# Pendulum craziness (LOGS)

1. Dec 7, 2009

### tinkeringone

Not sure, but apparently somebody threw some LOGS in my pendulum. HELP!

Trying to create a spreadsheet for calculating the friction coefficients in the pendulums I've been making. Here's the problem I just can't figure out.

μβ = ln (1+Z) − ln (1−Z)

where
Z = Δθ(L/d)
Δθ = 0.50 degrees
L = 1 meter
d = 18 mm, i.e. .018 meters
Z = 27.7777

I'm thinking that I already know that μβ = .34, but I can't figure out how to get Excel to calculate that right side of the formula, i.e. "ln (1−Z)", because 1-Z creates a negative value. So it's like you have to create some kind of imaginary number or something like Euler's?

Can anybody tell me how I can get Excel to calculate this, so I can enter my other pendulums into it?

Thanks,
Tinkeringone

2. Dec 9, 2009

### Redbelly98

Staff Emeritus
EDIT: Ignore this post. See Bob S's explanation in Post #3.

The ln of a negative number is a complex number,

ln(x) = ln(|x|) + iπ , . . x < 0​

(There is an ambiguity though, since you are free to add or subtract multiples of 2iπ to this result.)

However, I'm not aware of anything physically meaningful about a complex friction coefficient. You might want to check your equations or parameter values to see if there is an error somewhere that leads to taking the log of a negative value.

That being said, you have two choices for getting Excel to handle complex numbers. Either use two cells to store the value (1 cell each for the real and imaginary parts), or you can install the Analysis Toolpack add-in which will allow Excel to handle complex numbers.

Last edited: Dec 10, 2009
3. Dec 9, 2009

### Bob S

I would like to know the source for your equations μβ and Z. The log arguments 1+Z and 1-Z have to be unitless. For this reason, the angle delta θ has to be in radians, not degrees.
Bob S

4. Dec 9, 2009

### Redbelly98

Staff Emeritus
Bob, excellent point. If we use Δθ in radians, we get

Δθ = 0.50° = 0.0087 rad

Z = Δθ L/d = 0.0087*1/0.018 = 0.48 < 1

In that case, the log arguments are positive and there is no issue.

5. Dec 10, 2009

### tinkeringone

Bob – here’s the source of my information you requested. It’s a very interesting source. He’s loading up a simple pendulum with lots of mechanical friction to the point where the friction losses overwhelm the air drag losses, making it easier to calculate the friction loss coefficient.
It loses .5 degrees per half period, i.e. 1 degree per period from 30 degrees down to about 10 degrees. I think that the friction coefficient is the combined amount for the aluminum wire and the groove in the stationary 18 mm rod the wire pivots on. And I think that the combined friction coefficient he shows is .34.
The 27.7777 value I posted for Z was a value I came up with by trial and error.
The rest are based on data from the source.
The equation I initially posted was number (11 & 12) in the link. That is obviously a rearranged version of number (10) before those for the purpose of determining the friction coefficient rather than the loss in degrees per cycle.

Last edited by a moderator: May 4, 2017
6. Dec 10, 2009

### Bob S

Hi-
Looking at Eq 4 and 5 in your link, it is apparent that theta is in radians, not degrees (the small angle approx for cos(theta)). The pendulum in the link looks like a compound pendulum. Check the design before you build it.
Bob S

7. Dec 10, 2009

### Redbelly98

Staff Emeritus
Also, using radians exactly reproduces the results shown in Table 2, using their parameters:
L = 1 m
d = 18 mm
β = π
Looking at Fig. 3, it appears that the system behaves as a simple pendulum. Apparently the lower wire is stiff enough for this to be the case.

Interesting article, thanks for sharing!

Last edited by a moderator: May 4, 2017