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Proving Centripetal Acceleration - Lab

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data
    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. [itex]\mu[/itex] (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: [itex]\frac{Mg}{k}[/itex]. 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 [itex]\mu=5[/itex]. 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 [itex]r_{0}[/itex] 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 [itex]\frac{r_{0}}{r}[/itex] and [itex]f^{2}[/itex]. We plotted this data and took the slope. The slope was -0.1 This is where I am struggling...I dont understand what this slope represents.

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

    Here are some formulas we were given for part 2:

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

    *kx = [itex]mrω^{2}[/itex] = [itex]mr4\Pi^{2}f^{2}[/itex]

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

    3. The attempt at a solution

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

    I know that [itex]\frac{Mg}{k}[/itex] = 5.0mm from the first part of the experiment. Also, I know that [itex]\mu[/itex] = 5 (this was a given). g = [itex]9800\frac{mm}{s^{2}}[/itex]. So, when I plug in my frequency-squared, along the the numbers above this formula gives me the exact ratio of [itex]\frac{r_{0}}{r}[/itex]. I still can't connect the dots on how all this above help me to mathmatically verify that [itex]a_{c}=r\omega^{2}[/itex]. Can someone please help me to understand this? Thanks
    Last edited: Feb 3, 2012
  2. jcsd
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