Lab 7: Modeling Friction

Fortunately, we have not had to simulate class tension or class friction to model them. Instead, as physicists, we let small objects do the work for us and we stand above the fray. The only difference with this lab over the others is that we'll be doing 5 parts of relatively similar things. Tedium is a good way to model friction, indeed.

 

SECTION 1: Static Friction of Table

In this first part, we will fill a cup partially with water, and allow it to dangle over the side of the table with a string attached. The string will go over a pulley, and on the other side: blocks. We expect that the normal force between the blocks and the table will be the weight of the blocks. We also expect that the maximum static friction force will be the weight of the cup as it starts to gain enough leverage to move the blocks. Therefore, we expect a linear relationship as we start to add more blocks to the amount of water necessary to move them. And since the only thing standing between friction force and weight is the coefficient of friction, that's what we'll find.

The process involves slowly and carefully filling up that Styrofoam cup with a dropper, with a hand under the cup, and observing when the block starts to move. Ideally, the block moves just as it overcomes static friction. In practice, there are many flaws with this experiment. For one, the table and the blocks aren't of uniform smoothness, making it so that the placement of the blocks affect the results. Secondly, the angle of the blocks with the string changes the amount of force required to pull it. We also suspect that the table might be slightly slanted, perhaps with uneven legs. We found that pressing the blocks against the table slightly, then letting go, enables it to withstand far greater tension. These problems will influence our experimental results later on. We had to exercise our own experimenter's discretion in navigating these waters, and we find that our weeks of experience had not paid off. We were still raining wet behind the ears.

Nevertheless, we proceed to weight the cups.

And then weigh the blocks. And add more blocks, until we have 4.

Here's our data:

We launch LoggerPro to graph the maximum static force over normal force:

And since:

F_static = µ_s N
µ_s = F_static / N

The slope of the graph must be µ_s = 0.2188, as shown by the linear fit.

 

SECTION 2: Kinetic Friction of Table

If in the first part, we found static friction, it only makes sense that we now find kinetic friction. The process is similar. We hook up Logger Pro to a force sensor, which detects the weight that tugs on its little hook. We had a little trouble calibrating it, so that it zeroes the default atmospheric weight in both vertical and horizontal orientation. We use the same blocks used in Section 1, this time tied to the force sensor instead of the cup. As we pull on the block, LoggerPro should record the input from the force sensor, and we get a neat graph with force over time. The goal here is to find a relationship between friction force and mass, so we try to introduce little variance by pulling as steadily as possible. It is impossible to be perfectly steady, but that doesn't seem to affect our experiment much. Once again, we do this for 1-4 blocks.

We use statistical analysis in LoggerPro to calculate the mean force over a period of time. Since the blocks were pulled steadily, force should have been relatively constant. Then, we graph the force of the pull over the normal force of the blocks again. Once again, if:

 

F_tension = µ_k N
µ_k = F_tension / N

Thus, the coefficient of kinetic friction should be the slope of this graph:

Again, with a linear fit, we find that µ_k = 0.2395, as shown by the linear fit. Here, an close observation should reveal a fatal flaw in this experiment: The kinetic friction is larger than static friction, when it should really be a fraction of it. This result seems consistent across many groups, yet we don't know the exact cause.

 

SECTION 3: Static Friction of Track

The goal is slowly raise one end of the track, with a block on it, until gravity overcomes static friction, causing the block to slide. If we know the angle θ and the mass m of the block, then it should be possible to calculate friction. As you can see, the track is supported with a metal rod clamped to another one. The clamp can be loosened, allowing the support rod to be raised or lowered with ease.

Finally! By the way, this experiment suffers many of the same problems as the prior: uneven surfaces, block angle affects movement, air resistance might even play a part, vibrations of the block slipping could cause track to slip... We measured the height of our final track configuration to be approximately (22.6 ± 2) cm, and horizontal distance to be (98.0 ± 2) cm. Using the following formula, we could find the track angle, as well as a propagated uncertainty.

θ = tan^-1( y / x )
θ = tan^-1( 22.6 / 98.0 )
θ = 13.0º

dθ = | (∂θ/∂y) | dy + | (∂θ/∂x) | dx
dθ = | [1 / (1+y²)] * (1 / x) | 2 + | [1 / (1 + x²)] * y | 2
dθ = 2 / [x (1 + y²)] + 2y / (1 + x²)
dθ = 2 / [98 (1 + 22.6²)] + 2 (22.6) / (1 + 98²)
dθ = 0.00475º

We found our θ to be 13º ± 0.00475º, but it could even be give or take a few, since consecutive runs seem to give different results. As before, sometimes the block would "stick" until much higher angles, before sliding with gusto. This figure accounts for measurement error, but not all the other errors, which would make the problem too complex. The purpose of this lab, after all, is mainly to model frictional forces.

We also found the mass m of the block to be (0.160 ± 0.001) kg. That should be enough to solve for static friction, since the block has no acceleration before it overcomes static friction:

m g sinθ = µ_s N

m g sinθ = µ_s m g cosθ
µ_s = tanθ

µ_s = tan13º
µ_s = 0.231

dµ_s = | (∂µ_s/∂θ) | dθ
dµ_s = sec²(θ) dθ
dµ_s = sec²(13º) 0.0000829 rad
 dµ_s = 0.0000873

Coefficient of static friction is µ_s = 0.231 ± 8.73 * 10^-5, therefore, at least, the possible measurement error should play only a negligible role in the problem.

 

SECTION 4: Kinetic Friction of Track

We found that if we tilt the track high enough, gravity should over come static friction. So if we tilt the track higher, we should be able to find the kinetic coefficient if we know the acceleration of the block. Well, that's what the motion sensor's for!

We hook up LoggerPro with a motion sensor, which sends an infra-red signal to an object directly in front of it, and counts the length of time it takes to bounce back. As such, the motion sensor would not work if the object is too close, if the signal reflects before the sensor is ready to receive it, or too far, since the signal will be too weak. Luckily, if we place the motion sensor at one end of the track, then it works fine if the object runs from the middle.

So first the track is tilted up to an arbitrary angle higher than in Section 3. Then we use the cell phone app Clinometer, which allows us to tilt the phone on one side to measure the angle of the track. Since we have already, in the previous section, and in the previous lab, figured out the side of the phone we need to use to get an accurate measurement, we can just trust what the app outputs. A few tests on different parts of the track did give slightly different results, so we say that the angle is θ = 20.5º ± 0.5º. The mass m is, once again, (0.160 ± 0.001) kg, as the block hasn't changed.

We put the block somewhere half-way down the track, and capture the movement with the LoggerPro via the motion sensor. Here's the resulting graph of velocity vs time, and acceleration vs time:

Only a portion of this graph is relevant. We know that a falling object, subjected to theoretically constant friction and constant gravity, should accelerate linearly. The first part of the graph is either due to the object being let go, or the motion sensor software not working properly as it is turned on. The last half of the graph is due to the block hitting the bottom of the track, thus velocity reduces to 0 m/s. What we are interested in is the middle section, where velocity is linear and acceleration is constant, relatively speaking. Theoretically, if we find the slope of velocity during that section of time, and the mean average of acceleration, it should be the same. However, the real world isn't so clean most of the time: Our velocity slope turned out to be 0.8574 m/s², and our mean acceleration turned out to be 0.8527 m/s². So we split the difference and call it (0.855 ± 0.0028) m/s². Is it right to be so practical? The alternative is to do multiple runs and take a standard deviation, and that would have been beyond the time frame of this exercise.

Now we have all the variables we need. Since the acceleration must be the same as the difference between the forces of gravity opposite kinetic friction, the equation should be:

F_kinetic = m g sinθ - µ_k m g cosθ = m a

Once again, canceling out the mass, we get:

g sinθ - µ_k g cosθ = a
µ_k g cosθ = g sinθ - a
µ_k = (g sinθ - a) / g cosθ
µ_k = tanθ - [a / (g cosθ)]

Let's get this over with!

µ_k = tan20.5º - [0.855 m/s² / (9.8 m/s² cos20.5º)]

µ_k =  0.281

dµ_k = | (∂µ_k/∂a) | da + | (∂µ_k/∂θ) | dθ

dµ_k = | 1 / (g cosθ) | da + | sec²θ - (a / g) secθ tanθ | dθ

dµ_k = 0.0028 m/s² / (9.8 m/s² cos20.5º) + 0.00873 sec(20.5º)[1 - (0.855 m/s² / 9.8 m/s²) tan20.5º]
 dµ_k = 0.000305 + 0.00964
dµ_k = 0.01

Coefficient of kinetic friction is µ_k = 0.281 ± 0.01. Looks like the difference in acceleration had a minor effect compared to our careless measurement of the angle. Note, still, that our coefficient of kinetic friction is larger than our coefficient of static friction--it should be smaller! This could be for all the reasons above, but ultimately, we're not sure why this is.

 

SECTION 5: Kinetic Friction over Inclined Slope

This last section is similar to Section 4, except we're going to modify the experiment a bit. Instead of allowing the block to slide down the track, we're going to combine this with the first section and tie a weight to it over a pulley, so that the block slides upwards. To do this, we must move the motion sensor to the bottom of the incline, since the block will be moving to the top.

If the experiment is successful, then the coefficient of kinetic friction should be the same, or similar, to the previous part. However, given the large errors previously discovered, this may not be the case. As with before, our block is the same, so mass m is (0.160 ± 0.001) kg. The angle θ was measured again, and we found that the track has slipped a bit to 21.5º ± 0.5º. Oops! On the other end of the string, we used weights of known mass, which we'll designate as m₂, of 0.15 kg. It took a couple tries to get the timing down, since the block was getting pulled too fast, but pretty soon we got our graph, once again, of velocity vs time, and acceleration vs time:

As before, our velocity slope (0.5312 m/s²) is a bit different than our acceleration mean (0.5450 m/s²). So, once again, we split the difference and get a = (0.538 + 0.0070) m/s². The difference, this time, is that, with multiple masses, the situation is getting a bit hard to picture, so we'll be using free body diagrams.

This is a model of the experiment (made in Photoshop) showing all the forces. The pulley changes direction of the forces, so, to simplify our free body diagram, we could imagine the string straightening out, with the weights (wg in the model) pointing outward parallel to the slope. In real life, gravity points downwards, but insofar that the weights w interact with the system, such that it pulls the string in the direction before the pulley, we are justified in doing this. Next, we notice that there are 3 forces at an angle, and it would be much easier to set up the equation if we have less angular forces to break up into x and y components. Thus, we tilt the model, using different axes, such that f_k, T, and wg are all horizontal, and N is vertical, and mg is the only force at an angle. Once again, although gravity really points downwards, what matters is that the forces are preserved relative to each other, and not their absolute orientation. If we do this, our free body diagram looks like this:

We want to find µ_k, so we are only interested in the horizontal component:

F = (m + w) a = wg - µ_k mg cosθ - mg sinθ
µ_k = [(wg - mg sinθ - (m + w) a] / [mg cosθ]
µ_k = [(0.15 kg - 0.16 kg sin(21.5º))(9.8 m/s²) - (0.31 kg) 0.538 m/s²] / [0.16 kg (9.8 m/s²) cos(21.5º)]
µ_k = [0.8953 N - 0.1668 N] / 1.4589 N
µ_k = 0.7285 N / 1.4589 N
µ_k = 0.5

The result here is much different than the result in the previous part, despite measuring the kinetic coefficient of the same surface. We can speculate that the pulley might have its own friction. By eye, the block in this part seems to move faster than the previous part, but that is admittedly unreliable.

dµ_k = | (∂µ_k/∂a) | da + | (∂µ_k/∂θ) | dθ + | (∂µ_k/∂m) | dm
dµ_k = | (m + w) / (mg cosθ) | da + | [(wg - ma - wa) / mg] secθ tanθ + sec²θ | dθ + | (w / m²) secθ - wa / m²g cosθ | dm
dµ_k = 0.2125 s²/m (0.0070 m/s²) + 1.507 (0.05) + 5.952 (1/kg) (0.001 kg)
 dµ_k = 0.0014874 + 0.07535 + 0.005952
dµ_k = 0.0828

So it turns out that the track slipping in between experiments made quite an effect! Despite our unreasonable results, we have found that it is possible to model friction, and have identified some areas that deserve closer attention if we want more accurate results.

Lab 6: Propagated Uncertainty

The propagation of uncertainty is how various errors affect overall uncertainty. Previous labs have examined the differences between variance, average deviation, and standard deviation, but have invariably limited error to like variables. A problem occurs, given previous understanding, when multiple sources of error affect a single value. In order to contend with this, calculus is necessary, specifically partial derivatives: We single out each variable in a formula and take their derivatives with respects to the function assuming all other variables are constants, and we multiple each derivative by their respect errors, summing them up in the end.

So, assuming function p(x₁, x₂, ..., x_n):

dp = | (∂p/∂x₁) | dx₁ + | (∂p/∂x₂) | dx₂ + ... + | (∂p/∂x_n) | dx_n

In some cases, this could be tedious to compute, so we have a couple ways to acceptably simplify this. One way is to take ln of both sides and separate complex terms using logarithm rules. The second way is to divide both sides by the original function p, which essentially cancels out all the constants in each term.

We practice propagating uncertainty by calculating something of multiple measurements: the density of various metals. We have 3 small cylinders of metals: steel, copper, and brass, that will act as our objects to be measured.

In order to find density, we need their masses and volumes, and consequently their heights and diameters (because we cannot directly measure radius), applying the formula:

ρ = m / v
ρ = m / [π r² h]
ρ = m / [π (d² / 4) h]
ρ = 4 m / [π d² h]

Measuring the weight is easy. We use a scale to measure the normal force, and spit a value back at us in grams.

We use a caliper to measure the height and diameter. A caliper consists of a wrench-like clasp which takes the object. It has little marks on it, like a ruler, to tell the dimensions in mm, and then then a different set of marks which magically gets tenths of mm accuracy. The first mark from the second set from the left that closely matches a mark on the first set is the number for that figure.

Got all the measurements! (Note: We've since updated some of these to make them more accurate. We read the caliper wrong the first time.)

We'll first find the density of each metal, then propagate their uncertainties. Exciting!

ρ = 4 m / [π d² h]

ρ_steel = 4 (49 g) / [π (1.26)² cm² (5) cm]

ρ_steel = 196 g / 24.938 cm³

ρ_steel = 7.86 g/cm³

ρ_copper = 4 (58.4 g) / [π (1.28)² cm² (5.14) cm]

ρ_copper = 233.6 g / 26.457 cm³

ρ_copper = 8.83 g/cm³

ρ_brass = 4 (80 g) / [π (1.60)² cm² (4.80) cm]

ρ_brass = 320 g / 38.604 cm³

ρ_brass = 8.29 g/cm³

Comparing to ideal densities, according to Wikipedia (http://en.wikipedia.org/wiki/Copper), the density of copper is 8.96 g/cm³. Our value is off by 1.5%. Taking into account measurement errors, this isn't too bad--it's within range of variation. Brass, on the other hand, is an alloy of copper and zinc, making it impossible to find an exact density since we don't know the composition of our cylinder, but a good average seems to be 8.55 g/cm³. This gives us a 3% error. An average density of steel, also an alloy, is 7.85 g/cm³ (http://en.wikipedia.org/wiki/Steel). The error in this case is just over 0.1%.

Our numbers are relatively accurate, but let's propagate the uncertainty and see how reliable our numbers are without comparing them to other peoples' experiments. First, we simplify as much as possible:

dρ = | (∂p/∂m) | dm + | (∂p/∂d) | dd + | (∂p/∂h) | dh
dρ = | 4 / (π d² h) | dm + | -8 m / (π d³ h) | dd + | -4 m / (π d² h²) | dh
dρ / ρ = (dm / m) + (2 dd / d) + (dh / h)

Notice that by dividing our uncertainty by the density ρ, we end up with relative uncertainty. We can get our absolute uncertainty back again by multiplying it by ρ. This is much easier to do since we have already solved for density! So it turns out that:

dρ_steel / ρ_steel = (0.1 g / 49 g) + (2 * 0.01 cm / 1.26 cm) + (0.01 cm / 5.00 cm)
dρ_steel / ρ_steel = (0.002041) + (0.015873) + (0.02)
dρ_steel / ρ_steel = 0.038 = 3.8%
dρ_steel = 0.30 g/cm³

dρ_copper / ρ_copper = (0.1 g / 58.4 g) + (2 * 0.01 cm / 1.28 cm) + (0.01 cm / 5.14 cm)
dρ_copper / ρ_copper = (0.001712) + (0.015625) + (0.001946)
dρ_copper / ρ_copper = 0.019 = 1.9%
dρ_copper = 0.17 g/cm³

dρ_brass / ρ_brass = (0.1 g / 80 g) + (2 * 0.01 cm / 1.60 cm) + (0.01 cm / 4.80 cm)
dρ_brass / ρ_brass = (0.00125) + (0.0125) + (0.002083)
dρ_brass / ρ_brass = 0.016 = 1.6%
dρ_brass = 0.13 g/cm³

Our density with uncertainty is:

ρ_steel = (7.86 ± 0.30) g/cm³ (3.8%)

ρ_copper = (8.83 ± 0.17) g/cm³ (1.9%)

ρ_brass = (8.29 ± 0.13) g/cm³ (1.6%)

For further practice, we will take measurements to solve for an unknown mass, and express it with uncertainty.

Two spring scales are hung from long metal rods, carrying a single weight. The spring scales use displacement to measure the force exerted by the mass m on each rod, and also the tension of the strings connecting the mass. These were set up around our lab room; we picked the nearest station #8. θ₁ and θ₂ represent the angles under the strings and a horizontal line perpendicular to where m is hanged.

We use cell phone app Clinometer to measure the angles of the strings by tilting the cellphone on top of it. The app measures a different angle depending on the side the phone is turned, so we use an analog leveler to correlate. Clinometer measures to within 0.1º, but we can only be certain of ± 1º since the meter falls between 2 lines on the leveler.

And of course, we eyeball the tension forces from the spring scales, however glass distortion makes it hard to get a clear reading, so we can only say that we are confident within ± 0.5 N.

Our measurements are as follows:

F₁ = (7.5 ± 0.5) N
F₂ = (7 ± 0.5) N
θ₁ = 44º
θ₂ = 40º

In order to calculate for the mass, we apply Newton's 2nd Law, summing up the vertical components of the forces, then dividing by the gravity constant:

F_{y TOT} = m g
m = F_{y TOT} / g
m = F_TOT sinθ / g
m = (F₁ sinθ₁ + F₂ sinθ₁ ) / g
m = (7.5 sin44º + 7 sin40º) m kg/s² / 9.8 m/s²
m = 9.71 m kg/s² / 9.8 m/s²
m = 0.99 kg

To propagate the uncertainty, we must be careful to change the units for dθ into radians. The remaining process is the same:

dm = | (∂m/∂F₁) | dF₁ + | (∂m/∂F₂) | dF₂ + | (∂m/∂θ₁) | dθ₁ + | (∂m/∂θ₂) | dθ₂
dm = | sinθ₁ / g | dF₁ + | (sinθ₂ / g) | dF₂ + | F₁ cosθ₁ / g | dθ₁ + | F₂ cosθ₂ / g | dθ₂
dm = [(sin44 + sin40) m kg/s² / 9.8 m/s²] 0.5 + [(7.5 cos44 + 7 cos40) m kg/s² / 9.8 m/s²] 0.017453
dm = [ 0.136474 kg ] 0.5 + [ 1.097690 kg ] 0.017453
dm = 0.087 kg

If we divide the uncertainty by the mass, then we get a little more than 8.8%. Given our relatively uncertain measurements, this isn't too surprising. We express our final result as (0.99 ± 0.087) kg. With this, we have concluded our lab on how to propagate uncertainty. We have seen that loose measurements lead to larger relative uncertainty, and vice versa. We have also considered various ways in which we could simplify the calculation, and how they may not be beneficial in every instance.

Lab 5: Trajectories

We are starting to explore the more calculus intensive aspect of multidimensional kinetics. This experiment serves as sort of a review. Essentially, we will a let a ball roll down a ramp to hit some point on the floor, and then attempt to use math to find where the ball trajectory crosses certain points.

Involved in this experiment is a ring stand, which together with a clamp serves as a stand to raise an aluminum v-channel to an incline. Two wooden boards with a channel in them holds the second v-channel stably. A steel ball is placed near the top and allowed to roll down to hit the floor, where a carbon paper is placed.

The carbon paper will be marked upon impact, keeping track of where the ball ends up. We'll do it 5 times just to make sure it ends up reasonably in the same place, otherwise note its uncertainty.

Looking closely, there is a mark in the center of that carbon paper.

The goal now is to calculate v₀, the speed just as the steel ball leaves the ramp. To do this, we need to find the height of the table, and the distance of the landing point from the edge of the table. Since it's difficult to get a ruler perfectly perpendicular to the floor, we use a self-fashioned plumb bob, a length of string tied to a weight that will utilize gravity to make sure it goes straight down. We then measure the length of string. And from where the plumb bob meets the ground out to the carbon paper. In this manner, we found the height h to be 94 cm, and distance out from the table d to be 75.5 cm.

One way of solving the velocity is to first find the time it takes for the steel ball to reach the ground, and since horizontal and vertical components are independent, we only need to consider the vertical in account of gravity. We assume that the v-channel we laid on the table is perfectly horizontal, thus leaving the ball with no initial vertical velocity. So we set the equation up as such:

 h = 0.5 a t²

0.94 m = 0.5 (9.8 m/s²) t²

t² = (9.4 m) / (4.9 m/s²)

t² = 0.1918 s²

t = 0.328 s

Now, we plug this figure in the horizontal components to find velocity, such that:

v₀ = d / t

v₀ = (0.755 m) / (0.438 s)

v₀ = 1.7328 m/s

We will now place a wooden plank over the carbon paper where the ball lands, such that it forms an angle α with the floor, leaned up against the table, and figure out the distance d down this ramp that the ball will travel.

If we know how to calculate the trajectory of projectiles, then it should be possible to predict the path theoretically, such that our prediction should correlate with experimental results, give or take a margin of error. We could express this error as uncertainty. We first note that the point where the ball lands on the ramp must satisfy two conditions: It must be within the trajectory of the ball, and it must intercept the plank. We express the x and y coordinates in terms of the relevant variables:

Let's use the x-component to solve for time t, then plug it into the y-component to solve for d:

d cosα = v₀ t

t = (d cosα) / v₀

d sinα = 0.5 a t²

d sinα = 0.5 a (d cosα / v₀)²

d sinα = 0.5 a d² cos²α / v₀²

sinα = (d² / d) 0.5 a cos²α / v₀²

v₀² sinα = d 0.5 a cos²α

d = ( v₀² sinα ) / ( 0.5 a cos²α )

Plugging the values in, we get:

d = [ (1.7328 m/s)² sin47.4º ] / [ 0.5 (9.8 m/s²) cos²47.4º ]
d = [ (3.0026 m²/s²) 0.7361 ] / [ (4.9 m/s²) 0.4582 ]
d = (2.2102 m²/s²) / (2.2450 m/s²)
d = 0.9845 m

 Since we haven't learned how to propagate uncertainty yet, we can round this to a reasonable 0.98 m.

Okay. We are ready to set up the plank. We do one run with the steel ball, then tape carbon paper in its vicinity in order to mark the exact landing point, as we've done before. Then, we measure the distance and obtain our experimental value to see how close it is to our theoretical value.

No, those aren't termite holes. They are the marks left by the carbon paper. Measured from where the top of the plank meets the v-channel, we get (1.01 ± 0.02) m. This is fairly close to our theoretical value. Our variance is 1.01 m - 0.98 m = 0.03 m. We can divide it by our theoretical value to get a simple relative uncertainty, such that 0.03 m / 0.98 m = 3.1%.

There are conceivably many places where this error could have been introduced. Even with the plumb bob, our process involved holding a meter stick against a length of string, or the plank. This is admittedly inexact. The plank had also barely reached the v-channel, requiring props to hold it in place. This might have caused small shifts after the ball hit it for the first time. It is also possible that the cell phone app used to measure the angle was inexact, either inherently or because of the contours of the phone or the plank. The analog angle would have required even more guesswork. Our calculations also assume that the table and floor are flat (and not slanted), whereas this is not necessarily the case. Nevertheless, that our theoretical and experimental values are within 3.1% of each other can be considered a relative success, and shows some basis to the formulas.

Lab 4: Capturing Air Resistance on Tape!

Today, we set up sneaky webcams to try to catch naked air work its magic against coffee filters. We hope to see air push up against them. So we all march to the Design Technology building like a tabloid firm and hang around the staircase. But what we're doing is not so insidious, so we say. All we're doing is finding the hidden relationship between air and speed...

The process is that by dropping 1-5 coffee filters off that balcony (look up!), and capturing it with a camera, we could use video analysis software to discover its terminal velocity--the velocity of an object in free fall caused by the air resistance working against gravitational force. We also measure some reference point in the same depth so we know how many pixels the camera records per some unit of measurement. And we could do this while annoying a few bystanders, shooting two birds with one stone.

 

Because air resistance counters gravity, we know that it is a force, such that:

F_resistance = kvⁿ

And therefore, we find should find the terminal velocity and the force. Let's get the velocity out of the way first. The velocity is found by going through the video frame by frame and noting the change in position by the coffee filters. LoggerPro has a function to do just that; it also plots a position graph with that data, and we use a linear best-fit line to get an equation. We expect that the filters will fall some distance before it hits terminal velocity by air resistance, but we're not interested in the acceleration under normal gravity in the beginning. We want to find the slope of the linear displacement, which occurs after terminal velocity. This took some manipulation in LoggerPro.

As they all look exceedingly similar with slightly different slopes, here is just one example of a graph with the best fit line. We have 5: one for each number of coffee filters.

Here's what a data table looks like. The graph shown above is for 2 filters. Here's the data table for 3 filters:

Now we take all the tables generated by LoggerPro, and export it into a standardized CSV (comma separated values) file, which will be easier (in my opinion) to look at, and possibly work with in the future. Although we could import CSV directly into a spreadsheet program, we did not do this for this lab. Here's a photo showing what the CSV looks like, although as you can probably surmise, the data table for higher number of filters take too much space.

Anyways, from the data table, and consequently the graph, we get the terminal velocities for each number of filters:

• 1 Coffee Filter: 0.9222 m/s
• 2 Coffee Filters: 1.272 m/s
• 3 Coffee Filters: 1.641 m/s
• 4 Coffee Filters: 2.129 m/s
• 5 Coffee Filters: 2.525 m/s

Now that we got our independent variable, v on the formula above, we need to find F_resistance. Since it's a force, which is mass * gravity, we need to find the weight of the coffee filters in kilograms, and then multiply them by the gravitational constant to get force in Newtons. With a small digital scale, we find the weight of a single coffee filter is 1.035 g, which is 0.010143 N.

Using a curved fit line, we find k and n based on experimental data.

k = 0.01420 ± 0.00185

n = 1.390 ± 0.1071

Since we have working numbers now, we are ready to model the fall of the filters in Excel. This time, we look for the terminal velocity by looking at the velocity when acceleration goes to 0 m/s². We begin with a time interval of 30 Hz, because that is the refresh rate of the camera.

We can see from the spreadsheet that acceleration seems to be moving too quickly, and doesn't approach 0 m/s² close enough, and as such won't provide us with very accurate data. Therefore, we change the time interval to 200 Hz.

There, better. Pull the data down and find the terminal velocity.

As acceleration approaches 0 m/s², the velocity seems to settle around 0.785 m/s. When we look at our experimental data, we see that the terminal velocity we recorded is 0.9222 m/s. Not very close. However, we also see that our first sample point didn't really fall on the fit-line. If we choose a point that is closer to the line, then our modeled data should be closer to the experimental data (since the model is based on the fit-line, its data points should conform to it).

We choose the point for 2 filters, which sits just above the line, which means we expect that our experimental data will be just under the model. So we change the mass in the acceleration equation to reflect the mass of 2 filters, from 0.00135 kg to 0.0027 kg.

And, pull the data down, once again...

The terminal velocity with 2 filters seems to be approaching 1.292 m/s. While with 1 filter, the difference between our experimental and modeled data is approximately 0.14 m/s, with 2 filters, the difference is only about 0.02 m/s. And it is a little larger than our experimental data as predicted!

So in conclusion, we've proved that by looking at the model, we can predict things about what happens in experiments. The model allows us to interpolate to any mass and find its terminal velocity.

Lab 3: The Non-Constant Acceleration of a Terrorist Elephant

An elephant wearing a rocket pack and futuristic frictionless roller skates could only be an invention out of a dystopian novel. What would the elephant's excuse be to skate around hills using rocket power? Where does it get such equipment? No state in its right mind would sanction rockets to an elephant. The only explanation are IED, or improvised rockets. Which means our elephant is either a terrorist, or it stole them from terrorists armed with frictionless roller skates. Either way, the thought of it is terrorizing.

In this lab, we compare the results of solving a messy calculus problem involving elephants of questionable motivation analytically, or iteratively (or numerically) using Excel, as we have done before. The elephant in question has a mass of 5000-kg, and is rolling down a hill, going 25 m/s when it reaches the bottom and arrives at level ground (remember: ignore friction). Being unable to stop itself, the elephant employs its rockets to fire in the reverse direction, generating a constant 8000 N of thrust. The rocket burns fuel, leaving its mass to be m(t) = 1500 kg - 20 (kg/s) * t.

Find the displacement from the bottom of the hill to when the elephant comes to a stop.

The first step to any such problem analytically is to gather the known variables, then figure out what we need to do head towards a conclusion. Since the elephant and rocket are braced together, it makes sense to treat them as one system, so that the total mass becomes:

m_tot(t) = 6500 kg - 20 (kg/s) * t

We also know that the final velocity will be at a standstill, while the initial velocity is given, as such:

v = 0 m/s

v₀ = 25 m/s

Finally, since horizontal movement isn't typically acted on by any force, the only force in the picture comes from the rocket, firing in the opposite direction as the elephant's movement, therefore:

F_net = -8000 N

Looking at the pieces we have, there seems to be two possible relations to make: power and momentum, and Newton's Second Law. The latter is more familiar to us at this point, so we have:

F = ma

a = F / m

At this point, the plan is obvious. With acceleration, we could take 2 integrals and get the position:

a(t) = F_net / m_tot(t) = -8000 N / (6500 kg - 20 (kg/s) * t)

Divide by 20 to get:

a(t) = [-400 / (325 - t)] m/s²

Integrate from 0 to t with the given v₀ as the constant C. From this point, we do a series of calculations that is, frankly, too difficult to type neatly in a blog. Thankfully, Professor Wolf has generously provided the calculations. Hooray!

So after a flurry of numbers, the answer appears like a magic number crashing through the heavens. Behold: 248.7 m!

To be sure, although math is a huge part of physics, the purpose of this lab is not merely to demonstrate prowess in manipulating numbers, but to further reinforce how to use the iterative (numerical) process to model analysis, and let the data lead the way. Despite this being a relatively difficult calculus problem to solve on our level, the real world is much more complex, with much more known and unknown variables that make it exceedingly hard to predict everything deductively. Thus, it is necessary that we get used to modeling with data and finding approximate solutions. To do this, we open up trusty Microsoft Excel, again on the company's arch-nemesis, an Apple computer. This dichotomous combination seems to capture the worst of both sides.

Although we did this in class with Excel, I am posting pictures of Apache OpenOffice Calc from a personal computer, since I had neglected to collect all the relevant data. It is important to keep as many variables as possible when modeling, and avoid magic constants, so that we could tweak values and let the spreadsheet formulas reflect them automatically. In this experiment, we create 7 columns, for each variable, and define each of them with formulas that relate them to each other. Then we could change the time interval, for example, and see how the data changes.

We begin with an 0.1 s interval. We want to find the distance before the elephant stops, so in numerical terms this is the distance x when the velocity v reaches 0 m/s. Since these values are based on calculations of acceleration and time, they will not reach 0 m/s perfectly; we compensate by finding the row corresponding to a v closest to 0 m/s.

Here, at a v reasonably close to 0 m/s, the distance x is 248.697 m (taking more than significant digits to make a point). But what if we increase the time interval 1.0 s--how will the data look?

We can see that with a larger time interval, while the data is less accurate (our v is further from 0 m/s), we are able to see our result within 20 rows of data. Here, our distance is 248.379 m, which is further than our analytic ideal. That means that by decreasing our time interval, to say 0.05 s, we should be able to get a better answer, right?

Notice here that our v is closer to 0 m/s still, but it doesn't affect the x that much more--248.699 m as opposed to 248.697 m! The downside is that it requires 395 rows of data as opposed to 197. There is always a tradeoff in experiments when it comes to practicality versus accuracy. A spreadsheet running on 64-bit architecture could conceivably run tens of thousands or more rows of data, but once again the real world is much more complex, and accuracy increases costs in equipment, as well as time eliminating sources of uncertainty. It would be useless to articulate minute differences if much of it is caused by error. So the way we can tell how much accuracy we need is to note the uncertainty in the problem. We then manipulate the time interval such that the value v is within the standard deviation of the ideal value, in this case 0 m/s. We note that even with 0.1 s time interval, if we round the x we get, it becomes the same as the analytical answer of 248.7 m. Therefore, 0.1 s should have been sufficient, but not 1.0 s, which would have yielded 248.4 m. By getting a good grasp of the accuracy needed, we save both time and money--two things we need the most!

Lab 2: Free Fall and Uncertainty

The first thing to note is that there is nothing uncertain about free fall. Jump off a building and the conclusion is almost certain, unless you experience the impossible chance that your every particle simultaneously go through quantum entanglement and transfer you to a safe place. But outside of the 11th dimension, uncertainty in science almost certainly refers, not to ontological indetermination, but to random or systematic measurement error. Quantifying error is important in experiments because they express the context and confidence of the data. Figures without error specified are often hard to make sense of. Thus, although on the surface, the second lab is about measuring gravity, in between the lines what we are actually doing is determining error in the rather rudimentary apparatus and process. After all, even a high school physics student knows the gravitational constant is 9.8 m/s²

There are two kinds of error that we look at: random and systematic error. Random error is unpredictable and caused by limited precision of apparatus, or uncontrolled factors in the environment that affect the data. Systematic error is human error, incorrectly selecting apparatus or inaccuracies in measurement. It is systematic error that we focus on, because although there are minor fluxes in gravitational field, or perhaps increased temperature causes the tape to expand as the spark generator continues, or the power supply might have experienced unstable voltage, it is much more likely that measuring the distance between marks on a tape with an old wooden meter stick, that might even be slightly bent, introduces much more.

But before we continue discussion, I introduce the almighty spark generator. This 1.86 meter long stick features an electromagnet at the top to hold an object in place. When dropped, the two cables on each side of the object generate spark at a regular time interval (in this case, 60 Hz), burning marks into the spark-sensitive paper. We then use a meter stick, a wooden flat rectangular cuboid with lines on them, to measure the distance between each spark mark. The increasing distance between each subsequent mark indicates positive acceleration. The result, the total distance from a first significant mark, is then recorded into the exotic mythological software known as Microsoft Excel.

 The above spreadsheet was painstakingly produced--but only because Mac computers are a pain in the ass to use. If we had used PC's in class, I would have been able to reproduce that while mastering quilting with the other hand. I digress... The first and second columns show the time in 60 Hz increments and the measured distance from the first point on the spark paper, respectively. The third is the change in x, or the distance between each recorded distance and the one prior. The mid-interval time is, of course, the half-way point between each recorded time, and mid-interval speed represents the change in the distance between each measurement over the change in time.

So why do all this? The progress of computer technology has changed the way we approach experiments, allowing us to make use of an iterative computational process, and modeling approximate formulas, instead of deducting them through mathematical analysis. With Excel, we are able to see thousands of data points with a drag of the mouse. And then graph them to see the relation.

The first graph shows the relation of time against distance. The ideal formula is such that:

x(t) = v₀t + 0.5at²

That means that the experimental value shows the initial velocity to be 0.52 m/s, and the acceleration to be 2 times the first coefficient 4.79, or 9.58 m/s². The initial velocity could be due to that the first few values on the spark paper were ignored because they were too close together to measure.

The second graph shows the relation of the derivative, speed against time. This is given by the formula:

x'(t) = v(t) = v₀ + at

In this case, we could see that the initial velocity is 0.54 m/s, and the acceleration is 9.53 m/s². The values are a little different compared to before because our data points are not perfect, thus the best-fit line equation that models the the data are calculated a bit differently. However, as they are within a difference of 0.02 m/s and 0.05 m/s² respectively, the R² figure (which Microsoft describes as the "square of the Pearson function moment correlation coefficient") is pretty high.

From these graphs, we note that speed and time have a linear relationship, which proves that acceleration (the derivative of speed, and the slope of this graph) is constant. The linearity also proves that the instantaneous velocity in any time interval (with constant acceleration) is the same as the average velocity of that time interval. This is because the average of a straight line is its center. Or more rigorously, we say that, where a is constant:

 (x_f - x_i) / (t_f - t_i) = (v_f + v_i) / 2

(v₀t₂ + 0.5a(t₂)² - v₀t₁ - 0.5a(t₁)²) / (t₂ - t₁) = (v₀ + at₂ + v₀ + at₁) / 2

v₀ (t₂-t₁) / (t₂-t₁) + 0.5a (t₂ + t₁)(t₂ - t₁) / (t₂ - t₁) = v₀ + a (t₂ + t₁) / 2

v₀ + a (t₂ + t₁) / 2 = v₀ + a (t₂ + t₁) / 2

Now that we've made sense of our data, we compare our experimental value with the accepted value. We have derived the gravitational constant from our graphs to be 9.53 m/s² or 9.58 m/s². However, we expect the theoretical gravitational constant to be 9.8 m/s². The discrepancy here can be expressed by uncertainty -- ah-ha! We're back a full circle to the start of our journey. If we take a middle point from our 2 values, say 9.555 m/s², and compare it to the accepted value , we find that our value deviates by 2.55%.

Uncertainty is scalar, and any error introduced can only deviate the experimental value further. Since it doesn't make sense that error cancels each other out, people have devised methods to capture uncertainty which involve taking the absolute values of each factor or trial. The first method, the average deviation from the mean, adds the deviation from from each trial and divides the total by the number of trials. A second, better, method, called the standard deviation (also called the RMS, or root mean squared), takes the squares of each deviation, dividing the total, then taking the root--in the opposite order of its name. The standard deviation is the accepted method in science because it is better at quantifying the spread. With the average deviation, it is possible to have a wide variance in data come to the same deviation figure. Standard deviation better represents outliers, and also makes it possible, provided the data set is large enough, to plot a distribution curve.

To demonstrate, we take the experimental gravitational constant from each trial (i.e. from each group) in class and calculate a standard deviation.

The first 2 columns indicate that there are 8 trials, which different gravitational constants concluded from each. That they are all under the expected value shows that it is likely a systematic error--most likely air resistance. The average is 948.41 cm/s². Deviations are taken from the individual trial values from this average, then squared. By dividing this number by 8 trials, then taking the square root, we arrive at the standard deviation value of 8.71 cm/s². 75% of the trials fall within one standard deviation, which is better than the theoretical 68%. Standard deviation is less than 1% (0.92%) from the mean. We express our final conclusion as (9.48 ± .0871) m/s².

 If we plug our numbers into the normal distribution function (using 953 cm/s² as x), we seem to be within 39.8% of normal distribution.