Units of slope in a tension vs θ (theta) graph

[seanmyers23]

Homework Statement

In a physics lab we were finding slope on semi-log graph paper, with the y (up) axis representing tension in Newtons, and the x (sideways) axis representing θ in radians. Sadly, the prof introduced the new concept of natural logs to me - I'd never heard of these until this day. He told us to use the natural logs of the y coordinates to find rise, and I have next to know idea what this means, other than pushing the ln button prior to inputting a number in my calculator. What I'm trying to find is not only how to do the natural logs, but also what unit my answer for slope will be in. Any help would be appreciated.

Homework Equations

The Attempt at a Solution

If rise/run = y2 - y1/ x2 - x1, my equation looked like this: 5.00N - 9.67N/ 13∏/4 - 5∏4. This seems like complete nonsense to me.

 

[Chestermiller]

Homework Statement

In a physics lab we were finding slope on semi-log graph paper, with the y (up) axis representing tension in Newtons, and the x (sideways) axis representing θ in radians. Sadly, the prof introduced the new concept of natural logs to me - I'd never heard of these until this day. He told us to use the natural logs of the y coordinates to find rise, and I have next to know idea what this means, other than pushing the ln button prior to inputting a number in my calculator. What I'm trying to find is not only how to do the natural logs, but also what unit my answer for slope will be in. Any help would be appreciated.

Homework Equations

The Attempt at a Solution

If rise/run = y2 - y1/ x2 - x1, my equation looked like this: 5.00N - 9.67N/ 13∏/4 - 5∏4. This seems like complete nonsense to me.

Natural logs are just logarithms to the base e =2.718 rather than 10. The natural log of a number is just the power that you have to raise 2.718 to in order to get the number. Just like logs to the base 10, logs to the base e are such that ln(AB) = ln(A) + ln(B). In your problem, if T is the tension, and you are evaluating he slope over the θ interval from T₁ to T₂, the slope on this "semilog" plot is given by:

slope = ( ln(T₂) - ln(T₁) )/(θ₂ - θ₁)

but ln(T₂) - ln(T₁) = ln(T₂/T₁)

so, slope = ln(T₂/T₁)/(θ₂ - θ₁)

The units of the ratio T₂/T₁ are dimensionless, and, for small tension changes, ln (T₂/T₁) can be interpreted is the fractional change in the tension form θ₁ to θ₂.

I hope this helps.

 

[seanmyers23]

Thanks, that did help a bit, but I'm still concerned about having a dimensionless value divided by radians...in this case, our slope is supposed to represent μ, which from my understanding does not have any units. So does this mean that radians isn't really a unit either?

 

[Chestermiller]

Thanks, that did help a bit, but I'm still concerned about having a dimensionless value divided by radians...in this case, our slope is supposed to represent μ, which from my understanding does not have any units. So does this mean that radians isn't really a unit either?

Yes.

 

[seanmyers23]

Great, thanks a lot.