Quadratic air resistance on a ramp

In summary, the problem involves a puck of mass m being kicked up an incline with initial speed v0 and angle of slope θ. There is no friction between the puck and the incline, but there is air resistance with magnitude f(v) = cv2. The solution involves using Newton's second law and separating variables to obtain an equation with the variables v and t. The final step is to solve for v as a function of t, with v=0 for the time to reach the highest point.
  • #1
Habeebe
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.
 
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  • #2
Habeebe said:
[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?
 
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  • #3
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 to Quadratic air resistance on a ramp

1. What is quadratic air resistance on a ramp?

Quadratic air resistance on a ramp is a phenomenon that occurs when an object is moving down a ramp and experiences air resistance, which increases quadratically with the object's velocity.

2. How does quadratic air resistance affect an object's motion on a ramp?

Quadratic air resistance on a ramp causes the object to slow down as it moves down the ramp, due to the increasing resistance force acting against its motion.

3. What factors influence the strength of quadratic air resistance on a ramp?

The strength of quadratic air resistance on a ramp is influenced by the object's velocity, the shape and size of the object, and the density of the air surrounding the object.

4. How is quadratic air resistance on a ramp different from other types of air resistance?

Quadratic air resistance on a ramp is different from other types of air resistance because it is specifically related to an object's motion on a ramp and follows a quadratic relationship with velocity, rather than a linear one.

5. How can the effects of quadratic air resistance on a ramp be minimized?

The effects of quadratic air resistance on a ramp can be minimized by reducing the object's velocity, making the object more streamlined and aerodynamic, and reducing the density of the air surrounding the object.

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