ball in flight it going to decrease, and obviously, it won't be a perfect (symmetrical) parabola after that. Am I clear?

Answer 8

this is a good depiction of the 1st situation

Answer 9

It's clear...but I'm rusty on how to account for drag. I believe you'll need the surface of the golf ball that's exposed to the wind though so you can calculate the x/y force of drag.

Answer 10

Since \[F_{drag}=\frac{1}{2}\rho v^2C_dA\]

Answer 11

should I apply this in both x and y?

Answer 12

Yes

Answer 13

so the v^2 will have to be once the the velocity of the ball in the x direction, and then in the y direction?

Answer 14

I have the area of the ball, where do I get the p and the Cd from?

Answer 15

p is the density of the air which varies with temperature of course. A quick google search is giving me: \[\rho=1.2 kg/m^3\]

Answer 16

Cd is the drag coefficient...I need to see what that's supposed to be.

Answer 17

I have the temperature, it's 41 degrees celsius.. how can I calculate the density of the air using it? density = mass/volume.. in this case I need the mass and volume of the air?

Answer 18

Yea...or just find a table. It's going to be about 1.25

Answer 19

At 41C

Answer 20

Conflicting information on that...gimme a few minutes to see if we can just calculate it

Answer 21

cool thanks

Answer 22

@Shane_B great help dude
@Boushra Please award the server by a medal... click on best response

Answer 23

If you are satisified with the answer of the person ..

Answer 24

yeah.. he's doing great ^_^

Answer 25

I find a table on wikipedia saying that at 40 clesius, density = 992.2 kg/m^3

Answer 26

Alright, let's calculate the air density:

\[d=\frac{1atm}{RT}\]Where R is the specific gas constant of dry air: 287.05 J/kg Kelvin. T is in Kelvin also. Using that formula, I'm getting 1.124 kg/m^3.

Answer 27

and 1atm is obviously 101,325pa :)

Answer 28

ok, so you're converting 41 C to K. Hold on, I'll check

Answer 29

\[d_a{ir}=\frac{101325pa}{287.05*(41+273.15)}=1.124kg/m^3\]

Answer 30

yep.. got that correct as well.

Answer 31

This is getting deep :P

Answer 32

yeah.. :) I'm loving it :) now we're left with Cd

Answer 33

im trying to find an equation for it

Answer 34

I have the equation...but it depends on the Fd! Stuck at the moment...1 equation with 2 unknowns.

Answer 35

oh oh! there has got to be someway..

Answer 36

(back to the internet!)

Answer 37

;)

Answer 38

I know it's somewhere between 0.1 and 0.3...that's a start. But it would be nice to know for sure

Answer 39

agreed

Answer 40

Ok, from what I've read, it has to be a measured value (wind tunnel)! I'd say go with 0.2 since it's in the middle.

Answer 41

wind tunnel?

Answer 42

It's a room where they blow high-speed winds over objects and calculate the air resistance. Used for cars, planes, rockets, etc.

Answer 43

certainly can't get hold of that ... but, where did you get 0.1 to 0.3 from if I may know?

Answer 44

At any rate there are tables of values for common objects like a sphere (which is 0.47). However, a golf ball is dimpled (to reduce the drag) so it's going to be lower. From what I found online it's somewhere between 0.1 and 0.3...depending on the velocity used in the formula.

Answer 45

http://scienceworld.wolfram.com/physics/DragCoefficient.html

Answer 46

I saw the 0.1 on another site...already closed that one.

Answer 47

perfect, it seems approximate.. it would work even if the velocity of the ball is only 6 m/s, right?

Answer 48

Since the 0.3 is for speeds > 54 m/s...I go with something lower but between 0.1 and 0.3. Probably 0.1 since 6m/s is nothing really.

Answer 49

relatively

Answer 50

What are you using for the exposed surface area of the ball?

Answer 51

um... actually I need to figure that out I guess. all I have is the diameter of the ball

Answer 52

42.67 mm

Answer 53

I looked that up...the diameter is a range but at a minimum it's 42.67mm

Answer 54

I got 0.00285999427 m^2 for the exposed surface are of the ball.

Answer 55

:)

Answer 56

what's the equation for that please? ^^

Answer 57

Well, I just used the surface area of a sphere:\[A=4\pi r^2\]Cut that in half since only half the ball is exposed to the wind...then converted it to m^2 from mm^2

Answer 58

Basically: \[2\pi(\frac{0.04367m}{2})^2=0.0030m^2 \space(rounded)\]

Answer 59

awesome!

Answer 60

Now I think we can actually work on the real problem :)

Answer 61

I guess we have everything to figure out Fd

Answer 62

just one question, how will i be able to use 6 m/s (winds speed) here? i don't think it can be used in this equation.. thoughts?

Answer 63

I think you have to add the ball's x/y speed to the wind speed and then calculate the drag from those values. It makes sense to do that because it should be additive (regardless of the -x you got way earlier).

Answer 64

Thinking of it this way: If I'm moving 40 mph the wind will hit me at 40mph. If the wind is blowing in my direction at 20mph, then wind will be hitting me at 60mph.

Answer 65

Make sense?

Answer 66

hold on:)

Answer 67

it does. so, if the ball is moving at ux and the wind is hitting it at vx , then the v of the ball is vx - ux?

Answer 68

which I will substitute in the formula?

Answer 69

For the drag formula in the x direction, it would be vx - ux...and ux is negative so it would be vx-(-ux). This makes sense to me since it's blowing against the ball...so it should be combination of both magnitudes.

Answer 70

ok. but it would be vy - uy, correct?

Answer 71

yeah, things are starting to make sense, thanks @Shane_B :) .. i think it should be easy from here?

Answer 72

I don't know if it gets easier :) In the y direction, when the ball is going up the wind is helping it...when it's going down, the wind is working against the ball.

Answer 73

the wind in the y direction I mean in that last part.

Answer 74

oh I didn't think of it that way.

Answer 75

I'm thinking we must have way-overcomplicated this. I just found this:

http://www.phy.davidson.edu/stuhome/jocampbell/projectile/projectile.htm

Scroll near the bottom :)

Answer 76

The applet itself is on the main page...having played with it yet:

http://www.phy.davidson.edu/stuhome/jocampbell/projectile/projectile.ProjectileControl.html

Answer 77

*haven't

Answer 78

KE it says?! haha.. we've really complicated this

Answer 79

but they still use the Fd in x and y

Answer 80

After reading it, I'm pretty confident that page is oversimplifying it...for educational purposes. I don't see any real mistakes on what we've done so far.

Answer 81

I assume this question is for a class...if so, which one are you taking?

Answer 82

it's physics class, i'm taking it as an elective in my psych degree

Answer 83

We didn't even cover drag in Physics I/II/III !!!

Answer 84

I think this is more of a Dynamics class question.

Answer 85

are we able to assume that the velocity of the wind only blew when that ball was the the top of its trajectory? that way only the displacement in the x direction of the ball would reduce?

Answer 86

are you doing uni? which year?

Answer 87

I finished my EE stuff this past year...all graduated :)

Answer 88

I don't know if that's a safe assumption...the problem does say much about the wind or at what elevation it will come into play.

Answer 89

*doesn't !

Answer 90

well actually, we didn't cover that yet, I have an assignment on observation of underlying physics of a real world situation, taking into account air resistance. Obviously, I need to go a bit outside the book

Answer 91

the problem is something I came up with, I can alter in order to make things a little easier :), but I still need to apply air resistance

Answer 92

I think you/we are on the right track here...but I also think you have to factor in that the wind is opposing the ball on the way down.

Maybe a better question would be to calculate the air resistance in 1 dimension...ie: a car :)

Answer 93

My guess is that you're one of the over-achiever types :)

Answer 94

true, but I really want to involve projectile, and since golf is my favorite sport, I would like to give this assignment the best I can

Answer 95

I try :)

Answer 96

Well, I have some stuff I need to get to....but I think you're well on your way here. Maybe someone else will come in and simplify all this in like 5 formulas :)

Answer 97

see you are physics problems @Shane_B