Hovercraft (kleppner 2.37)

  • #1
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Homework Statement


[/B]
The "Eureka Hovercraft Corporation" wanted to hold hovercraft races as an advertising stunt. The hovercraft supports itself by blowing air downward, and has a big fixed propeller on the top deck for forward propulsion. Unfortunately, it has no steering equipment, so that the pilots found that making high speed turns was very difficult. The company decided to overcome this problem by designing a bowl-shaped track in which the hovercraft, once up to speed, would coast along in a circular path with no need to steer. They hired an engineer to design and build the track, and when he finished, he hastily left the country. When the company held their first race, they found to their dismay that the craft took exactly the same time T to circle the track, no matter what its speed. Find the equation for the cross section of the bowl in terms of T.

Homework Equations



Newton's law of motion

The Attempt at a Solution



Hard to start!
I wanted to work in polar coordinates to deal with circular motion, and at the end, once I find ##\theta##, go to cartesian coordinates so that ## y(t) = x(t) \tan(\theta(t)) ##.
Since the instructions say that for any time ##t##: ## T = \frac{2\pi R}{v(t)} ##; and since in circular motion ##v(t) =R \dot\theta ##, I am tempted to say that ##\theta(t) = 2\pi \frac{t}{T}##, but it does not explain why the engineer left the country.
I tried to draw forces in order to find a relationship between them, ##T##, and ##\theta## (or ##\tan(\theta)##).
I think there should be a reaction force ##\vec N## from the track radially, a propulsion force ##\vec F## tangentially, and a projection of the weight ##\vec W## tangentially and radially. With Newton's law of motion, I get :
##ma_r = -N - W\sin(\theta)##
##ma_\theta = F - W\cos(\theta)##
After reducing, I get
## F = mg \cos(\theta) ##
## N = m (\frac{4R\pi^2}{T^2} - g\sin(\theta)) ##

and now I'm lost.
 

Answers and Replies

  • #2
CWatters
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Not sure what you mean by "radially" or "tangentially"? Best draw a cross section of the track and show the forces.

The reaction force will be normal to the surface at that point. The exact direction depends on the slope of the track which may vary with height.
There will be a centripetal force because it's a circular track.
Weight acts vertically.
There is no friction.

The hovercraft doesn't slide up or down the banking so the sum of these forces must add up to zero.
 
  • #3
CWatters
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PS: He left the country because he realised his track design created very boring races.
 
  • #4
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Hello,

Sorry, I don't know how to make drawings on my PC, and I think I misunderstand the problem:
Why do you say the hovercraft is not sliding? I thought that the purpose was to make it slide circularly like a roller-coaster.

PS: By "radially" I mean along the radius, and by "tangentially", I mean along the tangent.
 
  • #5
CWatters
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Consider a car going around a banked track at a fixed speed. A component of gravity tends to make it slide down the banking. At the same time a component of centrifugal force (yes I know it's fictional/virtual) tends to make it slide up the banking. If one is greater than the other the driver will have to turn the steering wheel to compensate. Normally there is one optimal speed at which these forces are in balance and at that speed the driver can take his hands off the steering wheel.

In the case of the hovercraft there is no steering wheel so it must always go around the track at this optimum speed or it will slide up or down the banking.

Edit: The problem states that the lap time is always the same so the speed must depend on the circumference. If the speed varies with circumference then the bank angle may also have to vary with circumference and the equation for the cross section won't be a straight line.
 
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  • #6
CWatters
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PS: By "radially" I mean along the radius, and by "tangentially", I mean along the tangent.
Ok but the normal force only acts radially when viewed from above the track. When viewed from behind the hovercraft it's normal to the surface of the track which is banked.

Banked_Curve.png
 
  • #7
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Thank you soooo much, I had completely misunderstood the problem and would have never got it without your drawing !

Therefore you have a radial acceleration which is ##a_x = \frac{4\pi^2}{T^2}x##, and a vertical constraint ##\ddot y = 0 ##.
By your explanation and Newton laws you get that ##\tan(\theta) = \frac{4\pi^2}{gT^2}x##. In the end, since ## \tan(\theta(t)) = \frac{dy}{dx} ##, we get by integration

## y(t) = \frac{2\pi^2}{gT^2}{x(t)}^2##

Right?
 
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  • #8
CWatters
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oops posted work in progress.

Back soon.
 
  • #9
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why not by
##- m a_x = \frac{mv^2}{x} = F sin(\theta) ##
## 0 = F\cos(\theta)-mg ##
??
 
  • #10
CWatters
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Thank you soooo much, I had completely misunderstood the problem and would have never got it without your drawing !
Therefore you have a radial acceleration which is ##a_x = \frac{4\pi^2}{T^2}x##, and a vertical constraint ##\ddot y = 0 ##.
By your explanation and Newton laws you get that ##\tan(\theta) = \frac{4\pi^2}{gT^2}x##.
Yes. Not sure how you got the expression for "radial acceleration" but I got there from..

##mgSin(\theta) - \frac{mv^2}{x}Cos(\theta) = 0##.....................(1)
and
##v = \frac{2x\pi}{T}##....................(2)

Rearrange (1) to give
##\frac{Sin(\theta)}{Cos(\theta)} = \frac{v^2}{gx}##
Now..
##\frac{Sin(\theta)}{Cos(\theta)} = Tan(\theta)##
so
##Tan(\theta) = \frac{v^2}{gx}##
substituting for v using (2) gives same equation as you..
##Tan(\theta) = \frac{4\pi^2}{gT^2}x##

In the end, since ## \tan(\theta(t)) = \frac{dy}{dx} ##, we get by integration
## y(t) = \frac{2\pi^2}{gT^2}{x(t)}^2##
Right?
It's been years since I last did any integration but I don't expect the equation for the curve to be a function of time?
 
  • #11
CWatters
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I believe that should be just..

## y(x) = \frac{2\pi^2}{gT^2}{x}^2##
 
  • #12
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Thanks for your help!
My integration skills are low these days :-)
"radial acceleration" is the acceleration in radius direction. It is ##-\frac{v^2}{x}## because the motion is circular.
 

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