How close can we get to acceleration due to gravity g=9.80665 m/s^2 at home, in an hour?

Theory

Quick - let’s first solve it analytically.

Second law says

$$\tau=I\alpha$$ $$-mgL\sin\theta=I\frac{d^2\theta}{dt^2}$$

With small angle approximation ($\sin\theta\approx\theta$), this gives

$$-(mgL/I)\theta=\frac{d^2\theta}{dt^2}$$ $$\omega=\sqrt{\frac{mgL}{I}}$$

and since $T=2\pi/\omega$, the period is $T=2\pi\sqrt{\frac{I}{mgL}}$.

The problem says to use simple pendulum (point mass as the bob), so $I=mL^2$. Substituting, we get

$$T=2\pi\sqrt{\frac{L}{g}}$$

Now solve for $g$:

$$g=\boxed{\frac{4\pi^2L}{T^2}}$$

So all we need to do is measure the string’s length $L$ and the period $T$. Simple, right?

Attempt 1: Collecting data (running into reality)

The only materials we need are a string, a pendulum bob, and a stopwatch.

I didn’t have any ideal object with simple moment of inertia, so I thought I’d first try to just use a longer string and let the $d$ in $I=I_{cm}+md^2$ dominate.

And for the bob, I needed the densest thing around to remove the effect of string’s mass: my character this lotion bottle.

I took 5 trials of 10 periods each (let nature do the summing to reduce uncertainty).

Trial	Time for 10 periods (s)	Period $T$ (s)
1	15.45	1.545
2	15.36	1.536
3	15.47	1.547
4	15.40	1.540
5	15.39	1.539
Average	15.414	1.5414

$$T=1.5414\text{ s}$$

I only had the edge of protractor as my ruler, and the bob was already tied up, so it was a real pain to measure $L$. It came out, after 3 back-to-back measurements, to be ~14 inches or 35.6 cm.

$$L=0.356\text{ m}$$

Substituting into the equation, we get.. *drumroll*

$$g=\frac{4\pi^2L}{T^2}=\frac{4\pi^2\times 0.356}{1.5414^2}=\boxed{5.925\text{ m/s}^2}$$

What??

Attempt 2: Debugging, irl

What could be wrong?

• Friction from the string: probably not much—10 oscillations don’t show noticeable damping
• Air drag: same as above, though drag would increase period, and $g\propto\frac{1}{T^2}$, which is in the direction of the error. hm.
• String’s mass: definitely negligible compared to the mass of the bob.
• Small angle approximation: angle was already very small.
• Lotion’s moment of inertia NOT a point mass <- this is probably it; $I_{CM}$ is still pretty large compared to $md^2$.
• L is too uncertain <- I need a better way to measure than back-to-back guesstimates.

Time for round two.

This time I marked string’s length exact with the edge of a sheet of paper (11 x 8.5in) and accounted for a little 0.5cm leftover when tying:

Also adding in an extra ~1cm from the string to the paperclip’s CM, we get

$$L=11\text{in}+0.5\text{cm}+1\text{cm}=0.2944\text{ m}$$

To get a smaller $I_{CM}$, I used a small binder clip on one end. Voila:

After 10 minutes of eye-drying focus, I finally got 5 trials of 10 periods each:

Trial	Time for 10 periods (s)	Period $T$ (s)
1	11.07	1.107
2	11.07	1.107
3	11.12	1.112
4	11.10	1.110
5	11.16	1.116
Average	11.104	1.1104

Now do I just use the average period $T=1.1104\text{ s}$?

I tried playing the reaction time game a few times and found my delay to be ~0.2s. Since I start the stopwatch right as I let go, there’s no delay there. It’s not symmetric. Let’s subtract $0.2\text{ s}/10=0.02\text{ s}$ from the average:

$$T=1.1104\text{ s}-0.02\text{ s}=1.0904\text{ s}$$

Ok, but before we calculate the new $g$, let’s find its uncertainty.

Uncertainty analysis

How far off are we?

The uncertainty in $L$ is $\Delta L=0.01\text{ cm}=0.0001\text{ m}$. The uncertainty in $T$ is my reaction time of $\Delta T=0.2\text{ s}/10=0.02\text{ s}$.

$$\Delta g=4\pi^2 \times \Delta\left(LT^{-2}\right)$$

We can write $LT^{-2}$ as $\exp\left(\ln(L)+\ln(T^{-2})\right)$ and repeatedly apply $\Delta (f(x))\approx |f'(x)|\Delta x$, like a chain rule.

$$\Delta g=4\pi^2 \times \Delta\left(LT^{-2}\right)=4\pi^2 \times LT^{-2}\sqrt{\left(\frac{\Delta L}{L}\right)^2+\left(\frac{-2\Delta T}{T}\right)^2}=0.016\text{ m/s}^2$$

Results

Putting all that together…

$$g=\frac{4\pi^2L}{T^2}=\frac{4\pi^2\times 0.2944}{1.0904^2}=\boxed{9.7752\pm 0.016\text{ m/s}^2}$$

huzzah it worked!! sort of. it doesn’t exactly contain $g=9.805\text{ m/s}^2$, but not too bad.

(maybe things are just further from Earth’s core where I live :P)

Wow it’s been so long since I’ve built something physical. I think I’ve gotten so addicted to the braincandy of reading new theory and so used to just manipulating equations / imagining on paper that I’d forgotten how fun it is to actually get outside and poke reality.