Cavendish experiment to determine G

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In summary: The difference between the equilibrium positions is the arclength the laser draws out when rotating the large spheres.
  • #1
SuchBants
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I'm doing the experiment for a project using a setup as so:
https://imgur.com/a/7vfg2
In the derivation of the formula I used the amplitude of the oscillation on the ruler of the laser, basically taking it as the arclength of the arc drawn by the laser spot as the bar rotates inside the apparatus and got the same formula as given to me by an instructor:
https://imgur.com/a/ynAlx
He has written the value I used as the arclength of the laser as the "difference of scale readings of the equilibrium positions of the spot corresponding to the two arrangements (A) and (B) "

Positions A and B being the two possible positions of the large spheres.

Can anyone explain what this means? Why is there a difference between the equilibrium positions and how does that equal the arclength the laser draws out?
 
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  • #2
Imagine the large spheres oriented in an intermediate position between (A) and (B) so that their supporting rod is perpendicular to the rod supporting the small spheres (see below left). In this case, because of symmetry, the net torque on the small sphere assembly is zero. When you rotate the large sphere assembly so that A gets closer to the small sphere on the left and B gets closer to the small sphere on the right (see below right), will the small sphere assembly rotate or not? If so, in what direction, clockwise or counterclockwise? What about if the large spheres are rotated the other way?

Cavendish.png
 

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  • #3
kuruman said:
Imagine the large spheres oriented in an intermediate position between (A) and (B) so that their supporting rod is perpendicular to the rod supporting the small spheres (See below left). In this case, because of symmetry, the net torque on the small sphere assembly is zero. When you rotate the large sphere assembly so that A gets closer to the small sphere on the left and B gets closer to the small sphere on the right, will the small sphere assembly rotate or not? If so, in what direction, clockwise or counterclockwise? What about if the large spheres are rotated the other way?

View attachment 220119
So the small balls would rotate clockwise initially, then anticlockwise due to the restoring torque which gives it its oscillatory motion. And the reverse would happen if the large balls were in the other position.
I can picture that this would produce two arcs where the equilibrium positions are separated but can't understand how the difference of the equilibrium positions fits in...
 
  • #4
The oscillations are useful in finding the period which can be linked to the torsional constant of the filament suspension. Eventually, the amplitude of the oscillations will die out regardless of which configuration the large spheres are in. If the "zero" (no gravitational force) is when the supporting rods are perpendicular, then the effect of gravity can be calculated by measuring the angular displacement Δθ of the equilibrium position when the large spheres are brought closer than the perpendicular or zero configuration. If you think a bit about it, you will see that the difference between the equilibrium positions is 2Δθ. Considering how small Δθ is, you get better accuracy by measuring twice that much.
 
  • #5
I think I've understood it now. Ignoring the oscillations as just a result of it being a twisting of a wire, then the wire rotates θ to the rest position due to the gravitational attraction of the two large spheres. And it rotates θ to the other rest position when the balls are moved - therefore the difference in the rest positions being 2θ. Thank you!
 
  • #6
SuchBants said:
I think I've understood it now.
Yes you have.
 
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Related to Cavendish experiment to determine G

1. What is the Cavendish experiment to determine G?

The Cavendish experiment is a famous and influential experiment conducted by Henry Cavendish in 1797-1798 to determine the gravitational constant (G), which is a fundamental physical constant that describes the strength of the gravitational force between two objects.

2. How does the Cavendish experiment work?

The experiment involves measuring the tiny gravitational attraction between two large lead spheres using a torsion balance. The balance consists of a horizontal bar suspended by a thin wire, with two smaller lead spheres attached to each end. When the larger lead spheres are brought close to the smaller ones, the gravitational attraction causes the bar to twist slightly. By measuring the amount of twist, the gravitational force between the spheres can be calculated.

3. Why is the Cavendish experiment important?

The Cavendish experiment is important because it provided the first accurate measurement of the gravitational constant, which is essential for understanding the laws of gravity and the behavior of celestial bodies. It also paved the way for future experiments in physics and helped to confirm Isaac Newton's theory of universal gravitation.

4. What was the result of the Cavendish experiment?

The result of the experiment was a value for the gravitational constant of 6.754 x 10^-11 m^3 kg^-1 s^-2, with an uncertainty of about 1%. This value is still considered very accurate and is used in many calculations and experiments in modern physics.

5. Are there any modern versions of the Cavendish experiment?

Yes, there have been several modern versions of the Cavendish experiment, including those using more advanced technology and equipment. These experiments have confirmed the accuracy of Cavendish's original result and have been able to measure G with even greater precision.

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