Relationship between Excess Pressure and Surface Tension for a Drop on a Fibre

[Raihan amin]

Homework Statement

Think of a drop of radius $R$ deposited on a fibre of radius $b(b<<L)$.Find ${\Delta{P}}$

Homework Equations

The relationship between excess pressure and the surface tension is given by Yong-Laplace equation,
$$\Delta{P} = \sigma{(\frac{1}{R_1}+\frac{1}{R_2})}$$

The Attempt at a Solution

[/B]
Here i can't figure out what will be $R_1$and$R_2$ .I know that the curvature of radius is defined as $$R=\frac{y''}{(1+y'^2)^\frac 3 2 }$$ if $y'$is the derivative of y with respect to $x$.But in the solution, they have used two different formula for $R_1$ and $R_2$.
Can anyone help me understand these?

 

[haruspex]

Homework Statement

Think of a drop of radius $R$ deposited on a fibre of radius $b(b<<L)$.Find ${\Delta{P}}$

Homework Equations

The relationship between excess pressure and the surface tension is given by Yong-Laplace equation,
$$\Delta{P} = \sigma{(\frac{1}{R_1}+\frac{1}{R_2})}$$

The Attempt at a Solution

[/B]
Here i can't figure out what will be $R_1$and$R_2$ .I know that the curvature of radius is defined as $$R=\frac{y''}{(1+y'^2)^\frac 3 2 }$$ if $y'$is the derivative of y with respect to $x$.But in the solution, they have used two different formula for $R_1$ and $R_2$.
Can anyone help me understand these?

You need two radii of curvature lying in planes at right angles to each other and at right angles to the tangent plane of the surface.
The formula you quote is correct for the XY plane. The other plane is a bit tricky. A vertical slice gives a radius of y, but that plane is not at right angles to the tangent plane. To get the correct plane you have to tip it over at angle θ, as shown. The centre of curvature is still on the x-axis (is this obvious?) so the radius is y sec(θ).

 

[Raihan amin]

The centre of curvature is still on the x-axis (is this obvious?) so the radius is y sec(θ).

Can you elaborate it,please ?

 

[haruspex]

Can you elaborate it,please ?

Are you asking for proof that the centre of curvature is still on the x axis? If not, what?

 

[Raihan amin]

Are you asking for proof that the centre of curvature is still on the x axis? If not, what?

Yes,i am asking for that

 

[haruspex]

Yes,i am asking for that

I can't think of a simple argument... looks reasonable but I need to think some more.