Determining Viscosity of Oil Using a Dropped Ball Experiment

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Homework Help Overview

The discussion revolves around an experiment aimed at determining the viscosity of oil using a dropped ball. The position of the ball is described by a specific formula, and various parameters need to be solved for, including the value of Y in MKS units.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the behavior of the exponential term as time approaches infinity and its implications on the velocity equation. There are attempts to derive equations based on given conditions, and questions arise regarding the validity of assumptions made about the exponential function.

Discussion Status

The discussion is ongoing, with participants offering insights into the implications of the exponential term in the equations. Some guidance has been provided regarding the interpretation of limits as t approaches infinity, and a new equation has been suggested to help solve for the unknowns.

Contextual Notes

Participants are working with four unknowns and have derived three equations, but the discussion highlights the challenge of resolving these equations, particularly as they relate to the behavior of the exponential term over time.

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Homework Statement


In an experiment to determine the viscosity of some oil, a ball is dropped into some oil. The position of the ball is given by the formula y=Y+At+Be-Ct. At t=0, y=0.070 m, the velocity dy/dt is 0 and the acceleration is 0.0983 m/s2. As t→∞ the velocity approaches 0.480 m/s. What is the value of Y in MKS units?

Homework Equations


y=Y+At+Be-Ct
vy(t)=A+-CBe-Ct
ay(t)=C2Be-Ct

The Attempt at a Solution


With four unknowns (Y,A,B,C), I came up with three equations:
B+Y=0.070m
A=CB
BC2=0.0983

I wasn't sure what to do as t→∞
 
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Looking at your velocity equation, what happens to the exponential term as t approaches infinity? What does the exponential function look like as its exponent approaches infinity? What if the exponential is in the denominator? Try using a graphing program or wolfram alpha to plot it.
 
as t→∞ the exponent would turn to negative, therefore it would go to 0. But that would give me
vy(∞)=A-CBe-C∞
0.480=A-CB(1)

if A=CB

0.480=CB-CB

and

0.480≠0
 
You said the exponential term would go to zero but you plugged a 1 into your equation. Yes [itex]e^{-t}[/itex] goes to zero as t approaches [itex]\infty[/itex].
 
I skipped a step there, sorry. Yea I figured e^-Ct would go to 0 as t goes to infinity, but the math did not work out. Any more suggestions?
 
If the exponential term goes to 0 in your velocity equation, then you end up with 0.480=A correct? This gives you a 4th equation to solve for your 4 unknowns. Now you just have to do some algebra and solve the system of equations for Y.
 

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