Quadratic air resistance on a ramp

  • Thread starter Habeebe
  • Start date
  • #1
38
1

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.
 
Last edited:

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,194
7,293
[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
38
1
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.
 

Related Threads on Quadratic air resistance on a ramp

Replies
1
Views
1K
Replies
0
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
3
Views
3K
Replies
4
Views
9K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
3K
Top