# Year-Long Physics Project

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• November 7th 2006, 08:29 PM
CaptainBlack
Quote:

Originally Posted by Quick
hmmm, I got an interesting suggestion from my teacher on how to do the project without using a catapult or just plain dropping it.

He suggested attaching the ball to a string (like a pendelum) and releasing it at a certain degree and then find the air resistance from how long it takes to get to a certain degree above the base (because if I waited for the ball to stop I would run into human error) and then record the height and time.

The pendulum will be opening up a whole new can of worms.

1. You will have to worry about non-simple harmonic motion.
2. Damping other than air resistence that you don't know the nature of.
3. Its a low speed device for long pendulums where air resistance will be
small, short pendulums can be higher speed, but will leave you with timing problems.

and so on.

Quote:

How would I calculate air resistance? I don't need you to type the whole thing if you don't want, but I do need help seeing the general idea.
Build a mathematical model (or models) of the system with free
parameters representing the unknowns. Use model to predict the
result of experiments, vary the parameters until model and
experiment agree.

Don't worry about being able to "solve" the model all that is necessary
is that you can simulate the behaviour of the experiment to obtain the
results that you could have measured of off the experiment itself.

Quote:

My solution: All I could think of was that the highest point the ball reached on each side per swing and the time it takes to get there, along with the times to the base, are the things I need to record.
Video it using a digital camera of some sort, extract the frames onto
your computer and take the measurements off of the frames, using the
frame count as a timer.

You will need to calibrate the frame rate rather than rely on its nominal
value. Do this by filming a digital clock with a second hand and say counting
the number of frames for it to sweep through 60 seconds.

I would still go for filming a dropped projectile myself.

RonL
• November 8th 2006, 04:00 AM
topsquark
An additional possible problem with your pendulum experiment is that, to the level of accuracy you need, it might not be a "simple pendulum" anymore. (That is to say, you may need to consider that the string is also oscillating back and forth.) The change in the equation isn't that bad to make, but learning the theory of such will take a bit more effort.

-Dan
• November 8th 2006, 11:14 AM
Quick
Using the multiquote feature.
Quote:

Originally Posted by ThePerfectHacker
I hope you starts getting afraid of you.

Me too :rolleyes:
Quote:

Originally Posted by CaptainBlack
The pendulum will be opening up a whole new can of worms.

1. You will have to worry about non-simple harmonic motion.

:confused:
Quote:

2. Damping other than air resistence that you don't know the nature of.
such as friction on the string?
Quote:

3. Its a low speed device for long pendulums where air resistance will be
small, short pendulums can be higher speed, but will leave you with timing problems.

and so on.
Another problem...

Quote:

Build a mathematical model (or models) of the system with free
parameters representing the unknowns. Use model to predict the
result of experiments, vary the parameters until model and
experiment agree.

Don't worry about being able to "solve" the model all that is necessary
is that you can simulate the behaviour of the experiment to obtain the
results that you could have measured of off the experiment itself.
If only I knew how :o but the problem is that the main thing I have to do is find the "coefficient of drag" (so that I can compare my equation to the established equation) which is almost always determined experimentally.

Quote:

Video it using a digital camera of some sort, extract the frames onto
your computer and take the measurements off of the frames, using the
frame count as a timer.

You will need to calibrate the frame rate rather than rely on its nominal
value. Do this by filming a digital clock with a second hand and say counting
the number of frames for it to sweep through 60 seconds.
I was planning to film it and have a timer in the picture, which seems to be a little easier.

Quote:

I would still go for filming a dropped projectile myself.

RonL
I'm using everyones input to decide which one I should do. As a year long science fair project, the teacher thinks that that experiment is too easy and will make it hard to get a good grade. And the problem with the trebuchet is that I will need a reasonable sized one and my adviser at the school science seminar thinks it might not get approved by the science fair people, whoever they are :(.
• November 8th 2006, 01:07 PM
CaptainBlack
Quote:

Originally Posted by Quick

I was planning to film it and have a timer in the picture, which seems to be a little easier.

Filming on film?

I have on this machine the tools needed to take a avi file apart to get at
the individual frames, it is a relatively simple process.

Filming the experiment and the timer simultaneously seems to me a bit
complicated as you will need to have some complex optical arrangement
to get a close to timer and a distant experiment in focus and properly exposed
on the same frame.

RonL
• November 8th 2006, 01:22 PM
CaptainBlack
Quote:

Originally Posted by Quick

If only I knew how :o but the problem is that the main thing I have to do is find the "coefficient of drag" (so that I can compare my equation to the established equation) which is almost always determined experimentally.

For the dropped projectile you will have an equation of motion something like:

$
\ddot{x}=-g+C \frac{A}{m} |\dot{x}|^k
$

Where $C$ absorbs $C_D$ and $\rho$ and some other numerical constants
(because we have an unknown power of $\dot{x}$ the normal coeff of drag
is not relevant here)

There is already a gain over the free projectile experiment as you will now have reduced
the problem to a 1-D case.

Now such a model cannot be integrated in closed form, but it can be integrated
numerically to give a height against time curve which you will then need to adjust
$C$ and $k$ to get it to fit the experimental results.

With luck you should find $k \sim 2$ as most of the time you will
be in the high Reynolds number regime, then you can then convert from $C$ to $C_D$.
• November 8th 2006, 06:40 PM
Quick
Quote:

Originally Posted by CaptainBlack
For the dropped projectile you will have an equation of motion something like:

$
\ddot{x}=-g+C \frac{A}{m} |\dot{x}|^k
$

Where $C$ absorbs $C_D$ and $\rho$ and some other numerical constants
(because we have an unknown power of $\dot{x}$ the normal coeff of drag
is not relevant here)

There is already a gain over the free projectile experiment as you will now have reduced
the problem to a 1-D case.

Now such a model cannot be integrated in closed form, but it can be integrated
numerically to give a height against time curve which you will then need to adjust
$C$ and $k$ to get it to fit the experimental results.

With luck you should find $k \sim 2$ as most of the time you will
be in the high Reynolds number regime, then you can then convert from $C$ to $C_D$.

The point of the experiment is to create an equation from scratch to model Drag of a sphere through the density of the ball, the air density, and the size of the ball. Not to use previously made mathematics :(
• November 8th 2006, 09:07 PM
CaptainBlack
Quote:

Originally Posted by Quick
The point of the experiment is to create an equation from scratch to model Drag of a sphere through the density of the ball, the air density, and the size of the ball. Not to use previously made mathematics :(

You will still be better of with the 1-D dynamics you will get from the dropped
projectile though.

RonL
• November 9th 2006, 03:27 AM
topsquark
Quote:

Originally Posted by Quick
The point of the experiment is to create an equation from scratch to model Drag of a sphere through the density of the ball, the air density, and the size of the ball. Not to use previously made mathematics :(

Just a thought. You might get better results by dropping your ball down a column of water as opposed to air. The equations are essentially the same (though you now have to take into account a constant bouyant force) and the drag is much larger, making your time measurements more accurate. (Just make sure the radius of the column is significantly larger than the radius of the balls so "edge" effects don't come in. You don't want to have to think about vorticity, trust me!)

It would have a much larger Reynold's number so the equations would be significantly messier than the air drop, but if you don't need to model the equations this won't be a problem.

-Dan
• November 9th 2006, 04:29 AM
CaptainBlack
Quote:

Originally Posted by topsquark
Just a thought. You might get better results by dropping your ball down a column of water as opposed to air. The equations are essentially the same (though you now have to take into account a constant bouyant force) and the drag is much larger, making your time measurements more accurate. (Just make sure the radius of the column is significantly larger than the radius of the balls so "edge" effects don't come in. You don't want to have to think about vorticity, trust me!)

It would have a much larger Reynold's number so the equations would be significantly messier than the air drop, but if you don't need to model the equations this won't be a problem.

-Dan

Except that you are heading in the direction of low Reynolds numbers and
so the equations may well change form. Though the differences between air and
water may be worth reporting.

RonL
• February 14th 2007, 04:30 PM
Quick
This is the last thing I expected to have trouble on :( Density.

Anyway, I've got everything done except now I need to plot the graphs according to density, and it just so happens that I feel like I'm getting ridiculous numbers.

Example:

I have a sphere with a circumference of 39 centimeters, which I then changed to 0.39m

Therefore it's diameter is $\frac{0.39m}{\pi}$ (I know latex doesn't work, but you guys can understand it.)

Therefore it's radius is $\frac{0.39m}{2\pi}$

Which is about: $0.0621$ (My actual work is done to around 10 decimal places, just so everything works)

I cube that, then multiply by pi, and finally multiply by 4/3 to get the volume, which comes out to : $V=0.001$

Now I happen to have a ball that size with a weight of 687g, or 0.687kg

Then : $Density (\rho?)=\frac{0.687}{0.001}=687$

I don't know why, but that seems like too large a number, is it? (I'm not good with SI units)
• February 14th 2007, 08:08 PM
CaptainBlack
Quote:

Originally Posted by Quick
This is the last thing I expected to have trouble on :( Density.

Anyway, I've got everything done except now I need to plot the graphs according to density, and it just so happens that I feel like I'm getting ridiculous numbers.

Example:

I have a sphere with a circumference of 39 centimeters, which I then changed to 0.39m

Therefore it's diameter is $\frac{0.39m}{\pi}$ (I know latex doesn't work, but you guys can understand it.)

Therefore it's radius is $\frac{0.39m}{2\pi}$

Which is about: $0.0621$ (My actual work is done to around 10 decimal places, just so everything works)

I cube that, then multiply by pi, and finally multiply by 4/3 to get the volume, which comes out to : $V=0.001$

Now I happen to have a ball that size with a weight of 687g, or 0.687kg

Then : $Density (\rho?)=\frac{0.687}{0.001}=687$

I don't know why, but that seems like too large a number, is it? (I'm not good with SI units)

Look at the units of the density, its kg/m^3. In these units the density of
water is 1000 kg/m^3, so the relative density is 0.687. It will float in water.

RonL
• February 25th 2007, 06:43 AM
Quick
Results
I found out that I did not do the project correctly, and ended up finding that the force of air resistance is proportional to the force applied.

I didn't see that in my research, what would it be called?
• February 25th 2007, 07:00 AM
CaptainBlack
Quote:

Originally Posted by Quick
I found out that I did not do the project correctly, and ended up finding that the force of air resistance is proportional to the force applied.

I didn't see that in my research, what would it be called?

At this point I must say that I am confused:confused: .

I can't find a concise description of the project/experiment, but I don't
recall seeing an applied force mentioned.

You could post your set up and measurements here for us to look at

RonL
• February 25th 2007, 09:17 AM
Quick
Quote:

Originally Posted by CaptainBlack
At this point I must say that I am confused:confused: .

I can't find a concise description of the project/experiment, but I don't
recall seeing an applied force mentioned.

You could post your set up and measurements here for us to look at

RonL

I ended up dropping them. Gravity is the applied force.

Most of the time drag is measured without force, that's why I haven't seen such a thing in my research...
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