Terminal Velocity of a ping pong ball

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SUMMARY

The discussion focuses on calculating the terminal velocity of a ping pong ball using Euler's Method for solving ordinary differential equations (ODEs). The calculated terminal velocity for the ping pong ball is 28 m/s, while the expected value is approximately 8.9 m/s. Participants emphasize the importance of correctly applying the formula for acceleration, which is given by a = g − CρA/2m * v², and using a spreadsheet or programming tool to iteratively compute velocity over small time intervals (deltaT = 0.1s). The conversation highlights common pitfalls, such as miscalculating mass and misunderstanding the application of kinematic equations in non-constant acceleration scenarios.

PREREQUISITES
  • Understanding of basic physics concepts such as forces, acceleration, and drag.
  • Familiarity with Euler's Method for numerical solutions of ODEs.
  • Proficiency in using spreadsheet software like Microsoft Excel for calculations.
  • Knowledge of the equation for terminal velocity and its components, including drag coefficient (C), air density (ρ), and cross-sectional area (A).
NEXT STEPS
  • Learn how to implement Euler's Method in programming languages such as Python or MATLAB.
  • Research the impact of varying drag coefficients on terminal velocity calculations.
  • Explore advanced numerical methods for solving ODEs, such as Runge-Kutta methods.
  • Investigate the physics of drag forces in different mediums beyond air, such as water.
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in computational methods for solving differential equations related to motion and forces.

phil ess
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Homework Statement



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The Attempt at a Solution



I'm at a complete loss here. Part A was simple enough, and I calculated a terminal velocity of 28 m/s for the ping pong ball and 34 m/s for the golf ball. But I don't even know how to begin Part B. Wikipedia says that Euler's Method is for solving ODE's, which I don't think is first-year material. In any case, I don't even see where time comes from in v(t), here's what I tried:

Fnet = Fg + Fdrag
ma = mg - 0.5pv2AC

Then trying to figure time in somewhere: a = v/t

mv/t = mg - 0.5pv2AC

Then isolate v? Now i have a v(t) function I guess but i guarantee this isn't what I'm supposed to be doing! Please help!
 
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To calculate the speed at any time you just need to know how much it has acclerated, which depends on the forces acting on it = weight down and drag up. But the drag depends on the speed and so you are back to step 1.
You could write an ODE describing the acceleration and solve it analytically - but as you said this might not be first year work.

Euler's method is really just calculus by spreadsheet.
For each small increments of t (eg 0.1s) just calculate what the speed is at the end of that time and then what the drag is for that speed. Then use that drag to calculate the force for the next 0.1 seconds and so the speed and so on.

Hint - at time=0 there is no speed and so no drag.
 
Check your math for the terminal velocity of the ping pong ball in part A. I think you may have missed the decimal place in its mass (you may have used 25 g instead of 2.5 g).

For part B, you seem to be on the right track using Fnet = Fg + Fdrag

I would divide through by mass to get anet = ag + Fdrag / m

Then apply this to vfinal = a * t {use vinitial to determine acceleration, where vinitial = vfinal from the previous iteration.

Can you use a computer program for Euler's Method? Either a simple program or even a spreadsheet program will make this fairly easy to compute. Otherwise this may be fairly tedious by hand.
 
Ok here's the equation for acceleration of the falling ball, if down is positive:

a = g − CρA/2m * v2

So I'm in Excel now, and I'm supposed to use deltaT = 0.1s.

So what I did was this (note the calculation bar to see what i did):
Untitled-1.jpg


Now from my Vterm calculation I expect this to approach v = 8.9 m/s (thanks StoveBolt), but it breaks down badly for some reason. the way its calculated is as follows:

CρA/2m = 0.125206285 <-- that's where that comes from

Then multiplying that by v2 from the previous interval

Subtract this from g = 9.81

multiply by the current time interval as in v = a/t.

What am I doing wrong here? I'm still kind of confused so please be patient!
 
phil ess said:
Ok here's the equation for acceleration of the falling ball, if down is positive:

a = g − CρA/2m * v2

So I'm in Excel now, and I'm supposed to use deltaT = 0.1s.

So what I did was this (note the calculation bar to see what i did):
Untitled-1.jpg


Now from my Vterm calculation I expect this to approach v = 8.9 m/s (thanks StoveBolt), but it breaks down badly for some reason. the way its calculated is as follows:

CρA/2m = 0.125206285 <-- that's where that comes from

Then multiplying that by v2 from the previous interval

Subtract this from g = 9.81

multiply by the current time interval as in v = a/t.

What am I doing wrong here? I'm still kind of confused so please be patient!

Try adjusting the spreadsheet using this formula:
Vfinal = Vinitial + a * t

Remember, for each step you are only multiplying by the delta t of 0.1 seconds.

Let me know if you still have any trouble after this.
 
I see what you're doing, but regular kinematics equations only work with constant acceleration, and here that's not the case. Acceleration also changes depending on the speed, which is where I am having trouble. I'm not sure which speed values to use for each specific time interval.
 
You assume that over the small time (0.1s) the acceleration is constant. So you can calculate the change in speed over the time interval.
 
phil ess said:
I see what you're doing, but regular kinematics equations only work with constant acceleration, and here that's not the case. Acceleration also changes depending on the speed, which is where I am having trouble. I'm not sure which speed values to use for each specific time interval.

Regular kinematics equations do work in this case. This is where Euler's Method comes in handy - you assume constant acceleration over each small time interval. This allows you to use a close estimation of the velocity to compensate for the changing acceleration due to increases in drag. You may not get the exact answer, but you will come very close.

Allow me to clarify:

Using the formula:
Vfinal = Vinitial + a*t,

substituting
a = (9.81 - 0.125206*Vinitial2)
t = 0.1
Vinitial = Vfinal from each preceding step.

For each time interval,

Vfinal = Vinitial + (9.81 - 0.125206*Vinitial2)*0.1



Try it out and see.:wink:
 

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