Astronuc,
Many thanks for taking the Newtonic Oath initiative seriously and providing the link to APS discussion. To me, that discussion clearly shows the need for something like Newtonic oath. But I believe that the recommendations are too vague. And also I would say that data fabrication is...
Thanks everyone for comments. Thanks also to Doc Al. His comment makes me reevaluate what I promote in my blog. The main idea that I promote is scientific skepticism and questioning of authority. I also believe that physics experiments are important and they lose their experimental value when...
Hello,
Even though physics is a professional field practiced by licensed professionals there is no document establishing an ethical code of conduct. For instance, a medical doctor cannot just decide that he wants to use 18th century methods and starts treating his patients with bleeding! But...
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...
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...
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.
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...
Ok. Thanks. From hyperphysics I got 1/12 M L^2 and added it to the previous result:
I = 2 m r^2 + 1/12 M L^2
L = length of the rod
r = half length of the rod
So,
Total moment of inertia for the Cavendish experiment = r^2 (2m + 1/3M) = 13,137,851.84 cm2 g
m = weight of each ball =...
Thanks for your reply. Here's the drawing from Cavendish's original paper:
http://www.alphysics.com/cavendishexperiment/CavendishSchematic11.jpg
And here's a detail
http://www.alphysics.com/cavendishexperiment/CavendishSchematic111.jpg
Cavendish also gave the weight of the bar and...
Hi,
I am trying to compute the moment of inertia of the Cavendish pendulum. I used
I = 2(m r^2)
r = gyration arm
m = the weight attached to the pendulum
But this formula is for a dumbell type of torsion pendulum where the weights are attached to the bar.
Does anyone know the...
Yes, this is the part I am hoping to express as a mathematical expression.
In the absence of gravity we write that acceleration is proportional to displacement angle. Or
alpha = -(k theta)/(I L)
I = moment of inertia of the pendulum
alpha = angular acceleration
L = the length of...
Ok. Net force is zero at two points but how do we know from this equation that the arm is stationary at points where net force is zero? We know that at the first point (r = 0.13) the net force was zero but the arm was in motion because it moved to the next zero point. How do know that it is...
Sorry my mistake. I plotted the difference of gravity and torsion. You can see the spreadsheet http://spreadsheets.google.com/pub?key=pVujjCJGUv-mKnkJozlf3Rw.
By your numbers I mean G=1, k=10 and L=1. Given these numbers there are two points where gravity and torsion are equal. These are...
Doc Al,
Thanks very much for this post. I understand that it is wrong to say that if G/(l-r) is greater than kr it will always be greater than kr. Thanks for correcting this.
I hope you will help with a couple of other questions.
When I plot the curve for the difference (gravity -...
Ok. Thanks. I looked at the torsion pendulum. I also looked at ftp://ftp.pasco.com/Support/Documents/English/AP/AP-8215/012-06802b.pdf. I believe that the equilibrium you mention is represented by figure 14 in Pasco manual. Is this correct?
Then, let me use their notation S1 for this...