# Calculating gravity

• B
Hi everyone,

First of all, I am new to this thing and I hope you guys can help me with this.
So I am Ruzanna, a freshman,(I am 14 years old) and live in the Netherlands.

So, I have this project for physics where I had to make my own pendulum, and use it to calculate the gravity here.
Well I did that. Another part of this project is to talk to someone outside of the Netherlands, who wants to help me and also make a pendulum. Which is basically a rope with a mass on it.
I need you measure the rope. After that you need to hang it somewhere. Then you lift the mass about 10 cm and then let go so it moves, you don't need to push so only gravity is acting. Next thing is you measure how long it takes for the pendulum to make 10 periods (a period means the thing going back and forth ones) and i need that time. With that I would be able to calculate the gravity. It would help me like a lot!
Also it would be amazing if you could film it cause mij teacher wants "prove".

Hope to hears from you guys!

Ruzanna

## Answers and Replies

sophiecentaur
Science Advisor
Gold Member
Hi @Ruzana Welcome to PF.
That is one of the cheekiest OP's I have ever read but 10/10 to you for initiative.
I don't think you will get many takers for this experiment. Why don't you just do it yourself?
Just a couple of points; it may have been due to a translation error but "rope" is not good for this sort of measurement. It needs to be a thin thread - strong enough to support the mass with a mass of around 0.5kg. It should be supported on a very firm fixing which cannot move at all as the pendulum swings.
Good luck.

Drakkith
Staff Emeritus
Science Advisor
I don't think you will get many takers for this experiment. Why don't you just do it yourself?

Isn't her teacher requiring someone else to do the experiment with her? Granted that seems a bit strange...

@Ruzanna do you know why your teacher wants someone else outside of your country to build a pendulum and perform the experiment with you? To be honest, your post gives me the impression that you're cheating. I'm not accusing you of cheating of course, but your post does come off as a bit odd.

jtbell
Mentor
It sounds like the objective is to measure the acceleration of gravity, g, in different locations, and compare them.

@jtbell Yes indeed!

@sophiecentaur I did do it myself but the thing is, I can't go out of the country just to measure the acceleration of gravity there. And indeed it was a translation mistake :)

@Drakkith okay so, I kinda do get your point but I will explain.

I am doing this thing called TTO which basically means bilingual education, which means I have about 80% of my subjects in English. Talking about math, physics, chemistry, history etc.
And this TTO thing is also about internationalisation. Which is why I need to basically 'talk' to someone and make sure I develop my english and have contact with someone out of the country.
If you still think I am cheating I could send you this assignment, no problem.

Last edited:
Drakkith
Staff Emeritus
Science Advisor
@Drakkith okay so, I kinda do get your point but I will explain.

I am doing this thing called TTO which basically means bilingual education, which means I have about 80% of my subjects in English. Talking about math, physics, chemistry, history etc.
And this TTO thing is also about internationalisation. Which is why I need to basically 'talk' to someone and make sure I develop my english and have contact with someone out of the country.
If you still think I am cheating I could send you this assignment, no problem.

Ah, I see. No worries then.

OmCheeto
Gold Member
I get 9.8 m/sec^2 in Portland Oregon USA.

Data
29 cycles
54 seconds
0.86 meter length string​

@OmCheeto Thank you so much! Is it possible for you to send a picture of it? So I could show it to my teacher?

sophiecentaur
Science Advisor
Gold Member
It sounds like the objective is to measure the acceleration of gravity, g, in different locations, and compare them.
@Ruzanna It would be a good idea to do some research about how much g varies over the globe. Then do the experiment several times with the basic same equipment at your own location - but with different string and perhaps a different mounting. Only a few re-runs would be needed. This will give you an idea of the accuracy of this technique. Compare the spread of your results (include the results from @OmCheeto - who has been very helpful, I must say ) with the variation that can be found with more accurate methods. Ideally, you would get results from someone in Alaska or Finland and also somewhere Equatorial. You are looking for a consistent spread of only 0.5%, which could be hard with simple equipment.
PS When you come to present your results, it could impress your teacher if you quote the results from your internet research and make some comment, comparing your results with these. (=good Science)

@Ruzanna It would be a good idea to do some research about how much g varies over the globe. Then do the experiment several times with the basic same equipment at your own location - but with different string and perhaps a different mounting. Only a few re-runs would be needed. This will give you an idea of the accuracy of this technique. Compare the spread of your results (include the results from @OmCheeto - who has been very helpful, I must say ) with the variation that can be found with more accurate methods. Ideally, you would get results from someone in Alaska or Finland and also somewhere Equatorial. You are looking for a consistent spread of only 0.5%, which could be hard with simple equipment.
PS When you come to present your results, it could impress your teacher if you quote the results from your internet research and make some comment, comparing your results with these. (=good Science)

Yeah so what I first did was measure 1 period but because there is a thing called reaction time I came out on a g of 12. something. after that I measured the time of 10 periods and then divided the time by 10 so 1 got the average time of 1 period. After calculating g with the new number I got a g of 10. something. Still not what I wanted. So I decides to measure that 10 period thing 10 times. so at the end I got 10 times a 1 period time. Out of these 10 numbers I took the average when calculating g I got 9.91 m/s ^2 which is pretty close to 9.81 m/s^2. So i could really see how much 1 second would matter. Also I already did some reasearch (that was part one of this project)
Thanks for your help!

Merlin3189
CWatters
Science Advisor
Homework Helper
Gold Member
Just done the experiment here in the UK with first thing that came to hand, an old tin of paint. Video on youtube below.

The tin of paint is old and set solid. The length of pendulum to the estimated centre of mass was about 180cm. Time for 10 periods was 27 seconds or T=2.7. Gives a value of g of about 9.75 m/s/s

Not sure how accurate my estimate of the centre of mass is. You might care to redo the sums assuming I was 1cm out either way.

PS: There is a lot wrong with the way I did the experiment!

Ruzanna and Drakkith
Just done the experiment here in the UK with first thing that came to hand, an old tin of paint. Video on youtube below.

The tin of paint is old and set solid. The length of pendulum to the estimated centre of mass was about 180cm. Time for 10 periods was 27 seconds or T=2.7. Gives a value of g of about 9.75 m/s/s

Not sure how accurate my estimate of the centre of mass is. You might care to redo the sums assuming I was 1cm out either way.

PS: There is a lot wrong with the way I did the experiment!

Thank you so much!! (All of you though) Thanks for your time and effort :D

OmCheeto
Gold Member
@OmCheeto Thank you so much! Is it possible for you to send a picture of it? So I could show it to my teacher?
I've already disassembled it. Sorry!
It was just some fishing pole line duct taped to a doorway with a 70 gram lead weight at the end.

I may redo the experiment in the morning with a longer line and a video camera.
I attempted to do the experiment by watching some youtube videos, but youtube measures time in full seconds, so I don't think it was very accurate.

And I think some of the numbers listed in wikipedia's article about Foucault pendulums are either wrong or calculated. They are either way off, or too accurate.
The one here in Portland has a listed length of 27 meters, but I measured the period at 9.1 seconds, which gives a "g" of 12.8.
video:

I also did this for the Griffith Observatory in Los Angeles:
and the Chicago Museum of Science and Industry:

Those numbers came out much better using my periods, and the wiki listed lengths. (g of 9.6 & 10.1 respectively)

I think the main problem with doing the experiment with Foucault pendulums is that they are usually VERY heavy.
The one here in Portland weighs 400 kg, so I imagine the cable has to be very strong and HEAVY which will probably throw off the center of gravity, and make the maths not work out correctly. I may modify the experiment by using heavier rope to see how that affects the experiment.

It's indeed really accurate. Though I didn't expect the weight of the cable/rope could affect it too. So it's pretty interesting to know. The thing I saw with the measurement of the UK was, the actual period you needed to calculate the actual gravity was 2.69116 seconds. Which would then be 9.8119 m/s^2. The time measured by @CWatters was 2.7 seconds and g would be then 9.75 m/s^2 so those milliseconds mattered a lot.

If you would like to do it, sure. If not, it doesn't matter, also because I also have the numbers from UK now. Thanks anyway!

sophiecentaur
Science Advisor
Gold Member
I got 9.91 m/s ^2 which is pretty close to 9.81 m/s^2.
This has been a very useful exercise for you. I am impressed that you actually got some practical help from PF members. I am wondering if any of them would come round to my house and help me with the gardening, as well!
You found how much better it is to use several periods and your result shows that you have been careful. There are a number of other factors that can affect the result - the stiffness and mass of the string and the precise position of the Centre of Mass of the 'weight' itself and, as was mentioned earlier, the CM of the whole thing. Expressing the errors as a percentage of the 'accepted value' is useful and easier to appreciate when comparing one set of experiments with another. It's interesting that the results from all contributors fall on either side of published g. Compare that percentage range with the 0.5% spread between Equator and Pole. You can see how much harder it would be to refine your experiment to obtain that sort of accuracy.

Edit: PS When I suggested doing the experiment several times I meant several times, each one using ten or more swings. Every good experimenter does as many re runs a they possible can.

Last edited:
CWatters
Science Advisor
Homework Helper
Gold Member
It's indeed really accurate. Though I didn't expect the weight of the cable/rope could affect it too. So it's pretty interesting to know. The thing I saw with the measurement of the UK was, the actual period you needed to calculate the actual gravity was 2.69116 seconds. Which would then be 9.8119 m/s^2. The time measured by @CWatters was 2.7 seconds and g would be then 9.75 m/s^2 so those milliseconds mattered a lot.

If you would like to do it, sure. If not, it doesn't matter, also because I also have the numbers from UK now. Thanks anyway!
+1

The formulae means g is proportional to the square of the period so errors in measuring the period are magnified.

OmCheeto
Gold Member
Ugh!
The batteries in my camera just died, and I think it will take a couple of hours to recharge them.

Anyways, here are the results of this mornings experiment:

Length: 1.831 m

and

cycles: 59
time: 2:40.10 (m:sec) = 160.1 sec

g: 9.817 m/s^2

#### Attachments

• 2018.05.21.top.of.pendulum.experiment.png
136.9 KB · Views: 543
• 2018.05.21.bottom.of.pendulum.experiment.png
87.3 KB · Views: 561
mfb, sophiecentaur and gmax137
Wow so that is like really close to the 'official' g. So also the amount of periods (cycles) really matter. Well actually I start to believe like every single thing matters in this.
The thing I am still wondering about. I thought the mass would be a very important thing to calculate g. So I was kinda suprised when mass wasn't in the formula.

Drakkith
Staff Emeritus
Science Advisor
The thing I am still wondering about. I thought the mass would be a very important thing to calculate g. So I was kinda suprised when mass wasn't in the formula.

Mass gets canceled out since it is in both equations, the one for the gravitational force and the one that determines acceleration:

##F=MA##
##F=G\frac{M_1M_2}{r^2}##

Replacing F in the 2nd equation and using ##M_2## as the mass we're interested in:
##M_2A=G\frac{M_1M_2}{r^2}##

Dividing ##M_2## from both sides yields:
##A=G\frac{M_1}{r^2}=g##

This is 'little' g. The only things left are the mass of the large body, Earth in this case, the gravitational constant, and the distance from the object.

Mr Wolf
So basically it is there, just canceled. So they basically 'mixed' the 2 formulas to create a formula for g and because of that the mass got canceled out. So if you would calculate the gravitational force mass is important.

sophiecentaur
Science Advisor
Gold Member
So basically it is there, just canceled. So they basically 'mixed' the 2 formulas to create a formula for g and because of that the mass got canceled out. So if you would calculate the gravitational force mass is important.
You can look at gravity in two ways. As well as being Newtons / kg, g is also expressible as an acceleration metres / second squared and that doesn't contain mass.

Mass gets canceled out since it is in both equations, the one for the gravitational force and the one that determines acceleration:

##F=MA##
##F=G\frac{M_1M_2}{r^2}##

Replacing F in the 2nd equation and using ##M_2## as the mass we're interested in:
##M_2A=G\frac{M_1M_2}{r^2}##

Dividing ##M_2## from both sides yields:
##A=G\frac{M_1}{r^2}=g##

This is 'little' g. The only things left are the mass of the large body, Earth in this case, the gravitational constant, and the distance from the object.

To make things a "little" more complicated, it's interesting to point out that, when you compare the two laws, M1 and M2 that appear in Newton's law of universal gravitation are conceptually pretty different from the M2 that appears in Newton's second law of motion.

Drakkith
Staff Emeritus
Science Advisor
To make things a "little" more complicated, it's interesting to point out that, when you compare the two laws, M1 and M2 that appear in Newton's law of universal gravitation are conceptually pretty different from the M2 that appears in Newton's second law of motion.

How so?

Well, M1 and M2 which appear in Newton's law of universal gravitation are the "property" (gravitational mass) of a body so that two bodies can actract each other, while the M2 which appears in Newton's second law of motion is a different "property" (inertial mass) that causes a body to oppose to a force applied to it, and, for some reasons, they are both called simply "mass". It's quite "prodigious" you can compare the two laws.

Last edited:
The only things left are the mass of the large body, Earth in this case, the gravitational constant, and the distance from the object.
The "distance from the object" is actually the distance between the centres of mass of the objects. This means that the altitude of the pendulum has a small effect and should be a part of the variable data considered by the experiment. The effect is probably too small to be measured using the simple equipment available to the young experimenters,, but is worth a bit of thought in order to better understand the scientific method.
I believe Ruzanna has made a wise choice by posting in this forum.

https://en.wikipedia.org/wiki/Gravity_of_Earth#Altitude

Drakkith
While working on a pendulum clock I found a way to determine g (gravity) with very good precision. The principle is the following:
1. Set the pendulum so the clock it is exact within 10 seconds in 7 days, using a digital watch.
2. Using the adjusting nut at the end of the pendulum, lengthen the pendulum let say 2.5 mm (The thread of the adjusting nut is generally 0.50 mm and extremely precise, then turning it 5 turns will enlarge the pendulum in 2.5 mm.
3. Now measure in exactly 7 days (take the time at the same time you did the adjustment) measure how many seconds the clock is delayed.
Let us say that it delayed 51 minutes or 3060 seconds. Then the period of this length is:
Seconds in 7 days 604800 sec T1 = 1 sec
Delay of 51 minutes 3060 sec
T2 = (604800 + 3060)/604800 = 1. 005059523 T1 = 1 sec since the clock is exact in 7 days
Using the simplified pendulum formula T = 2 * PI sqr (L/g)

then g = L *(2 *PI/T1 )squared =( L + 2.5)*(2 *PI/T2 )squared

Resolving for L = 2.5 /(T2squared -1 ) Remember T1 = 1 then L = 247.979 mm

Then g = L * (2 *PI/T1) squared = 978.93 mm/sec2 = 9.7893 m/sec2

In dealing with the pendulum adjustment, I found in Google a formula which worked quite well for any latitude and altitude above sea level and in few steps I could precisely adjust the clock to less than 1 minutes in 7 days.
It is g = 9.780327 * (1 + A *(sin L ) squared - B *(sin (2*L)) squared - .C * HASL
Where L Latitude of the place in degrees
HASL altitude above sea level in meters
A = 0.0053024
B = 0.0000058
C = 0.000003086

sophiecentaur
Science Advisor
Gold Member
That's a very well thought out approach and a good bit of experimenting. It would be difficult to produce a piece of home equipment that could touch it for precision.
A possible problem with that could be that the pendulum bob weight can't always be approximated to a point mass. This could produce a small error. What was the latitude where the measurement was carried out and what would be the % difference between your value and the 'official' one?

mfb
Mentor
The pendulum was a fun lab project.

The length measurement will typically be the largest source of uncertainty when the experiment is done carefully - you can reduce the uncertainty in time measurements, but the length is harder to get accurately. Meanwhile there are a couple of other effects that can be taken into account to avoid a systematic bias:

- the pendulum bob is not a point mass, as mentioned. This makes it a physical pendulum. In addition to the motion of the mass the mass also rotates, this has an influence on the pendulum frequency.
- It is only an approximation that the period does not depend on the amplitude. It actually does, but the impact is small for small angles. It can be taken into account, however. The error is 1% at an amplitude of 23 degrees and still 0.2% at 10 degrees.
-- In addition, the amplitude decreases over time.
- If the pendulum swings long enough: One measurement of 100 oscillations is better than 10 measurements of 10 oscillations each.

As mentioned, the difference in gravitational acceleration is small. To see it, you would have to take both the size of the bob and the amplitude-dependent period of oscillation into account - or make sure everyone uses the same setup to have the same bias.

You have to have a clock with a 1 sec period. And then work adjusting the length with the adjustment nut until the clock is precise to 10 sec in a week. Measuring the length is not an issue as long as the pitch of the thread of the adjustment nut is known.

sophiecentaur
Science Advisor
Gold Member
You have to have a clock with a 1 sec period.
Why? (Apart for convenience) Any two numbers would suffice as long as the formula that's being used is accurate enough. The point mass assumption seems to have been pretty good.
Did you compare your result with the 'official' value at your latitude?

Reply to Sohiecentaur: Having a 1 sec period clock simplifies calculations. (most wall clocks are were designed with 1 sec period) The point mass is not important since pendulum length is not measured but calculated from the pendulum formula. I know the formula for gravity works, because I used it to set the clock to 10 second precision in a week after being repaired. I live In Mendoza, Argentina where latitude at my home is 31.919859 degrees South and at HASL = 821 m, then g is far from 9.806.m/sec2 (Actually it is 9.76462 m/sec2.. This g gave me a theoretical value for L = 247.34 mm which I never measured but allowed me to calculate how many turns to give to the adjustment nut once I have the delay/advance of the clock in 7 days. You can use other number of days, but for my clock it was the period for recharge the spring motors (is it the correct name?) It was really more fun than engineering task!!!!!