How Long Does It Take for a Water Droplet to Move Radially on a Turbine?

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The discussion revolves around calculating the time it takes for a water droplet to move radially on a turbine from its formation point to a specified radius. The initial conditions include a droplet mass of 0.002 kg, starting at r0 = 0.08 m and moving to r = 4.13 m, with a turbine speed of 402 rad/s. The equation for velocity is given as V = w√(r^2 - r0^2) = dr/dt, leading to the integral for time calculation. A participant struggles with integrating the function and questions the placement of the square root in the equation. Clarification on the integration process and proper notation is sought to resolve the calculation errors.
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Homework Statement


A water droplet condenses at r0=0.80 m. The turbine is rotating at a fixed speed of N RPM.

You can treat the water droplet as a point mass of m=0.002 kg. Starting from rest at its formation point, it starts moving outward on the radial vane as shown. Its mass stays constant during this motion (no new condensation nor evaporation).
Calculate the time (in seconds) it takes for the droplet to move from the point of formation (ro) to the radius r.
r0= 0.08m
r= 4.13m
w= 402rad/s

Homework Equations



V = w√(r^2-r0^2) = dr/dt

The Attempt at a Solution



∫dt = ∫(dr/w√(r^2-r0^2))

dt is from 0->t and dr is ro->r

I know that this is the equation for determining the time taken but I do now how to to integrate that function. I have been told it is t= 1/w*ln(r+√r^2-r0^2). But when i attempted to calculate the time it was always wrong.
 
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Why don't you show your calculations?
 
GaryTravis said:
∫dt = ∫(dr/w√(r^2-r0^2))
Is the square root term in the numerator or denominator? If the latter, then you need another set of parentheses to show it.
 

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