Lab 10: Work

I'm all for practical learning, but physics in itself feels like an exercise in reinventing the wheel--over and over and over again. Everything we prove is something that has mathematically modeled hundreds of years ago. That's why I aspire to be an engineer instead of a physicist--to create something of practical use to people who don't live inside an underground Switzerland lab. And honestly, we're too busy trying to jump through a bunch of hoops within a predefined time limit during a lab to actually sit down and think about all the relationships and why they are--much easier to see the symbolic relations without distractions like how to enter a constant in a proprietary software called Logger Pro.

But this lab isn't just work, it's about work. More specifically, it's about defining work in a physical context, and how physics work coincides with the common understanding of work. We measure work in three ways: walking up stairs, running up stairs, and pulling a weighted backpack up a distance with slippery rope. Since the physics definition of work is just:

Work = Force * Distance

... if we know the carrying weight and the displacement, then there should be no problem in calculating it. The hardest part is actually doing the experiment.

First we measured the vertical height of each step to be 17 cm, with a total of 26 steps, for a total height of 442 cm. The first task, then, is to walk up the stairs in an arbitrary pace. As with all labs before, we use the small plastic timer, which does it's job. What it doesn't capture is the precise time or place in which the stairs start or finish, as this requires a manual button-push, introducing human error. For the run--self-explanatory!

	Time (s)
Walk	16.08
Run	4.96
Rope	27.82

Next, we use a rope and pulley system tied to a balcony at the top of the stairs (same height) to lift backpacks weighted 9 kg.

If the incline of the stairs is 30º, then it should only take half the force to traverse it compared to climbing vertically.

Note that these are not the stairs we climbed, this is for illustration purposes only!

F = mg sin30°

However, note that the distance is doubled. Therefore, since work is force times distance, there should be no change in work over different angles:

(1/2)F * 2h = F * h

The difference then, comes down to mass. Climbing stairs is equivalent to carrying the entire body weight, which is much heavier than pulling up a 9 kg backpack. The work required for each activity (assuming 160lbs body weight, or roughly 72.6 kg) is:

 

	Mass (kg)	Force (N)	Distance (m)	Work (J)
Walk	72.6	711.48	8.84	6289
Run	72.6	711.48	8.84	6289
Rope	9	88.20	4.42	390

Therefore, while lifting things target a specific muscle for working out, running would be preferred for burning calories. Speaking of calories, apparently a calorie is defined as "the amount of energy required to warm 1 g of air-free water from 14.5 °C to 15.5 °C at a constant pressure of 101.325 kPa (1 atm)". The definition requires the ambient temperature to be at 15 °C, so given that assumption (the day was much hotter!), the conversion is approximately 0.2389 calories per joule of energy. We could also figure power using the formula:

Power = Work / Time

The results are:

	Work (J)	Time (s)	Calories	Power (w)	kcal
Walk	6289	26.08	1502.46	241.14	1.50
Run	6289	4.96	1502.46	1267.94	1.50
Rope	390	27.82	93.17	14.02	0.09

The last column is kilo-calories, which is commonly referred to as a "calorie" in nutrition, so one would need to burn 1000 actual calories in order to burn what is commonly referred to as a calorie. Considering that climbing these stairs only burnt 1.5 kcal, a person would need to climb the stairs roughly 634 more times in order to burn off the 952-calorie Kentucky Club Salad from the WOW Cafe. Wow. Looks like I need to stop ordering 1180-calorie Texas Toast Burgers.

Lab 9: Hurricane Speed and Angle

The Hurricane is a common amusement park ride created by the Allan Herschell Company in the 1940s. It consists of a central support pillar with levers extending out to rides, which will spin and, using the behaviors of centripetal acceleration, cause the rides to left up. The angle that the ride lifts, as it turns out, depends on the speed by which it turns. The real version also uses pneumatic cylinders to oscillate the height, but we are modelling this without the hydraulics.

Since we don't have enough rulers to measure the Hurricane itself, we use our own rotating apparatus: a tripod tied to a motor and a stick, and a string tied to a mass on the other end. It sounds dangerous, but it's quite impressive in person.

The motor can be controlled to spin at a certain volt, which implies a constant speed if unimpeded. As it spins, the mass, a rubber stopper connected by a string at one end of the horizontal bar, rises at an angle. The first step to setup any experiment is to measure all the relevant angles and hypothesize on the relationship, so we could have an idea on what to expect.

Now we draw a model with the relevant dimensions:

Using the model, we draw a free body diagram:

And we're set to calculate for the relationship between angular velocity ω and the angle θ. Because there is centripetal acceleration on the horizontal access, we say that (where r is radius):

F_x = m a = m r ω²

Resolving the x and y components of the forces, we have:

x: T sinθ = m r ω²

y: T cosθ = m g

Dividing them, we get:

tanθ = r ω² / g

A word about the radius r: Since the width of the top horizontal bar is the radius of the system with the mass at rest, and the mass only swings outward, we could say that the radius is the radius at rest r₀ plus the length that forms a triangle between the string at rest and the string that is swung out at a certain angle θ. Thus...

r = r₀ + L sinθ
tanθ = ω² (r₀ + L sinθ) / g
g tanθ = ω² (r₀ + L sinθ)
g tanθ / (r₀ + L sinθ) = ω²
ω = sqrt[g tanθ / (r₀ + L sinθ)]

With the theoretical relationship armed in hand, we are ready to start experimenting to see if it confirms our analysis. To find the speed of the mass, we use a basic timer to time 10 revolutions, to improve accuracy.

As the motor spins, we expect that the string will form an angle outward, but this also raises the height of the mass. If we know the difference in height of the mass compared to the position at rest, then it is possible to solve the horizontal component. To measure the height, we place a piece of folded paper clamped to a metal bar under the apparatus.

We slowly raise the height until the mass just hits it as it swings around.

Studiously measuring the height:

And we do this 8 times...

Time (s)	Rev	Height (cm)	Velocity (ω)	Angle (θ)
39.68	10	50.0	1.5835	9.0125
32.88	10	62.8	1.9109	24.6816
29.53	10	76.2	2.1277	34.3175
26.98	10	95.5	2.3288	45.0257
25.14	10	106.6	2.4993	50.3369
22.05	10	114.1	2.8495	53.7027
19.40	10	142.5	3.2388	65.3757
18.50	10	148.7	3.3963	67.7657

 

We get the velocity ω by dividing 2π radians per revolution by T, the period, or seconds per revolution. The angle θ is found by subtracting the measured height from the overall height h of the apparatus, then dividing it by the length L of the string, and taking an inverse cosine of the result.

A similar set of data derived in class, from the experimental angle θ, then using the analytic relationship calculated earlier to reach the velocity ω:

We graph the analytically derived velocity ω_theory against the experimental velocity ω_expt to see if there's a correlation:

The regression line coefficient is 0.96483. The slope is 1.0135, meaning that we are under 2% error. That isn't too bad considering all the measurements we had to take, each of which could introduce error. What's interesting is that we find that velocity depends wholly on radius, angle, and gravity, and not on mass. But perhaps that isn't too surprising given that the pendulum period equation also excludes mass, presumably for the same reason.

Lab 8: Centripetal Acceleration

The commonly believed centrifugal force has long been rendered fictitious. Although it is true that an object spun in circular motion tends to fly outwards, it must be made clear that this is not a force externally acted upon an object, but comes from the object's moment of inertia. Instead, an object confined into circular motion must be acted upon by an external force pulling it towards the center. As such, centripetal force is not some independent physical phenomenon, but the force named for causing circular motion; this force could be tension (when the object is spun by a string), or normal (for example, in banked curves). It might be more clear, then, to call it centripetal acceleration (and also centrifugal acceleration that an object causes about a system) caused by other forces. Since centripetal does not refer to an independent force, we conclude that there must be some relationship such that we could find centripetal acceleration without introducing any additional variables.

If we consider an object swung uniformly about with a string, its tendency to move in a straight line (in one direction) due to inertia is perpendicular to the string. In order to determine this relationship, we sample some initial point in the flight path, and call the velocity direction and magnitude v_i. Then, we sample another point and call this the final velocity v_f, such that the angle formed in between them is θ:

Note that both v_i and v_f have the same magnitude, but different directions. And since they are both exactly perpendicular to the points that form the angle θ, the angle between them must also be θ. We can place them together to calculate a vector difference, or Δv.

The change of this velocity over the change of time is the definition of acceleration:

 a = Δv / Δt

To find Δv, we consider that the vector diagram forms an isosceles triangle because both v_i and v_f have the same magnitude. Therefore, we can divide θ down the middle, forming 2 identical right angles, with v_i and v_f being the hypotenuses. Then, we can say that half of Δv is v*sin(θ/2), as illustrated by the following:

Therefore:

Δv = 2v sin(θ / 2)

And since time is the arc length divided by speed, we have (where r is radius):

Δt = r θ / v

We now have all we need to find the relationship. Acceleration restated becomes:

a = [2v sin(θ / 2)] / [r θ / v]

a = (v² / r) [sin(θ / 2) / (θ / 2)]

And since we're interested in instantaneous acceleration:

a = v² / r

This means that if we devise an experiment testing the acceleration and velocity squared of an object, we should note a linear relationship, and that relationship (slope) should be the radius. And we're going to do just that!

Pictured above is a heavy iron plate used for esoteric training by Shaolin monks. We attached an accelerometer to it, so that when we spin it, the accelerometer would pick up the acceleration and record it into Logger Pro. Also required are 5 pieces of masking tape and a messy desk. In order to get the velocity, we use rudimentary (compared to the accelerometer) handheld timers, timing 2-4 revolutions of the plate. If we divide the number of revolutions by the time, we should get the angular velocity (rev/s). On that note, we know that angular velocity ω is:

ω = v / r
a = r ω²

Here's the data from the timers, the calculations, and the data recorded by the accelerometer (wow, it's that much easier to copy your own spreadsheet directly into a blog instead of messing with copying and pasting from those Macbooks):

Time (s)	Rev	Velocity (ω)	Acceleration (m/s²)
4.4	2	0.455	1.40
2.5	2	0.800	5.00
3.7	2	0.541	2.20
1.4	2	1.429	12.90
1.6	4	2.500	37.41

 If we plot angular velocity against acceleration, this is what we get:

The linear fit has an pretty good correlation, meaning the experiment is pretty consistent. The slope is 0.1493, so we should expect the radius to be about 14.9 cm, if our calculations were correct.

Measuring the plate with a ruler, we find the radius to be about 15 cm. That confirms our results.

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.