# Measuring the value of g

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Thread starter 6 years ago
#1

I have difficulty understanding the answer to q 9b why the experiment is not measuring the actual g ???

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Thread starter 6 years ago
#2
Anyone.?

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6 years ago
#3
(Original post by Lamalam)
Anyone.?

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Perhaps you could post up the whole question then, starting from part A?
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6 years ago
#4
(Original post by Lamalam)
Anyone.?

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Edit - Ignore this, look at my other post
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Thread starter 6 years ago
#5
(Original post by Doctor_Einstein)
Perhaps you could post up the whole question then, starting from part A?
I attaches two pictures in this thread,so i did attached a complete question. what more info do you need>>[email protected]@
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Thread starter 6 years ago
#6
(Original post by Arithmeticae)
Because some of the force is used up to make the pendulum move, so not all of it will be left and you won't be finding the exact value of g.
i though g is actually the acceleration , isn't it?
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6 years ago
#7
(Original post by Lamalam)
i though g is actually the acceleration , isn't it?
The external forces mean that your calculation isn't 100% accurate, because they mean that you're not just measuring acceleration due to gravity (F=ma)

Edit - Seems a bit of a strangely worded question, you are still calculating g (just that your calculations may not be exactly right due to external forces from the rotation of the earth.)
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6 years ago
#8
(Original post by Lamalam)
I attaches two pictures in this thread,so i did attached a complete question. what more info do you need>>[email protected]@
Sorry, my bad. It was all on one line without a space so I thought it was just one link.
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6 years ago
#9
(Original post by Lamalam)

I have difficulty understanding the answer to q 9b why the experiment is not measuring the actual g ???

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The experiment is measuring the value of g. However the value of g is less at the equator than at the poles because of two effects.

1. Centrifugal force (a fictitious force resulting from the Earth's rotation)

2. The Earth bulges slightly at the equator

The combination of 1 and 2 gives rise to a reduced value for g at the equator compared to the poles.

So your teacher is wrong to say that the true value of g isn't being measured. It is being measured and objects here at the equator will accelerate to the Earth with an acceleration of g as derived from the pendulum experiment - but it is less than you would get at the poles. The true value of g however varies depending on where you are on the Earth's surface.

I can see what your teacher is getting at however. Because if you use the equation g = Gm/r^2, you will obtain a greater value than that obtained from the pendulum experiment. This is because part of the Gm/r^2 force goes into maintaining the orbit of the apparatus around the equator (centrifugal force), and only some of it contributes to downward acceleration. I however take g to be the downward acceleration due to gravity, not Gm/r^2, but definitions will vary person to person.
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