How Does Quadratic Air Resistance Affect a Puck on an Incline?

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SUMMARY

The discussion focuses on analyzing the motion of a puck on an incline affected by quadratic air resistance, modeled by the force equation F = -mg sin(θ) - cv². The user applies Newton's second law, resulting in the differential equation a = -gsin(θ) - (c/m)v². The integral solution involves arctangent functions, leading to a complex expression for velocity as a function of time. The key takeaway is that the user is on the right track but needs to simplify the expression to find the time to reach the highest point, where v = 0.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of quadratic air resistance modeling (f(v) = cv²)
  • Basic principles of kinematics on inclined planes
NEXT STEPS
  • Study the integration of differential equations involving arctangent functions
  • Learn about the effects of air resistance on projectile motion
  • Explore numerical methods for solving complex motion equations
  • Investigate the relationship between incline angles and motion dynamics
USEFUL FOR

Students in physics or engineering courses, particularly those studying dynamics and motion under resistance, as well as educators looking for examples of real-world applications of calculus in physics.

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*\frac{dv}{dt}

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

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=\frac{dv}{dt}=-gsin(θ)-cv2/m

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

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:

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

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 \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v on my original equation, ultimately getting:

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

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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Habeebe said:
\frac{ arctan(v*\sqrt{\frac{c}{ mgsin\theta }}) } {sqrt{\frac{cgsin(\theta)}{m}}} |^{v}_{v_0} = -t
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?
 
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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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