# Fres = -mgsinθ pendulum

1. Dec 5, 2012

### aaaa202

Okay I hope I can write this so it makes sense what I am thinking.
For a pendulum you have:

Fres = -mgsinθ
And Fres points along the circumference:
So:
mRθ'' = -mgsinθ

I wonna discuss this property that you can just express the acceleration as Rθ''. In a cartesian coordinate frame your acceleration would depend on the characteristic length scale of x. How do you know that the characteristic length scale of Rθ is the same as x? I know it sounds weird, but I hope you understand what Im going at.

2. Dec 5, 2012

### HallsofIvy

Staff Emeritus
Re: pendulum

I'm not sure what you mean by "characteristic length scale" but I suspect it has to do either with the units of measurement or a specific length in the problem.

The angle, $\theta$ is "dimensionless" so that "$R\theta$" has the same units as R. Whatever "length scale" you are using for R also applies to $R\theta$.

3. Dec 5, 2012

### Staff: Mentor

Re: pendulum

are you asking if X and Y are measured in meters then would r*theta be measured in meters?

4. Dec 5, 2012

### aaaa202

Re: pendulum

no rather something like this: If we imagine that at some point the x-axis of our coordinate system tangential to the arc then the pendulum will move a distance dx measured in cartesian coordinates. How can we know that dx=rdθ, I mean what is it that says that these differentials are comparable? After all what if we made r bigger?