weighing machine on another planet? (1 Viewer)

rkk

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Hi,

Just wondering if anyone knows why the mass (read on the scale of a weighing machine on the moon) becomes less than that on Earth? ....


I was told...
The machine has some calibration with the acceleration??

I don't get it !!

please help !!!
 

insert-username

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rkk said:
Hi,
Just wondering if anyone knows why the mass (read on the scale of a weighing machine on the moon) becomes less than that on Earth? ....
I was told...
The machine has some calibration with the acceleration??
I don't get it !!
please help !!!
You're confusing mass with weight. An object's mass is the amount of material it contains, and is measured in kilograms, grams, etc.

However, an object's weight is the force exerted on it due to its position in a gravitational field. Weight is measured in newtons. For example, on the Earth's surface, an object's mass might be 60kg. However, its weight is a force, and since force is equal to mass x acceleration, the object's weight is 60 x 9.8 (acceleration due to gravity) = 588 N.

All scales use the fact that on the Earth's surface the acceleration due to gravity is 9.8ms-1. What they scale is actually measuring is your weight force, which it then converts into mass (by dividing by 9.8). Thus, when you take the scales to the moon, you will appear lighter. This is because the acceleration due to gravity is less on the moon, but the scales are still using the value of 9.8 from Earth. When you step on the scales, your weight in newtons is reduced, but since the scales are still calibrated for the Earth's gravitational field, they give a lower reading.

The mass of an object is a constant, but the weight of an object changes as it changes position within a gravitational field.


I_F
 

rkk

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pendulum problem

Why is that we need a smaller angle for a better approximation when doing pendulum experiment??


(Sorry for the simple questions)
 

mitsui

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i donno if it is the exact answer
but if u swing the bob at a large angle, u will notice it actuall goes around in circles (i think they used to haf one at townhall QVB)

becoz the large distance the mass is travelling will b affected by the actual motion of earth, and will not swing bak and forth in a straight line.

=]
 

rkk

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Even with a small angle, wont the pendulum end up swinging in circles anyway? Im sure it ends up in circles that are really small??
 
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Mountain.Dew

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this may not be true, but

may want to consider air resistance. there is lower air resistance when oscillating at a smaller angle than a larger angle. this reduces the inaccuaries that may appear in ur results, from the reduction in air resistance. remember that this experiment THEORETICALLY ASSUMES that there is no air resistance --> the only force operating is the weight force, under the influence of gravity.
 

mitsui

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rkk said:
Even with a small angle, wont the pendulum end up swinging in circles anyway? Im sure it ends up in circles that are really small??
yea it does
but doing it in a smaller scale reduce the effect of it.

and wat mountain.dew said is true too. XD - cant believe i overlooked air resistance
 

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rkk said:
Why is that we need a smaller angle for a better approximation when doing pendulum experiment??
Mitsui and Mountain Dew basically have the right answers, air resistance and the force of gravity on the pendulum have more impact when the object is in flight for a longer period.
 

zeropoint

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I hate to burst your bubble, but you guys are way off :). Air resistance is absolutely negligible at the speeds of a typical plumb bob.

The so-called "small angle approximation" arises in the derivation of the equation of motion for a planar pendulum, where one assumes that the amplitude of oscillations is small. This means that the formula for the period

T = 2 * Pi * Sqrt [L / g]

is only a limiting case for small theta and breaks down at larger angles

I'll sketch the derivation here for anyone interested:

Resolving the forces on the bob, parallel and perpendicular to the string we obtain

T = m g cos (theta)

parallel to the string, and

m a = - m g sin (theta) --- (*)

perpendicular to the string, where T is the tension in the string, m is the mass of the bob, theta is the acute angle betweent the string and the vertical and a is the tangential acceleration of the bob.

A little mathematics will convince you that

a = s'' = L theta''

where L is the length of the string and the double-prime indicates the second derivative wrt time (recall the arc-length formula, then differentiate twice wrt time). So we can write Equation (*) as

theta'' = - (g/L) sin (theta)

This equation is too complicated to solve as it stands. But you may know that sin (x) is approximately x for small x (where x is measured in radians). Thus, for small angles, we can use the approximation

theta'' = - (g / L) theta

from which it immediately follows that the angular frequency is Sqrt[g / L], and hence the period is 2 * Pi * Sqrt[L/g].
 
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rkk

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yeah i checked up in wikepedia,

was very complex...well for me...



so if u were writing up a justification of using a small angle what would it be?

sorry for the trouble..
 
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Mountain.Dew

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rkk said:
yeah i checked up in wikepedia,

was very complex...well for me...

so if u were writing up a justification of using a small angle what would it be?

sorry for the trouble..
i suppose that zeropoint does have a point (??? lol) with the mathematics, but in the HSC physics exam, or any school exam, i dont think that there is any need for that mathematical proof to show the need for smaller angles for this experiment. however, it is interesting and useful. for that, i thank you, zeropoint.

i guess it is safe to say u use smaller angles to obtain more accurate results from reduction air resistance, reducing the motion of pendulum in 3-dimensions that may come from human error. best to be on the safe side.
 

zeropoint

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Mountain.Dew said:
i suppose that zeropoint does have a point (??? lol) with the mathematics, but in the HSC physics exam, or any school exam, i dont think that there is any need for that mathematical proof to show the need for smaller angles for this experiment. however, it is interesting and useful. for that, i thank you, zeropoint.

i guess it is safe to say u use smaller angles to obtain more accurate results from reduction air resistance, reducing the motion of pendulum in 3-dimensions that may come from human error. best to be on the safe side.
Hi Mountain Due

I doubt it's required either, but the fact remains that the small angle approximation is the most significant source of error. As I told you before, air resistance is utterly negligible at these speeds. I recall figures of less than 100 parts per million (I can check these if you like). Friction at the connection point and the small angle approximation are thousands of times more significant.

James
 

zeropoint

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rkk said:
yeah i checked up in wikepedia,

was very complex...well for me...



so if u were writing up a justification of using a small angle what would it be?

sorry for the trouble..
The justification of using small angles is that the motion of the pendulum is approximately simple harmonic for small angles, whereas at larger angles the behaviour is far more complicated (in fact no solution exists for the general case).

James
 

rkk

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I shall write all these things in my justification!! Thanks guys!
 

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