Lab 22: Physical Pendulum

We have previously worked with ideal pendulums, theoretically mass-less besides a point mass at the end. In reality, pendulum motion occurs across many different kinds of objects of different configurations. Today, we solve for the period of a semi-circle attached at the top and bottom, and swung sideways.

She even has a sexy mole. *Ahem* Anyways, here's an illustration of the different points on the circle. The radius of the circle is 14.6 cm.

To find period, we recall that finding the period before requires finding the angular velocity, which is found by acceleration (in this case, angular acceleration). One way of finding some relationship relating acceleration in this case is:

Therefore, we see that we need to find the moment of inertia and the center of mass. If we imagine it a whole circle in which the pivot is in the center, then this half circle pivoted at the top is just the circle with half the mass, so it is possible to infer that the moment of inertia at that point is just:

To find the center of mass, we must get our hands dirty with some calculus:

Now, we have what we need to analytically solve for the period, with the pivot on top. We also simplify our calculations by assuming small angles, replacing sinθ with θ.

Using a photogate, as pictured above, and Logger Pro, we experimentally find the period with similar results:

Ignore the large box on the bottom, which is inaccurate because it shows the period after running for longer than it should. We can look at the data in the table instead, and see that the period is roughly 0.86s before the effects of friction, among other things kicked in.

We can also turn the semi-circle around and pivot it at the bottom:

Analytically, this is a mess. We must first use the parallel axis theorem to find the inertia at the center of mass using our inertia at the top and then use this new center of mass inertia, and parallel axis again to find the inertia at the bottom:

Whew! With that done, we use the torque equation, as before, to solve for the period with the pivot at the bottom of the circle:

And experimentally:

Once again, we ignore the period box on the bottom and see that the period is actually around 0.83s. An overview of results is in order:

	Results		Error
	Analysis (s)	Experiment (s)	Absolute (s)	Relative (%)
Top	0.832	0.86	0.028	3.40%
Bottom	0.816	0.83	0.014	1.70%

Pretty close, I should say! And to round out the semester, here's a Dendrobium goldschmidtianum from my collection that just bloomed:

Thanks for the semester, Professor Wolf. It was hard, but you were reasonable, answered our questions when we had them, and even taught us a little about local Walnut politics.

Lab 21: Mass-Spring Oscillations

For this lab, we will measure these properties for different springs, and graph their relation to see if it matches up to what we know. Instead of having every group run experiments for every spring, we had every group do just one spring and share the results, because, according to game theory, cooperation saves time. But first, since the springs all had different masses, we needed to balance them by adding 1/3 of the difference between the springs' masses, so that their results are relevant to one another. This is because 1/3 of the spring's own mass contributes to kinetic energy, since parts of the spring moves slower than the end point, which is usually measured. So, after reporting our spring masses, we added mass in addition to a 100g mass that we will all be using, according to this formula:

m_added = 1/3 (m_{heaviest spring} - m_spring)

The heaviest spring mass is 27.7g, and ours, being the lightest, is 9.7g. With 18g of difference, we needed to add 1/3 of the difference, or 6g. The experiment set up is the same as Lab 12. To rekindle old memories (aren't all memories old?)...

The motion detector is positioned under the spring with masses as described above. Of course, Logger Pro is set up to capture the distance between the bottom of the mass and ground. If we add an arbitrary mass, then the spring should stretch some amount, modeled by the expression:

F = mg = kΔx

We ran Logger Pro first and it recorded x₀ to be 0.4953m. Note that this does not have to be absolutely accurate, we only need it to be accurate relative to further measurements, so that we get that change in x. With a 50 gram mass, the new x is 0.2945m, changing 0.24497m from x₀.

Solving for k, we get:

With mass and spring constant in place, we took off that arbitrary 50g of mass, leaving the 100g + 6g, and ran Logger Pro to capture several oscillations, the average of which gives us a reliable period.

Here is the data from every group:

SPRING	Spring Constant (N/m)	Mass of the spring (g)	Mass you have to add to the 100g hanging mass (g)	Period with the 100+ gram* mass attached (s)
1	2.44	9.7	6	1.60
2	18.52	19.9	3	0.47
3	6.50	11.1	5	0.81
4	23.30	27.7	0	0.42

Plotting the period and spring constant, we see:

The power of the best-fit curve is -0.6197. Ideally, it should be -0.5. Unfortunately, due to mix-up in computers (and incompetence in downloading data to our own), we weren't able to get our period and mass graph, but we see that other groups have gotten results with similar errors of ± 0.1 in the power. Ideally, the period-mass relationship should be 0.5.

Alternatively, we could derive this relationship using Newton's 2nd Law. At equilibrium position, the spring is at standstill, so the net acceleration is 0. Therefore we could derive the position in terms of the other variables, and define any other position in terms of this equilibrium position:

With the 2nd Law, we find that:

And plugging in our equation above, we get:

And this makes sense! More mass on a spring should increase the tension which stretches it further, increasing period. More spring constant makes the spring stiffer, so that it takes more energy to stretch it, decreasing period. Overall, although we weren't able to show our own period-mass graph, we can see that the empirical results do resemble the analysis.

Lab 20: Conservation of Linear and Angular Momentum

There comes a point when the basics have all been established, and the only thing left to do is to combine concepts and find new ways to apply them. This lab is like that, marking the end of a chapter with one final inelastic collision to slow us down, as if saying, "Hold yer' horses", while pulling on the reigns. However, this petty obstacle will be soundly defeated like Excalibur upon a vegetable.

In the first part of this lab, we have a steel ball being released from a ramp, leaving it horizontally at some velocity. Sounds familiar, right? And then hits a ball-catcher clamped on top of a pair of disks with diminutive friction. Still familiar? What is the final angular velocity? Well, there are two parts to this problem, clearly delineated, both of which have been covered in previous labs. The first part is almost identical to a part of the lab we've completed on trajectories (Lab 5), and the second is very similar to the past few labs which have used this "Rotational Dynamics Apparatus ME9279-A".

Thankfully, Professor Wolf has offered to do the experimental part of the lab--an excellent opportunity to show off his decades of classroom wisdom. First, he magically produced this ramp from the depths of the catacombs, clearly a sacred treasure of the Mt. SAC Physics Department, unavailable for use to the common practitioners, for example, in Lab 5.

Before anyone could take a good picture, he lays paper and carbon paper on the floor over the projected trajectory, as if he had inherited Goku's instant transmission. Then, with a measured hand, he centers the ball over the ramp, making sure not to exert any excess energy, so that the yang of the ball perfectly coexists with the ying of the ramp, creating a state of absolute universal synergy, in the absolute stillness which could only truly be achieved by accomplished classroom experimenters, called t₀.

The ball hits the paper, producing a round mark, indicating the lack of spin caused by nervousness when releasing the ball.

Measuring the distance with a dowsing rod plumb bob.

With all the dimensions measured, we can now figure out the velocity of the ball as it exits the ramp. Recall that we cannot exactly figure out the velocity from using the work-energy theorem because we lack information on the frictional coefficient of the ramp. Such an arcane artifact undoubtedly lack manufacture specifications, which would probably require top level clearance at the physics equivalent of Hogwarts. Instead, we can use kinematics to solve for air time and work backwards to get initial velocity. The ball is sufficiently heavy as to minimize the effects of air resistance. Of course, for Professor Wolf, solving physics problems of this level is a rote exercise, not even fit to be called manual labor. As such, the measurements were produced automatically, without any hint of hesitance that would plague the average student trying to figure out what to do next. And here are the measurements, ±0 (okay, maybe ±1 mm or so):

While the next part of the experiment is being set up, we solve for v₀, the speed of the ball as it exits the ramp:

Okay, so we need to figure out the inertia of the disks, again for this lab, if we are to apply the conservation of angular momentum principle to solve for angular velocity. Since the air pushing through the disks nearly eliminates friction, we are able to solve for inertia using the tried and true method of using a hanging mass and Logger Pro to record angular acceleration.

By the time we had solved for v₀ though, Professor Wolf had long finished measuring the diameter of the torque pulley (50 mm, so the radius is 0.025 m), and the hanging mass (0.0247 kg), and obtained the descending and ascending accelerations from Logger Pro (α_down is 5.934 rad/s² and α_up is 5.339 rad/s², so α_average is 5.6365 rad/s²).

All that's left to do is to plug in the numbers and get the inertia of the disks.

 

Due to the aforementioned instant transmission technique, we find that the ramp has been teleported, and screwed, next to the rotational disks!

Now for the final act--the Prestige. If we release the same ball off that, presumably, same ramp, then it should enter the ball-catcher attached to the disk at the same velocity we previously calculated, and due to conservation of angular momentum, we should be able to predict the angular velocity of the spinning disks.

But there are a few variables missing! On behest of an anonymous student amidst the crowd, who was probably planted there by the Secret Order of the Physics Department, Professor Wolf measures the diameter (19 mm, so radius is 0.0095 m) and mass (0.0283 kg) of the ball, and the distance from the center of the disks to the center of the ball (0.07 m). One could tell from the ruler already attached to the ball-catcher that this measurement was preordained, yet the suggestion comes from the crowd! This is a conspiracy no less than JFK!

We have previously determined the inertia of the disks. Now we can determine the inertia of the ball. With its radius, we needn't consider it as a point mass, but a solid sphere rotated about a parallel axis:

So the total inertia: I_total = 0.00121331888 kg m²

We have all the pieces. So, according to my calculations...

Muhahaha. What are Professor Wolf's experimental results, using the rotational sensor?

1.775 rad/s! Mine are faster, I win. Actually...

Results (rad/s)		Error
Analytical	Experimental	Absolute (rad/s)	Relative (%)
1.931	1.775	0.156	8.8

Almost 9% error. Drat! Well, given the multiple parts in this lab, there are actually many different sources of error. For one, the ball does not exactly leave horizontally from the ramp into the ball catcher. Also, the ball could have rolled inward after landing. The lab tables, as noted in prior labs, are not necessarily flat (the table could have been slanted to one side). The measurements for the height and distance the ball landed on the floor from the table could be affected by incongruences in the table, or the possibility that the ramp did not point perpendicularly out from the table. Also, since the ramp was not actually screwed into the rotational device, it could have shifted while the ball rolled. Also...

Ehh, time to call it a day.

Lab 19: Angular Momentum

If a ruler held from a fixed pivot swings down and hits a mass of putty, how high can it continue to swing? Or alternatively, how high can this shoe swing after picking up that ball of gum?

Or how wide can my mood swing after being denied this sandwich because the school store had a line and there was only one worker there?

These are things that may or may not involve angular momentum. For this lab, we will try to analytically predict the swing of the ruler, and eat that sandwich while no one's looking. Then, we will compare our predictions with results from a video capture. In order to figure this out expediently, I swiftly weighed our ruler and putty. We need to find the length of the meter stick... before the pivot.

So we used the ruler to measure itself recursively, and pondered whether it would be possible for it to measure it measuring itself, and so on... However, that idea had to be put to rest, for the sooner we set up the experiment is when I get to pass off the tough stuff to my math wiz lab partner and eat my sandwich, killing two physics problems with one putty. Fortunately, my appetite was sated. Unfortunately, my lab partner did not do the correct math the first time. In any case, I present our measurements:

Measurements
Axis shift from center of mass to pivot	0.493 m
Mass of ruler	0.088 kg
Mass of putty	0.015 kg

There are three steps to this problem. First, as the ruler is held up and starts its descent, gravitational potential energy converts into rotational kinetic energy. At the bottom of the swing is when the ruler moves the fastest, since all of the energy has been converted. Second, when the ruler hits the putty, angular momentum is conserved; however, since it's an inelastic collision, the angular speed changes. The lower speed clearly lowers the energy that could be converted back into gravitational potential. So our three steps are:

1. GPE → rot KE
2. L₀ = L_f
3. rot KE → GPE

The moment of inertia of the ruler is as a bar rotating about its center of mass, but shifted to one side (using the parallel axis theorem). Because the ruler is not swing exactly at an edge, it would be more reasonable to use the inertia formula for swinging about the center than swinging about an edge. We begin converting energy to figure out the speed of the ruler just as it reaches the bottom:

Then, we use angular momentum to figure out the speed of the ruler just after it makes contact with the putty:

Now we find the height from the kinetic energy of the ruler converted back into gravitational potential:

The numbers are plugged in. Voila!

To be honest, that seems a bit high, but we are assuming that there is no rotational friction in the pivot or air resistance, so it probably makes sense that the numbers do not match personal experience. There is usually friction, whether between objects or air, whenever an object swings about another, which is why a playground swing does not act as an ideal pendulum. How convenient would that be for the parents! Push once, and, "I'll be back in half an hour, son!" Anyways, now to set up and do this experimentally:

The clay is wrapped with some tape to aid it in sticking on the ruler.

There are only power strips on one side of the table, and we didn't want to move the entire set up, so we had to do some creative ad hoc wiring...

We use a video capture to record the swing of the ruler. Through a string of bad luck, we could never set up the camera correctly. For one, the preview didn't work, so we couldn't check the clarity beforehand. Second, higher definition seemed to cause problems. Maybe I should bring my own camera to physics classes next year!

As you can see, Logger Pro shows a paltry 0.1321 m for the final height, about a third of our analytic calculations. We should've brought some Viagra! If we count the blurry dots on the meter stick though, which presumably represent 5 or 10 cm marks, the blue dot waaay to the right is about 3-4 notches up, suggesting that the scale is probably calibrated badly. Also, since the capture was so blurry and overexposed, we had to make some guess-timates on where the end of the ruler was. These problems could contribute to the disparity, but I still think that friction isn't eliminated in the pivot. If we lubed it up, the stick would've probably gone higher. Just saying!

Lab 18: Moment of Inertia of a Triangle

In Lab 16, we used a rotary device that pushes air through a pair of disks, minimizing frictional torque, and allowing the disks to spin for a long time, almost unimpeded. We were able to discover torque from the tension of the mass and angular acceleration, setting up the equation to solve for inertia, but we did not go through with it. In Lab 17, we used a disk that did not push air through it, so we had to solve for frictional torque. Here, we will use the air disk device again to simplify calculations, and use measured angular acceleration to actually solve for inertia of the disks, and any object attached to them. In this case, we use a uniform metal triangular plate rotated about its center of mass.

There are two ways to affix the triangle on the rotating mechanism. We will call the first the "vertical" orientation:

And this we call the "horizontal" orientation:

Our model presents this triangle very graciously.

Before we experiment, it is possible to solve for the inertia of the triangle analytically. We find it easiest to solve for the inertia from one edge, then work backwards with the parallel axis theorem to find the inertia about the center of mass:

Whew! So, apparently, the inertia only depends on the mass and one of its dimensions. So let's get out those calipers and start measuring.

Length	0.149 m
Width	0.098 m
Mass	0.455 m

So analytically, our moments of inertia are:

I_vertical = 0.455 kg * 0.098² m² / 18
I_vertical = 0.000243 kg m²
I_horizontal = 0.455 kg * 0.149² m² / 18
I_horizontal = 0.000561 kg m²

Now that we have in mind what to look for, let's get the experiment underway. As before, we will first find the inertia of the disks by measuring angular acceleration with a mass hanging off a torque pulley. We need to make a few measurements:

Hanging Mass	0.025 kg
Pulley Radius	0.025 m

Everything is hooked up to Logger Pro. We let it record the rotational motion and use the slope of the velocity charts to figure out acceleration:

Disks by itself, without triangle.

Triangle in vertical orientation.

Triangle in horizontal orientation.

And as before, we average the descent and ascent acceleration to cancel out any remaining effects of friction that the disk has, even with the air blowing through it.

α (rad/s2)
Disk by itself.
down	2.1040
up	2.2780
average	2.1910
Vertical orientation.
down	1.9430
up	2.1020
average	2.0225
Horizontal orientation.
down	1.7350
up	1.9380
average	1.8365

Finally, we have all the pieces to calculate the inertia experimentally:

We can simplify this by replacing all the variables that remain constant:

Now, we can just substitute the angular acceleration we found earlier to find the total moments for each system. The inertia of the triangle by itself are then derived by subtracting the inertia of the disks.

The results are surprisingly similar. It's always daunting to set out to mathematically prove something, since any error either calls into question the entire concept in one's knowledge, or sends him in furious exercise to try to match what he knows as theory with reality. In this case, it worked out okay...

	Results (kg m2)		Error
	Analytical	Experimental	Absolute (kg m2)	Relative (%)
Ivertical	0.000243	0.000233	0.000010	4.3
Ihorizontal	0.000561	0.000540	0.000021	3.9