Help with mechanics problem-With my math in particular

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In summary, the conversation is about a person seeking help with a mechanics problem involving integrals and finding the velocity of a mass in a medium with drag force. The person shares their attempt at solving the problem and asks for feedback and advice.
  • #1
JDStupi
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Help with mechanics problem--With my math in particular

Hello all, I would just like to let you know that this is a fairly simple problem that is giving me trouble because of my sub par knowledge of integrals and I appreciate the help.

1.
A mass m has velocity v=v[itex]_{0}[/itex] at time t=0 and coasts along the x-axis in a medium where the drag force is F(v)=-cv[itex]^{}3/2[/itex]. Use the method of Problem 2.7 [seperation of variables] to find v in terms of the time t and the other given parameters. At what time (if any) will it come to rest?

2.t=m[itex]\int[/itex][itex]dv\frac{dv}{f(v)}[/itex] (and that integral is a definite integral from v to v[itex]_{0}[/itex] I just didn't want to mess up the rest.)


The Attempt at a Solution



Ok so the previous problem was simple enough (that was the problem this one was referring to) It was to simply use the separation of variables method to solve the special case that F(v)=F[itex]_{0}[/itex] a constant.

I went through the normal steps:

m[itex]\frac{dv}{dt}[/itex]=F

dv=[itex]\frac{F}{m}[/itex]dt

[itex]\int[/itex]dv=[itex]\int\frac{F}{m}[/itex]dt

v=[itex]\frac{Ft}{m}[/itex]+v[itex]_{0}[/itex]

and

x=[itex]\frac{Ft{2}}{m}[/itex]+v[itex]_{0}[/itex]t+x[itex]_{0}[/itex].


Now, here is the thing, because of my general lack of confidence in my mathematical abilities I am simply posting here so that you guys can review what I have done, point out the flaws in it and perhaps give me hints and/or other forms of help if you see something that I did wrong, or even if it could've been done better or what have you. I feel as though my answer is wrong though, because of its complexity.

m[itex]\frac{dv}{dt}[/itex]=-cv[itex]^{3/2}[/itex]

= m[itex]\frac{dv}{v^{3/2}}[/itex]=-cdt

= m[itex]\int[/itex][itex]^{v}_{vo}[/itex]=[itex]\int[/itex]-cdt

= 2m(([itex]\frac{1}{\sqrt{v_{o}}}[/itex])-([itex]\frac{1}{\sqrt{v}}[/itex]))=-ct

= (([itex]\frac{1}{\sqrt{v_{o}}}[/itex]))-(([itex]\frac{1}{\sqrt{v}}[/itex]))=[itex]\frac{-ct}{2m}[/itex]

[itex]\sqrt{v}[/itex]=[itex]\frac{\sqrt{v_{o}}2m}{((2m)+(\sqrt{v_{o}}ct))}[/itex]

v=[itex]\frac{4m^{2}v_{o}}{4m^{2}+4m\sqrt{v_{0}}ct+v_{o}c^{2}t^{2}}[/itex]


And so by the time I get to that point, I don't see any simplification that would make that look any nicer and am lead to believe that I did it wrong because of that messy answer, and so I turn to you guys for advice. Thank you very much to all who help.
 
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  • #2


I got the same thing without looking at your calc.
Looks like there is no t that makes v=0 other than infinity.
 

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