Dimensional Analysis: Period of a Pendulum to the length

[DracoMalfoy]

Homework Statement

Use unit analysis to show that the constant, 2π, is unitless.

Homework Equations

T=2π√L/g[/B]

The Attempt at a Solution

T= [T]
L= [L]
g= a= [L]/[T]^2= [L T^-2]

[T]= 2π√[L]/[L T^-2]
[/B]
Is this correct? I wasn't really sure how to do this. I'm using book examples to help me figure it out.
I already know that 2π is unitless by its own definition.

 

[haruspex]

Is this correct?

It would be correct, as far as you have gone, if you to use parentheses to show that the "/g" is inside the square root.
Next, simplify, treating the dimensionalities as normal algebraic variables.

 

[DracoMalfoy]

It would be correct, as far as you have gone, if you to use parentheses to show that the "/g" is inside the square root.
Next, simplify, treating the dimensionalities as normal algebraic variables.

So.. cross out the Ls and leave T^2? It kinda confuses me with the lettering.

 

[DracoMalfoy]

So.. cross out the Ls and leave T^2? It kinda confuses me with the lettering.

[T]^2= 2π⋅ [T]^2

 

[Ray Vickson]

[T]^2= 2π⋅ [T]^2

$\sqrt{[T]^2} = [T].$

 

[haruspex]

[T]^2= 2π⋅ [T]^2

One more simplification to make.
You can write a dimensionless term as [1].