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Skier Going Down Slope

  • Thread starter Mbaboy
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19
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There is a circular slope that a skier is going down. He is at the top. If he goes just fast enough to start down the slope, at what angle to the center of curvature does the skier go airborne? I'm not sure if its asking for an exact angle or what. Any help or shove in the right direction is appreciated.
 
Here's a suggestion: Think about the normal force.

Dorothy
 
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Well I know that in order to go airborn the normal force is equal to zero. But when it is not zero, the component of the weight towards the center of the slope must be greater than the normal. I'm stuck from there.
 

marcusl

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Mbaboy said:
Well I know that in order to go airborn the normal force is equal to zero. But when it is not zero, the component of the weight towards the center of the slope must be greater than the normal.
No, the component of weight towards the center of the circle is the normal force.
 
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I don't understand. The normal force is perpendicular to the surface. So if they are both the same, the skier will have no centripetal acceleration.
 

radou

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Mbaboy said:
I don't understand. The normal force is perpendicular to the surface. So if they are both the same, the skier will have no centripetal acceleration.
The normal force is acting towards the center of the circular slope, just as the centripetal force is. Gravity is acting downwards. The component of gravity which is of interest to you is in the direction opposite to the normal and centripetal force, and it equals G*cosA, where A is your angle of interest. (I assume you already drew a diagram.) Now, just solve Fcp + N = G*cosA, where the normal force N equals zero (the 'airborne' condition).
 
radou said:
The normal force is acting towards the center of the circular slope, just as the centripetal force is. Gravity is acting downwards. The component of gravity which is of interest to you is in the direction opposite to the normal and centripetal force, and it equals G*cosA, where A is your angle of interest. (I assume you already drew a diagram.) Now, just solve Fcp + N = G*cosA, where the normal force N equals zero (the 'airborne' condition).
Hi Radou.

I get mg sin A for the component along that vector. Is A the angle to the horizontal?

Dorothy
 
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I was always told the normal force is perpinducar to the surface, so it would be pointing out from the center. Maybe I didn't clarify the problem well enough, but it is as if the skier is skiing off the top of a sphere.
Can you explain how the normal is causing the centripetal acceleration?
 

radou

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Mbaboy said:
I was always told the normal force is perpinducar to the surface, so it would be pointing out from the center. Maybe I didn't clarify the problem well enough, but it is as if the skier is skiing off the top of a sphere.
Can you explain how the normal is causing the centripetal acceleration?
Oh, sorry, I thought it was like the skier was skiing inside the sphere. :smile: Well, in that case, the normal force is pointing 'outside' the surface of the sphere. The centripetal force is pointing towards the center, as stated before, and gravity is pointing down in the vertical direction. So, you can write, pretty much as I already suggested, G*cosA + Fcp = N. You know that N = 0, so, solve for the angle A.

P.S. The normal force is the reaction from the surface onto the skier.
 
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Why is it gcosA and not mgcosA. Also, is Fcp the centripetal force? If so what is causing it? How can you solve for it because A can't be in terms of Fcp.
 

radou

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Mbaboy said:
Why is it gcosA and not mgcosA. Also, is Fcp the centripetal force? If so what is causing it? How can you solve for it because A can't be in terms of Fcp.
I didn't write g*cosA, I wrote G*cosA, where G = mg, so yes, you can write mg*cosA. Fcp is the centripetal force caused by circular motion.

Well, you don't know the centripetal force, right. But, you can apply the energy consetvarion theorem from the point at the top of the sphere to the point where the contact will be lost, to obtain the velocity the skier has at that point. By plugging in that velocity into the expression for the centripetal force (the masses will cancel), and by returning to G*cosA + Fcp = N, where N = 0, you can calculate the angle A. I hope I didn't miss something again this time. :rolleyes:
 
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So what I'm doing is mgh=(mv^2)/2 and h can be represented as the chord length 2rsin(.5A) times cosA. So basically the answer is one huge expression that I don't feel like writing.

radou said:
Fcp is the centripetal force caused by circular motion.
I still don't understand what you are saying there. Circular motion can't cause centripetal force. Objects in motion keep a straight velocity unless acted on my a force perpindicular to that velocity.
 

radou

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Mbaboy said:
So what I'm doing is mgh=(mv^2)/2 and h can be represented as the chord length 2rsin(.5A) times cosA. So basically the answer is one huge expression that I don't feel like writing.



I still don't understand what you are saying there. Circular motion can't cause centripetal force. Objects in motion keep a straight velocity unless acted on my a force perpindicular to that velocity.
I hope this helps: http://www.mcasco.com/p1cmot.html" [Broken].

Regarding the skiing problem, I solved it, and it's no huge expression.
 
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