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Hello  Steve,  hope you're well,

I have two quick questions for you;

1.  In your calculations (bottom) you state that  r = .11  and that  2pir = .687  but  2pir  =  .6911     Did I make an error or why the difference ?

2.  We desperately need the Pendulum horizontal swing Radius of these six (pendulum Length to Time) entities;  (1 meter - 2.0069105645614025 seconds),  (2 meter - 2.8382001388725806 seconds), (3 meter - 3.476071064067089 seconds), (6 meter - 4.915906842576353 seconds), (7 meter - 5.309786257377708 seconds).

We'd be willing to pay you to do this for us as it is extremely important that we are able to obtain these radii.  Could you do it ?  (Even though you may not agree with the value, could you please use;  9.80175174 for the Acceleration of Gravity.)


You wrote;


The large amplitude swing you describe in the swinging back and forth case could be a problem. Normally a pendulum with a 2 second period has a length of 1 meter (approximately, depending on the local value of g). That's true for small displacements. There are errors in the relatively simple relationship between length and period when displacements are large. The 1 meter swing translates to 1 radian swings. I went to the site
and played with their calculator incrementally until I found that a pendulum 0.97 meters long, making 1 radian swings, has a period of 2 seconds. So the swing you described wasn't as big a deal as I feared.  The 2 second circular travel time information defines the value of the angular velocity, w.  w = 1 revolution/ 2 seconds = 2*pi radians/2 s = 1 pi radian/s  The system must have centripetal force given by the formula   Fc = m*w^2*r = m*w^2*r
(I'll leave the angular velocity as w for now.) The force Fc will come from Th, the horizontal component of the tension in the cord.

To find the tension, T, start with analysis of Tv, the vertical component of T. Tv must be equal and opposite the weight, m*g, of the mass, m.  Tv = T*cos(theta) = m*g
where theta is the angle between vertical and the cord. So
T = m*g/cos(theta).  We will need variable names for the sides adjacent and opposite to the angle theta. Let's use r for the radius of the orbital circle and y for the vertical distance between the upper end of the cord and the plane of the circular orbit.  Now that we have an expression for T, look at the horizontal component of T and the centripetal force.  Th = T*sin(theta) = (m*g/cos(theta)) * sin(theta) = m*g*tan(theta)  That expression for Th must be equal to the expression for centripetal force.
Th = Fc
m*g*tan(theta) = m*w^2*r
Solving for r
r = g*(tan(theta)/w^2)
So we have 2 unknowns in this expression, theta and r. The function tan(theta) is opposite/adjacent, or r/y in our case. Substitute r/y for tan(theta).
r = g*r/(y*w^2)
Solve for y.
y = g/w^2 = (9.81 m/s^2)/(pi radians/s)^2 = 9.81 m/pi^2 = 0.994 m

That says that if the plane of the circle is just a bit less than 1 meter (0.994 m), it will have a rotation time of 2 seconds. Assuming your pendulum is exactly 1 meter, again let theta be the angle between vertical and the cord.
theta = arccos(y/l) = arccos(0.994 m/1 m) = arccos(0.994) = 6.3 degrees.  To find the radius
sin6.3 = r/l
0.11 = r/1 m
r = 0.11 m

So the circumference is 2*pi*r = 0.687 m

Hello Sharice,

1. That discrepancy was the result of variations on rounding off an intermediate result or using the full number of digits that my calculator still held from the last calculation. I had no idea that you would be wanting the result carried out to so many places. Near the end where I calculated y, I rounded off the number my calculator got for y and entered = 0.994 in my explanation. At that point something interrupted me. Maybe it was time for lunch -- I don't know. Then when I continued, I entered that rounded off value, 0.994, and continued. If you do all the steps in the paragraph that starts "That says that if the plane" without resetting your calculator, you will get 0.687255257 m. I rounded that off to 0.687 m. Remember, r = 0.687255257 m was obtained after I entered the rounded off value 0.994 m for y. So 0.687255257 m is not accurate to the level you are looking for. I'll do it as accurately as I can as part of my answer to #2.

Note: my calculator only carries 10 places. You gave me times to 16 places. Your pendulum lengths and the local value of g should be measured as finely if you really need the circumference to as many digits as the time data you gave me implies.

2. I combined all the individual steps I showed you previously into one formula that you can use for any pendulum length and circle time you like. That formula is

Circumference = 2*pi*L*sin(arccos[g*T^2 / (4*pi^2)])

I'll use your value of g as I use that formula to determine the circumference for the original case you gave me in October:
g = 9.80175174 m/s^2   L = 1 meter   T = 2 seconds
Starting with the expression arccos[g*T^2 / (4*pi^2)]
Plugging in the data, I get 6.722337705 degrees. That is theta.
Sine of that is 0.117057932.
Multiplying by 2*pi*L, I get 0.73549668 m as the circumference.

I hope you can do the steps I just did and get the same result. And then I'm sure you can do the calculations for whatever combination of pendulum length and circle time you need.

I hope this helps,


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Steve Johnson


I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.


I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University

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