Quadratic air resistance on a ramp

  • Thread starter Habeebe
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  • #1
Habeebe
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Homework Statement



I kick a puck of mass m up an incline (angle of slope = θ) with intial speed v0. There is no friction between the puck and the incline, but there is air resistance with magnitude f(v) = cv2. Write down and solve Newton's second law for the puck's velocity as a function of t on the upward journey. How long does the upward journey last?

Homework Equations



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

According to Wolfram Alpha (I use this later):
[itex]\int \frac{dx}{a+bx^2} = \frac{arctan(\frac{\sqrt{b}x}{\sqrt{a}})}{\sqrt{ab}}[/itex]

The Attempt at a Solution



I set the axes so x is along the ramp in the direction v0 and y is normal to the ramp upwards. This gives force and acceleration in the x direction only.

F=weight+resistance=-mg*sin(θ)-cv2
a=[itex]\frac{dv}{dt}[/itex]=-gsin(θ)-cv2/m

Separation of variables gets me to:
[itex]\frac{-dv}{gsin(\theta)+\frac{cv^2}{m}} = dt[/itex]

I didn't know offhand how to do the integral, and it looked fishy, so I Wolfram Alpha'd it to see if I get something that makes sense before I figure out the method. Using that solution I with limits of v from v0 to v and t from 0 to t I get:

[itex]\frac{ arctan(v*\sqrt{\frac{c}{ mgsin\theta }}) } {sqrt{\frac{cgsin(\theta)}{m}}} |^{v}_{v_0} = -t [/itex]


And this is a jumbly mess. I can't really tell if I'm right or not because I can't identify intuitively what parts of the expression on the left stand for what. My gut feeling is that this can't be right because the answer is so absurdly ugly.

I also tried using [itex]\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v[/itex] on my original equation, ultimately getting:

[itex]\frac{cv^2/m+gsin\theta}{cv_{0}^{2}/m+gsin\theta}=e^{-2cx/m}[/itex]

But this is a function v(x(t)) and I'm not really sure how to go about solving that for v(t).


Thanks for the help.
 
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Answers and Replies

  • #2
haruspex
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[itex]\frac{ arctan(v*\sqrt{\frac{c}{ mgsin\theta }}) } {sqrt{\frac{cgsin(\theta)}{m}}} |^{v}_{v_0} = -t [/itex]
Looks fine to me. You are asked to obtain v as a function of t, so a few steps to go yet.
For the time to reach highest point, what will you put for v?
 
  • #3
Habeebe
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v=0. It looks really easy to solve for (atria), I just need to know that I'm on the right track. This is the first class that I've had where the answers come out this ugly.
 

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