Can a Skier Experience Weightlessness on a Parabolic Ski Slope?

[TyErd]

Homework Statement

The profile of a ski slope is shown below. The section QR inside the dotted box is parabolic in shape. The ski slope at point R is at 45degrees to the horizontal

A 75kg skier moves down the slope from P to Q and continues to R and beyond. Assume that the skier is a point mass, and air resistance and friction are negligible.



Question1. Is it possible for the skier to experience weightlessness while moving along the entire path QR. Explain.

Question 2. To achieve the situation in question 1, the skier must move at a certain speed at point Q. In terms of V, what is the magnitude of the vertical component of the skier at point R.

Homework Equations

W=mg and perhaps a=gsintheta

The Attempt at a Solution

acceleration due to gravity is 10m/s/s and has to equal gravitational field strength of 10 N/kg.
Using a=gsintheta, i substitue angle 45 to get 10sin45 which equals 7.07m/s/s. Not sure if this is correct so far. Dont know wehre to go now.

 

[Andrew Mason]

Homework Statement

The profile of a ski slope is shown below. The section QR inside the dotted box is parabolic in shape. The ski slope at point R is at 45degrees to the horizontal

A 75kg skier moves down the slope from P to Q and continues to R and beyond. Assume that the skier is a point mass, and air resistance and friction are negligible.



Question1. Is it possible for the skier to experience weightlessness while moving along the entire path QR. Explain.

Question 2. To achieve the situation in question 1, the skier must move at a certain speed at point Q. In terms of V, what is the magnitude of the vertical component of the skier at point R.

1. The question is asking whether it is possible that there would be no normal force acting on the skier during Q-R (ie. the skier is doing a jump and is in the air during Q-R). How would that occur?

2. Think of an object in free-fall. What is the initial vertical velocity at Q? What is it in terms of h - height fallen - after Q?

AM

 

[TyErd]

um, it asks for weightlessness for when it is moving along the path QR, doesn't that mean it should be in contact with the slope? correct me if I am wrong because I may have interpreted the question a different way.

 

[Andrew Mason]

um, it asks for weightlessness for when it is moving along the path QR, doesn't that mean it should be in contact with the slope? correct me if I am wrong because I may have interpreted the question a different way.

Whether he is in contact with the slope is the issue. It says is that he follows the slope from P to Q and continues to R and beyond. It does not say he follows the slope from Q to R. If he follows the slope from Q to R he obviously has a normal force acting on him so he is not weightless.

Since the second part of the question suggests that 1. is possible (ie. he is weightless for the entire Q-R portion) the interpretation I have given above must be correct. If your interpretation (that he follows the slope from Q-R) is correct then it is impossible that he is weightless during the Q-R portion.

As a further hint in solving this: assume that the skier at Q moving with a velocity v which is entirely horizontal experiences freefall from Q-R. What is the shape of his trajectory?

AM

 

[TyErd]

So the answer to question one is simply, it is only possible for the skier to experience apparent weightlessness is if their is no normal force thus he is in midair. While he is in midair the skier will accelerate at 9.8m/s/s which is equal to the Earth's gravitational field.

 

[TyErd]

for question 2, there is no initial vertical velocity so u=0, a=9.8m/s/s x=-2m and ten we find v. so the formula is v^2=u^2+2ax. Is this correct?

 

[TyErd]

x=2m sorry not -2m

 

[Andrew Mason]

for question 2, there is no initial vertical velocity so u=0, a=9.8m/s/s x=-2m and ten we find v. so the formula is v^2=u^2+2ax. Is this correct?

No. For the skier to be weightless from Q-R he has to land at R, not before. You have to work out the horizontal speed that will provide that horizontal range (4 m.) in the time it takes for the skier to fall the stated distance (you have to work it out from the fact that R is on a parabola - slightly less than 2 m.).

AM

 

[TyErd]

okay, so where would i begin, hmm..

 

[Andrew Mason]

okay, so where would i begin, hmm..

[Note: In looking at the diagram again I may have misled you. It looked to me like the curve turned up before R putting the apex after R. But I think it was intended to be a parabolic path downward with the apex at Q.

In that case, your original thought that the skier followed the curve will work if the free-fall path from Q to R is the same as the curve. There would be no normal force if the skier just about touches the slope, but not quite.

The analysis is the same as I suggested, it is just that he is in a free-fall path that is arbitrarily close to the slope. It can work because the free-fall path and the slope are parabolas opening downward.]

Write out the equation for the slope:

y = -kx^2 . Can you determine the value for k? Are you given the exact vertical distance below Q at R? It looks like it is almost 2 m. but not quite.

The freefall path is: y = -gt^2/2. What is t in terms of x and v (horizontal speed at Q)?

If you can determine k you can find v.

AM

 

[TyErd]

would the equation be y=-kx^2+2 ?

 

[TyErd]

and also why is it important that we find k?

 

[Andrew Mason]

would the equation be y=-kx^2+2 ?

Why +2? I would put the origin at Q.

and also why is it important that we find k?

Because the condition for weightlessness is $y = -gt^2/2 \ge -kx^2$ (ie. the free-fall path is at or above the slope. If it is below the slope, the slope pushes up on the skier (ie. preventing the skier from following the free-fall path).

AM

 

[Andrew Mason]

You can find k from the slope at R.

I get k = 1/8 and [itex]v =2 \sqrt{g} = 6.26 m/sec[/tex]

AM

 

[TyErd]

okay thankyou I get you, funnily enough I got the same answer using the method I described in post 6 however your way helped me understand the question so much more. Thankyou!

 

[Andrew Mason]

okay thankyou I get you, funnily enough I got the same answer using the method I described in post 6 however your way helped me understand the question so much more. Thankyou!

You are welcome. I just wanted to point out that the diagram is misleading.

The equation for the path in the QR rectangle is:

$$y = -x^2/8$$

so for x = 4 m, y = -2 m.

R has co-ordinates (4,-2) which is not how it is drawn.

AM