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I am taking an introductory lab course in which we've done an experiment on the physical pendulum.

drawing of physical pendulum

We've seen that for small oscillations, the period is

$$T=2\pi\sqrt{\dfrac{I_S}{Mgd_{cm}}}\tag{1}$$

where $S$ is the pivot point, $M$ is the total mass of the object, and $d_{cm}$ is the distance between $S$ and the pendulum's center of mass.

Now, varying $d_{cm}$, I've obtained seven different periods. I've calculated $g$ in terms of $T$ and $d_{cm}$ for those seven distinct values of $T$ and $d_{cm}$. So, $g$ can be expressed as

$$g=4\pi^2\dfrac{I_S}{T^2Md_{cm}}.\tag{2}$$

All the periods I've obtained were always greater than $1$, and the values of $g$ where between $10.15$ and $10.3$. I am trying to understand why is it that $g$ gave me always greater than $9.8$, the expected value, and not less than $9.8$. If I consider the air friction, I would expect the period to be greater; from what I've said and from equation (2), I would say that the values of $g$ should be less than $9.8$, contrary to the values I got.

Note that that $T$ is in seconds, $[g]=\dfrac{m}{s^2}$ and the angle from which the pendulum was released is of $25$ degrees approximately.

I would appreciate if someone could help me to understand the reason why I've obtained these values for $g$.

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    $\begingroup$ When you say the period is greater than 1, what units are you using? Similarly when you report your value of $g$, I'm assuming you're using meters per second squared? You should include that. $\endgroup$ – David Z Mar 12 '14 at 0:17
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    $\begingroup$ What's your release angle? That formula is only valid for small amplitudes. $\endgroup$ – Kvothe Mar 12 '14 at 0:36
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    $\begingroup$ @Kvothe: Yes, but wouldn't applying the formula for larger angles (where the pendulum will take longer to swing back and forth than the simple formula implies) yields a smaller value for the gravitational acceleration -- assuming accurate measurements? $\endgroup$ – Keith Thompson Mar 12 '14 at 0:41
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    $\begingroup$ @KeithThompson - You are correct. The first order correction for larger angles is $g = \frac{4\pi^2 L}{T^2} \left( 1 + \frac{1}{8} \theta_0^2 \right) + {\cal O} \ ( \theta_0^4 \ )$. Thus, the measured value using the first order formula is generically smaller. $\endgroup$ – Prahar Mar 12 '14 at 0:52
  • $\begingroup$ Thanks for all your comments, guys; to @David Z: sorry for not clarifying that information, I have added the units and the angle. So, I still don't have the slightest idea why I got greater values than $9,8$, I can't think of any possible mistakes I could have made during the experiment. $\endgroup$ – user100106 Mar 12 '14 at 2:01
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I) OP is using the period formula

$$\tag{1} T~=~2\pi\sqrt{\frac{I}{MgR}} $$

for a compound/physical pendulum (in the small amplitude limit) to estimate the gravitational acceleration constant

$$\tag{2} g~=~\left(\frac{2\pi}{T}\right)^2 \frac{I}{MR}. $$

Here $I$ is the moment of inertia around the pivot point; $R$ is the distance from CM to the pivot point; and $M$ is the total mass.

II) After doing the experiment OP finds values for $g$ that are 3-5% too big. (These results are close enough that OP likely did not make any elementary mistakes with units.) A finite amplitude of

$$\tag{3} \theta_0 ~\approx ~25^{\circ}~\approx~ .44~ {\rm rad}$$

makes the pendulum

$$\tag{4} \frac{\theta_0^2}{8}~\approx~ 2\%$$

slower, as compared to the ideal pendulum (1), cf. comment by Prahar. So correcting for a finite amplitude makes OP's estimates worse, 5-7% too big, as Keith Thompson points out in a comment above.

So the discrepancy is caused by something else. The culprit is likely that it is difficult to get a precise estimate for the moment of inertia $I$. All the other quantities $T$, $M$ and $R$ should be fairly easy to measure reliable. So OP's value for $I$ is likely too big. According to Steiner's theorem

$$\tag{5} I~=~MR^2+I_0,$$

where $I_0$ is the moment of inertia around the CM (and the actual quantity which is poorly known).

III) Below follows a suggestion. Plot OP's seven data points in an $(x,y)$ diagram with axes

$$\tag{6} x~:=~R^2 \quad\text{and}\quad y~:=~R\left(\frac{T}{2\pi}\right)^2.$$

Theoretically, the $(x,y)$ data points should then lie on a straight line

$$\tag{7} y~=~ax+b$$

with slope

$$\tag{8} a~=~\frac{1}{g}$$

and $y$-intercept

$$\tag{9} b=~~\frac{I_0}{gM}.$$

In other words, find the best fitting straight line. This method should hopefully produce a good estimate for $g$ without having to know $I_0$ a priori. (By the way, notice that we in principle also don't need to know the mass $M$, cf. the equivalence principle!)

IV) Finally, as always in experiments, estimate all pertinent uncertainties in the various measurements.

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  • $\begingroup$ Super clear and helpful answer, you're right about the moment of inertia, I had some difficulty calculating $I_0$ for the two "disks" and the rod. $\endgroup$ – user100106 Mar 16 '14 at 22:00

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