Divide by Zero Error: Solving Pendulum Radius

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The discussion focuses on calculating the radius of a pendulum using acceleration data in the x and y directions. The user encounters a divide by zero error when plotting the data, particularly at the extremes where the pendulum momentarily stops. This occurs due to near-zero tangential velocity and radial acceleration, leading to significant errors in calculations. A suggested solution is to filter the data to include only readings where x velocity is maximized, although this approach has yielded mixed results across different data sets. The user aims to cross-check the data to ensure consistent accuracy in their calculations.
Sam Smith
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I have data from a pendulum and I am using it to work out the radius of the pendulum. I have acceleration in the x and y directions and so thought this would be easy enough. Simply I determine the (velocity in the x direction)^2/acceleration in the y direction. However when I use python to give me a graph I notice that I get a graph with a large peak at the extremes (ie where the pendulum monetarily stops at the extremes and I am therefore dividing by zero) Any way around this?
 
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By "x" and "y", one assumes that this amounts to "tangential" and "radial" for an accelerometer mounted on the pendulum. Your complaint is that at the ends of the pendulum's arc you have [near] zero tangential velocity and [near] zero radial acceleration. You suffer from a loss of significance because the error bounds on the two values are as large as the quantities themselves.

One possibility is obvious. Filter your data to use the figures where x velocity is largest.
 
Yes I took this approach taking instantaneous readings. It was correct for some data but not another set. I am hoping to cross check them now so that I can be sure I can always get these values correct
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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