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.