Duplication of the Cavendish experiment of 1798

[Pioneer1]

A computation of G from the Cavendish experiment data

Hi,

I used original data from the Cavendish experiment to compute the value of G. My preliminary computation yielded a value for G 2.67 times greater than the recommended value:

G(Cavendish) = 1.78424*10^-7 cm^3 sec^-2 g^-1

G(recommended) = 6.67428*10^-8 cm^3 sec^-2 g^-1

I used the well-known method of equating the torque of torsion wire (k theta) and the torque of gravity (GMmL/s^2) and solving for G:

G = k theta s^2 / MmL

k = torsion constant = 724.68 g cm^2 sec^-2
theta = excursion angle from mid-point = 0.0054788 radians
s = distance between weights = 21.97 cm
M = big weight = 158100 g
m = small weight = 729.8 g
L = gyration arm = 93.09 cm

More information about the geometry of the experiment can be found in my wiki.

I would be grateful for any corrections and comments. Thank you.

 

[Pioneer1]

Sorry I didn't realize you deleted this thread. I thought it didn't go through. I just saw your message now.

Please be advised that this is a different topic than my previous ones. I would appreciate any comments regarding the computations. Thanks again.

 

[Pioneer1]

Why are you changing variables in the middle of the discussion? Realize the restoring torque can be written as:
$\tau = k' \theta$

Ok but, I realized that if we equate the restoring force kr to GMm/r^2 as we have been talking about dimensions do not match. We need to use torque not force.

k theta = GMmL/r^2

Is this correct?

This seems to relate to what I posted today about the calculation of G from the Cavendish data. (It was moved to this thread.) There is a discrepancy of about 2.5 and I think I am missing something relating to this issue. Would you be able to take a look at that computation?

Thanks again for helping with this.

 

[Pioneer1]

I've made some progress with this stuff with the help of folks at sci.math and sci.physics.research. I am grateful to everyone here as well for helping. I have another question relating to the solution of the equation of motion.

The equation of motion is

Iy'' + Ry' + ky = C/(a -yd)^2

Primes indicate time derivatives.

y = theta = the angle of excursion
I = moment of inertia
R = damping
k = torsion
d = moment arm
C = 2GMmd
a = distance between weights
yd = angular distance between weights in radians

The simpler solution is to linearize this by writing it as

Iy'' + Ry' + ky = C/a^2

Then the solution is

y(t) = A cos(wt) (e^-t/tau) + B/w

A = amplitude
w = omega_0^2
B = 2GMmd/Ia^2


I also got a numerical solution for the non-linear equation that can be seen in the sci.math thread. Now I want to compare the two to see if linearizing the equation is justified.

Two problems. The non-linear solution has the initial conditions y(0) = 0 and y'(0) = 0. And in the non-linear solution we neglected the damping term by setting it to zero.

But the linear solution includes the damping term and does not have the same initial conditions. So I cannot compare them at this point.

I was wondering if someone can help with solving the linear equation without the damping term

Iy'' + ky = 2GMmd/a^2

for initial conditions y(0) = 0 and y'(0) = 0.

The values of the constants are included in the sci.math thread.

I appreciate your help. Many thanks.