Standard deviation in Graviational force, g

[shyta]

I'm doing a lab report, but I don't need help for it. I just need the standard deviation in g.
I need to compute the theoretical value for centripetal force together with its uncertainty for my circular motion experiment.

That is, since my mass was measured to be 0.208536kg, standard deviation can be assumed to be 0. To compute the theoretical value for centripetal force, F, together with its uncertainty, I need the uncertainty in g.

i.e.
F=mg
dF/F = sqrt( (dm/m)^2 + (dg/g)^2 )

where dm is assumed to be 0

Correct me if I'm wrong. Anyone able to provide dg? I did a Google search and found something like g at equator to be 9.78 and 9.83 at the poles.

can i just assume dg to be 0.02 in this case?


thanks wise ones

 

[cjl]

Why are you assuming that your mass measurement is exact? There will be some error in any mass measurement method. Also, was the way you measured your mass really accurate to 1mg? That seems like a very large number of significant digits in your mass measurement.

As for deviation in g, it should not matter for a circular motion experiment. Why do you use g at all if the lab is about centripetal force?

 

[Bob S]

From LBL PDG table of constants

standard gravitational accel. g = 9.806 65 m s⁻² (exact)

gravitational constant
G_N = 6.674 28(67)×10⁻¹¹ m³ kg⁻¹ s⁻² (±0.1%)

Bob S

 

[Pengwuino]

Yes, I second the question: How did you get your mass to be measured so accurately?

 

[shyta]

The mass is measured by a balance. It's reading is in that amount of sig fig. There was no instructions in the manual to repeat the measurements of m, therefore there will only be 1 value. It's impossible to find a standard deviation for m. If so, should I assume a 0.000001kg uncertainty?

with regards to cgl 2nd question,
As for deviation in g, it should not matter for a circular motion experiment. Why do you use g at all if the lab is about centripetal force?

why doesn't my deviation in g matter? it should theoretically. g is used here because in the experiment we used the weight of a hanging mass to measure the centripetal force. it's hard to describe and i don't really wish to go into details of the experiment.

 

[Pengwuino]

You actually have a scale that has that kind of accuracy? I can't imagine that for an academic setting.

I don't think cgl knows the experiment you're talking about but I do. The gravitational acceleration will come in when you try to equalize the force from the spring creating the centripetal acceleration so you do need to find the uncertainty associated with it.

@Bob S: Why does the PDG call it exact?? I'd like to know the reasoning behind that seemingly arbitrary declaration.

 

[Redbelly98]

From LBL PDG table of constants

standard gravitational accel. g = 9.806 65 m s⁻² (exact)

Okay, but that's just a conversion factor for acceleration. Like many conversion factors, it is simply defined to be a certain value. It isn't the actual acceleration due to gravity at the OP's location, which is the quantity of interest here.

 

[Bob S]

@Bob S: Why does the PDG call it exact?? I'd like to know the reasoning behind that seemingly arbitrary declaration.

Here it is from NIST.gov (National Institute of Standards and Technology):

http://physics.nist.gov/cgi-bin/cuu/Value?gn|search_for=gravity

Bob S

 

[Vanadium 50]

Bob is answering an entirely different question. Yes, there is a defined "standard gravity", and depending on where you are on the earth, it's quite close to local gravity or not so close. But using it for local gravity - as it is clearly meant to be by the equation - is a misuse: it's equivalent to assuming you are at STP instead of measuring the temperature.

 

[shyta]

hahah I thought so. so anyone? should i assume a 0.000001kg uncertainty for my mass since that is to the accuracy of my balance?

i think i will just be assuming a 0.02m/s^2 uncertainty for my g.

 

[Vanadium 50]

You have a balance that can weigh half a pound to within a milligram?

 

[cjl]

OK, I think I understand the experiment better now. The local deviation in G is fairly small, and 0.02 m/s². should be a decent estimate (though you could get a lot more accurate than that if necessary).

I'm still surprised that you would have a scale available to measure half a pound to milligram precision. You might want to double check with your teacher/professor or the scale's manual to make sure it's really that accurate, but in the absence of any other information, assuming +/- one of the least significant digits is a decent assumption for scales (so +/- 1mg in this case).

 

[shyta]

thanks cjl. I'm pretty sure the scale is to that accuracy so no worries. :)

 

[Pengwuino]

thanks cjl. I'm pretty sure the scale is to that accuracy so no worries. :)

You should definitely check. And tell us what kind of scale it is because I'm sure most of the people here are interested in knowing. At that precision, the small changes in air pressure that occur in a normal lab room would have to have some effect on it!

 

[Redbelly98]

thanks cjl. I'm pretty sure the scale is to that accuracy so no worries. :)

Just wanted to comment, I think people's skepticism is not so much that such a scale exists; rather that the physics department would budget the huge amount of money required on equipment for a student laboratory.

Equipment in student labs tends to get abused over time. Not by everybody, but when hundreds of untrained people are using that equipment then some of them are not as careful as they ought to be.