Year-Long Physics Project

**Quick** · September 20th 2006, 04:56 PM

ARGGHHHH!!!!! For the past two weeks I have been trying to come up for idea's for my freshman year-long physics project, and all I've come up with is some golf thing!!!

Any suggestions to testable topics?

**topsquark** · September 20th 2006, 05:15 PM

Originally Posted by Quick

ARGGHHHH!!!!! For the past two weeks I have been trying to come up for idea's for my freshman year-long physics project, and all I've come up with is some golf thing!!!

Any suggestions to testable topics?

Have you thought about something with wave motion? I don't know what kind of budget you have, but a wave tank is easy to construct.

I once helped a student do an experiment with the center of mass motion of a dancer as she did some ballet moves (leaping, whatever they call that.) Figuring out the location of the CM of a human body isn't the easiest thing to do, so this one's a challenge.

Golf is good, you can also do air resistance using a tennis ball cannon, or that thingy they use in baseball to fire the baseballs at the batter.

That's all I can get off the top of my head. I'll try to think of more.

-Dan

**Quick** · September 20th 2006, 05:21 PM

Originally Posted by topsquark

Golf is good, you can also do air resistance using a tennis ball cannon, or that thingy they use in baseball to fire the baseballs at the batter.

What would I be testing? It reminds me of my plan to drop balls of different density from a point and see how air resistance affects the balls speed according to density (remember my acceleration question?)

**topsquark** · September 20th 2006, 05:47 PM

Originally Posted by Quick

What would I be testing? It reminds me of my plan to drop balls of different density from a point and see how air resistance affects the balls speed according to density (remember my acceleration question?)

You could do that. What I had in mind was finding a way to predict where the ball is going to land. Basically air resistance formulas take two forms, linear and quadratic. The problem is that in reality as the speed varies the equations of motion can transform from one type of resistance to the other. (And nothing says that the resistance is a simple function of speed anyway.) So neither model will reliably predict where the ball is going to land. It takes a lot of data to figure out how at what point the type of resistance changes. I don't know if this is too difficult for you to be doing or not. It depends on how much your teacher knows and is willing to help out with.

-Dan

**Quick** · September 25th 2006, 02:35 PM

Originally Posted by Quick

What would I be testing? It reminds me of my plan to drop balls of different density from a point and see how air resistance affects the balls speed according to density (remember my acceleration question?)

that will be my project.

Now my question: If it's falling with force 9.8m/s^2 and air is pushing upward at 2m/s^2, will the net force be 7.8m/s^2 (I've forgotten most of my physics)

**topsquark** · September 25th 2006, 03:58 PM

Originally Posted by Quick

that will be my project.

Now my question: If it's falling with force 9.8m/s^2 and air is pushing upward at 2m/s^2, will the net force be 7.8m/s^2 (I've forgotten most of my physics)

NOOOOOOOOOOOO!!! (Falls to the ground, hands clutching his throat gasping for breath and dies in great agony.)

9.8 m/s^2 is an acceleration, NOT a force!! So if the air resistance is providing an upward acceleration of 2.0 m/s^2 then the overall acceleration will be 7.8 m/s^2 downward.

And yes, forces add (vectorally) so you can do the same thing with forces.

-Dan

**Quick** · September 25th 2006, 04:12 PM

so let's say that I drop a ball from 100 meters up and it hits the ground in 5 seconds (theoretical). Now this is me trying to apply what I know, please correct me if I'm wrong (which I probably am)

So we have: $y=y_0+v_0t+\left(\frac{1}{2}\right)at^2$

substitute: $0=100+0(5)+\left(\frac{1}{2}\right)a(5)^2$

then: $0=100+\left(\frac{1}{2}\right)25a$

thus: $-100=\left(\frac{25}{2}\right)a$

therefore: $\frac{-200}{25}=a=-4$

thus the acceleration is -4m/s^2

the acceleration without air resistance should have been -9.8m/s^2

thus the acceleration of the air resistance is 5.8m/s^2

**topsquark** · September 25th 2006, 04:46 PM

Originally Posted by Quick

so let's say that I drop a ball from 100 meters up and it hits the ground in 5 seconds (theoretical). Now this is me trying to apply what I know, please correct me if I'm wrong (which I probably am)

So we have: $y=y_0+v_0t+(1/2)at^2$

substitute: $0=100+0(5)+(1/2)a(5)^2$

then: $0=100+(1/2)25a$

thus: $-100=(25/2)a$

therefore: $-200/25=a=-4$

thus the acceleration is -4m/s^2

the acceleration without air resistance should have been -9.8m/s^2

thus the acceleration of the air resistance is 5.8m/s^2

Correct! Though you should probably say that "the acceleration DUE to the air resistance is 5.8 m/s^2." The term "air resistance" usually denotes a force. (Yes, I'm a prig when it comes to terminology.

)

-Dan

**Quick** · October 9th 2006, 04:16 PM

Originally Posted by Quick

so let's say that I drop a ball from 100 meters up and it hits the ground in 5 seconds (theoretical). Now this is me trying to apply what I know, please correct me if I'm wrong (which I probably am)

So we have: $y=y_0+v_0t+\left(\frac{1}{2}\right)at^2$

substitute: $0=100+0(5)+\left(\frac{1}{2}\right)a(5)^2$

then: $0=100+\left(\frac{1}{2}\right)25a$

thus: $-100=\left(\frac{25}{2}\right)a$

therefore: $\frac{-200}{25}=a=-4$

thus the acceleration is -4m/s^2

the acceleration without air resistance should have been -9.8m/s^2

thus the acceleration of the air resistance is 5.8m/s^2

I thought Latex wasn't yet doing anything

**Quick** · October 29th 2006, 03:43 AM

Anyway, could you help me with my computations?

First off I plan to record how long it takes to hit the ground to calculate vertical air resistance, let's say it takes 8 seconds and it starts off going 4m/s upwards (which is negative for this instance) : $y=y_0+v_0t+\left(\frac{1}{2}\right)at^2$

Substitute: $0=2-4(8)+\left(\frac{1}{2}\right)5^2a$

Simplify: $0=2-32+\left(\frac{1}{2}\right)25a$

Simplify: $0=\left(\frac{25}{2}\right)a-30$

Add 30 to both sides: $30=\left(\frac{25}{2}\right)a$

divide both sides by 25/2: $\frac{60}{25}=a$

Then: $a=2.4$

(keep in mind I made the beginning numbers up)

Thus the average acceleration of air resistance is: $9.8-2.4=7.4$

Now to find the air resistance horizontally, the ball went 20m we'll say and was going 8m/s to begin with...

Start off: $x=x_0+v_0t+\left(\frac{1}{2}\right)at^2$

Solve for a: $\frac{2(\overbrace{x-x_0}^{\Delta x}-v_0t)}{t^2}=a$

Substitute: $\frac{2(20-0-8(8))}{8^2}=a$

then: $\frac{2(20-64)}{64}=a$

Therefore: $\frac{2(-44)}{64}=a$

Thus: $\frac{-88}{64}=a$

Finally the acceleration due to air resistance (for the horizontal movement) is: $-1.375=a$

Is that all correct?

**topsquark** · October 29th 2006, 04:47 AM

Originally Posted by Quick

Anyway, could you help me with my computations?

First off I plan to record how long it takes to hit the ground to calculate vertical air resistance, let's say it takes 8 seconds and it starts off going 4m/s upwards (which is negative for this instance) : $y=y_0+v_0t+\left(\frac{1}{2}\right)at^2$

Substitute: $0=2-4(8)+\left(\frac{1}{2}\right)5^2a$

Simplify: $0=2-32+\left(\frac{1}{2}\right)25a$

Simplify: $0=\left(\frac{25}{2}\right)a-30$

Add 30 to both sides: $30=\left(\frac{25}{2}\right)a$

divide both sides by 25/2: $\frac{60}{25}=a$

Then: $a=2.4$

(keep in mind I made the beginning numbers up)

Thus the average acceleration of air resistance is: $9.8-2.4=7.4$

Now to find the air resistance horizontally, the ball went 20m we'll say and was going 8m/s to begin with...

Start off: $x=x_0+v_0t+\left(\frac{1}{2}\right)at^2$

Solve for a: $\frac{2(\overbrace{x-x_0}^{\Delta x}-v_0t)}{t^2}=a$

Substitute: $\frac{2(20-0-8(8))}{8^2}=a$

then: $\frac{2(20-64)}{64}=a$

Therefore: $\frac{2(-44)}{64}=a$

Thus: $\frac{-88}{64}=a$

Finally the acceleration due to air resistance (for the horizontal movement) is: $-1.375=a$

Is that all correct?

Unfortunately, no, this isn't correct. The problem is that the air resistance will always be against the direction of motion (ie. opposite the direction of the velocity.) So as the object moves upward, the acceleration due to air resistance is downward. As the object falls, the acceleration due to air resistance is upward.

In addition, most air resistance models have the resistance proportional to either the speed or the square of the speed of the object, so the acceleration due to air resistance will not be constant. (Thus you can't use the equation you are using.)

If you are going to assume a constant acceleration due to air resistance (you could perhaps call it an "average acceleration") what you would need to do is assume a downward constant acceleration as the object rises, then assume the same acceleration as the object falls, but now in an upward direction.

So say we have your object being projected upward at 4 m/s from a height of -2 m above the ground. (Assume +x downward.) To find the time it takes to get to the maximum height:
$v^2 = v_0^2 + 2(g + a)(x - x_0)$ where a is your resistive average acceleration.

$0^2 = 4^2 + 2(9.8 + a)(x + 2)$

x will be the max. height. (NOTE: Given your coordinate system and choice of origin, the max height will be negative!)

Now:
$x = x_0 + v_0t + (1/2)(g + a)t^2$
gives you the time to get to max height.

Now the object falls from rest.

$x = x_0 + v_0t + (1/2)(g - a)t^2$
where x0 is now your max height, x is at ground level so x = 0 and v0 = 0 since it's falling from rest. (Remember that a is upward for this portion of the flight path since the object is falling.)

Add the two times and this gives you the time it takes to reach the ground after being projected upward at 2 m/s. If you measure this time your time equation will give you the value for the average acceleration over the path.

-Dan

**CaptainBlack** · October 29th 2006, 06:09 AM

Originally Posted by Quick

Anyway, could you help me with my computations?

First off I plan to record how long it takes to hit the ground to calculate vertical air resistance, let's say it takes 8 seconds and it starts off going 4m/s upwards (which is negative for this instance) : $y=y_0+v_0t+\left(\frac{1}{2}\right)at^2$

How are you proposing to collect the data.

If you have not selected a method might I suggest that you use the
movie function on a digital camera to video the experiment. Then
extract the frames of the movie and take measurements off the frames.

Typical frame rate is 15-30 frames per second, so you should get much
more data from the experiment than just the time to hit the ground.

A major problem that you would need to address of course is calibrating
the measurements. you might need a measuring rod in the movie at
the same distance from the camera as the dropped object. (main problem
might be that the image scale changes across the frames).

(Don't know why I am telling you this - It was to be the main idea in a
paper for the Mathematical Gazette on data analysis I was planning to
write)

RonL

**Quick** · November 4th 2006, 04:20 AM

1 last question (for now anyway)

Since I'm launching the balls out of a trebuchet, the starting velocity will be almost the same for each instance (all you have to do is change the counterweight), but the problem is that they aren't the same.

So my question: Could I just divide the average acceleration by the starting velocity to see the change between the different densities? Is there anything I could do to solve this problem?

EDIT: I just realized that since the starting velocities will be very close then I can chop off the decimals that don't match (like using two decimal places instead of four) but I would very much not like to do that because it gets rid of some sig figs, so if there is a way to avoid that that would be nice

**topsquark** · November 4th 2006, 04:46 AM

Originally Posted by Quick

1 last question (for now anyway)

Since I'm launching the balls out of a trebuchet, the starting velocity will be almost the same for each instance (all you have to do is change the counterweight), but the problem is that they aren't the same.

So my question: Could I just divide the average acceleration by the starting velocity to see the change between the different densities? Is there anything I could do to solve this problem?

EDIT: I just realized that since the starting velocities will be very close then I can chop off the decimals that don't match (like using two decimal places instead of four) but I would very much not like to do that because it gets rid of some sig figs, so if there is a way to avoid that that would be nice

A question and a comment:
1) Different densities? Are you talking about the density of the balls?
2) Acceleration divided by velocity gives units of 1/s, not density!

As long as the density of the balls are fairly close to one another you should be able to continue without any difficulties. What I would do is make a chart of the densities of the balls, then use the average density in your equations. You can list the average density and its standard error in your overall Data Chart and mention the differing densities as a possible experimental error. My rule of thumb is that if the standard error is less than 5% or so of the average value that the average is fairly good estimate of a constant.

-Dan

**Quick** · November 4th 2006, 04:55 AM

Originally Posted by topsquark

A question and a comment:
1) Different densities? Are you talking about the density of the balls?
2) Acceleration divided by velocity gives units of 1/s, not density!

As long as the density of the balls are fairly close to one another you should be able to continue without any difficulties. What I would do is make a chart of the densities of the balls, then use the average density in your equations. You can list the average density and its standard error in your overall Data Chart and mention the differing densities as a possible experimental error. My rule of thumb is that if the standard error is less than 5% or so of the average value that the average is fairly good estimate of a constant.

-Dan

I forgot I didn't explain my project very well, here is my experimental design:

Defining Question: By how much will a ball's speed change when its size, weight, and how hot it is outside changes?

Hypotheses: The size of the ball will be the greatest factor in deciding the average acceleration of air resistance. The second important factor is the density of the ball. Finally, the least affecting factor will be the temperature of the air.

Independent Variables:
The Density of the ball: 5-10 levels per ball size, control will be unaltered density (I won't fill the control with weight)

The Outside Temperature: 4-6 levels, control to be decided

The Size of the Ball: 4-5 levels, Control will be the middle sized balls.

Dependent Variable:
The average acceleration of air resistance

Number of Trials:
4 trials per density

Constants:
The Starting Velocity of the ball.
The Angle the ball is released.
The Ground Level
The Starting Location

Procedure:
1. Set up the trebuchet in a level area
2. Place the first ball in the trebuchet
3. Launch, makeing sure to measure the starting velocity of the ball
4. Time how long it takes for the ball to hit the ground
5. Measure the distance between the landing point and the point of release
6. Discover the average acceleration for that time (using equation a=2(d/t)-2v, where d is the change in distance in meters, v is the starting velocity in meters per second, t is the time in seconds, and a is the average acceleration of the ball)
7. Repeat until you have 4 tests of that ball
8. Use different density balls while repeating steps 3-7 until all densities have been used.
9. Change the ball size and repeat 2-8
10. Repeat step 9 until all ball sizes have been used
11. Repeat steps 1-10 when the temperature changes
12. Repeat step 11 until sufficient data is collected

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