Proving Centripetal Acceleration - Lab

[dbakg00]

Homework Statement

This is from a lab. Part 1 of the Lab: We started out by hanging a spring vertically and attaching different weights to the spring. The mass of each weight along with the relative mass were given. Once the spring reached equilibrium, we measured the elongation and recorded it. Once this process was repeated for severel different masses, we plotted the Δy vs. $\mu$ (which the teacher defined as (M/m) - he called the ratio "relative mass," but didn't explain what it was). Once graphed, we were instructed to take the slope and that this number would serve as the "calibration constant" for our spring, he identified it as: $\frac{Mg}{k}$. For our spring, it was 5.0mm

Part 2: During the second part of the experiment, we attached this spring horizontally to a turntable and attached a weight to the spring. It was given to us that the weight has a "relative mass" of $\mu=5$. The turntable has a readout of the frequency, f, and the radius measurement of the weight's position, r. When the turntable is at rest and everything is at equilibrium, the $r_{0}$ was equal to 2.5cm. We then proceeded to turn on the turntable and record the "r" for different frequencies (in intervals of .250Hz). We then calculated $\frac{r_{0}}{r}$ and $f^{2}$. We plotted this data and took the slope. The slope was -0.1 This is where I am struggling...I don't understand what this slope represents.

Homework Equations

Here are some formulas we were given for part 1:
ky = mg
or
y = (g/k)m = (Mg/k)µ

Here are some formulas we were given for part 2:

*stretch = $r - r_{0}$
*kx is the inward elastic force which is rewquired to produce centripital acceleration, $\frac{v^{2}}{r}$ or $r\omega^{2}$, where v=tangential velocity .
*angular velocity: $\omega=\frac{v}{r}$

*kx = $mrω^{2}$ = $mr4\Pi^{2}f^{2}$

thus: $\frac{r_{0}}{r}=1-4\Pi^{2}(\frac{\mu}{g})(\frac{Mg}{k})f^{2}$

The Attempt at a Solution

Well, regarding $\frac{r_{0}}{r}=1-4\Pi^{2}(\frac{\mu}{g})(\frac{Mg}{k})f^{2}$,

I know that $\frac{Mg}{k}$ = 5.0mm from the first part of the experiment. Also, I know that $\mu$ = 5 (this was a given). g = $9800\frac{mm}{s^{2}}$. So, when I plug in my frequency-squared, along the the numbers above this formula gives me the exact ratio of $\frac{r_{0}}{r}$. I still can't connect the dots on how all this above help me to mathmatically verify that $a_{c}=r\omega^{2}$. Can someone please help me to understand this? Thanks

 

[anathema]

Thank you for sharing your lab experience with us. It seems like you are on the right track with your calculations and understanding of the formulas provided. I can offer some clarification on the purpose of the slope in the second part of your experiment.

In the second part, you are measuring the change in radius (r) as the turntable rotates at different frequencies (f). The slope of your graph, which you calculated to be -0.1, represents the change in radius per change in frequency (Δr/Δf). This slope is related to the angular velocity (ω) of the turntable through the formula ω = Δθ/Δt = 2πf, where Δθ is the change in angle and Δt is the change in time.

Using the relationship between angular velocity and tangential velocity (v = rω), we can then calculate the inward elastic force (kx) required to produce the centripetal acceleration (v^2/r) of the weight attached to the spring. This is where the formula kx = mrω^2 comes in, with m representing the mass of the weight and r representing the radius at which the weight is attached.

Finally, by substituting in the values for r and ω (which can be calculated from the slope of your graph), we can verify that the calculated elastic force (kx) is equal to the centripetal force (mv^2/r) acting on the weight, thus confirming the relationship between centripetal acceleration and the inward elastic force.

I hope this helps to clarify the purpose of the slope in your calculations. Keep up the good work in your experiments!