Inconsistent problem about centrifugal/contact force

[FranzDiCoccio]

Homework Statement

A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of r = 36 m. Neglect friction and air resistance. What must be the height h of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

Relevant Equations

conservation of mechanical energy
Newton's second law

This is problem 49 in chapter 6 of "Physics - 9th edition". A similar question was asked here several years ago (although with a different height).
The figure is below. I added point A and angle $\theta$.
The solution is pretty easy. For the purpose of my discussion I'm assuming that the height is zero at the "center" of the rightmost hill, and consider a generic point A on such hill, instead of its crest. The radius going through A forms an angle $\theta$ with the vertical direction. The answer to the question can be obtained by setting $\theta =0$ in the solution.
The condition for losing contact at A is that the normal force there is null, which means that the centrifugal force equals the projection of the weight along the radial direction. This gives $v^2 = g r \cos \theta$.
Conservation of total mechanical energy requires that $2 g h_0 =2 g r \cos \theta + v^2$, where $h_0$ is the height of the leftmost hill wrt the center of the leftmost hill. Substituting the first equation into the second one we get
$$h_0 = \frac{3}{2} r \cos \theta$$
Since the height of point A is $r \cos \theta$, the starting point should be
$$h = \frac{r}{2} \cos \theta$$
above A for the skier to lose contact at A.
If $\theta=0$ one gets $h = r/2 = 18\, {\rm m}$, which is the solution given by the book.

The problem in my opinion is that $h$ is a decreasing function of $\theta$. This means that the height for losing contact at an angle $\theta>0$ is less than $18\, {\rm m}$. This in turn means that if the first hill is $18\, {\rm m}$ above the second, the skier will definitely lose contact before reaching the crest of the second hill.

So, if I'm correct, the proposed exercise is not very hard, and in a way it's nice, but it does not make a lot of sense.

 

[haruspex]

Nice point, but it does only say the crest is circular. In the limit, that reduces to saying merely what the radius of curvature is there; saying it is "circular" is meaningless without specifying some range, or saying something about the third derivative.
On that basis, the radius of curvature could be sufficiently greater at all previous points to retain contact.

 

[FranzDiCoccio]

I'm actually not very clear on what is meant here by "crest of the hill". I interpreted that as "the uppermost portion of the hill", not just its very top. That is, I assumed that the (section of the) top of the hill can be described as a circular arc, which is what the figure suggests.
In that case, the skier will jump at the starting point of that circular arc (at the start of the range, if I get your meaning for this word right).
In a situation like the one depicted in the figure (which is not so unrealistic for a ski run, by the way) the skier will jump well below (and before) the top of the hill.

It would be interesting to find the function describing the convex part of the hill such that the contact is lost only at its highest point. I am not able to picture it. I cannot shake the idea that the contact would be lost below the top anyway. But of course I can be wrong.
I'll think about that.

 

[Lnewqban]

I believe that you are correct.
The value of the tangential velocity of the skier at point A is greater that at the top of the second hill, and being square, it has an important influence.
If we imagine the angle being at 90 degrees, the velocity of the skier would be greater at the h+r point, having no reason to deviate from a vertical trajectory, regardless of the r value.

 

[haruspex]

I interpreted that as "the uppermost portion of the hill", not just its very top.

Quite so, but my point is that in the absence of any specified length of that portion it can be arbitrarily short.

It would be interesting to find the function describing the convex part of the hill such that the contact is lost only at its highest point.

Consider the boundary case, where it just remains in contact over some section. If you compare it with a certain other scenario you can see the answer immediately.

 

[FranzDiCoccio]

first of all, thanks for your time and help.

Quite so, but my point is that in the absence of any specified length of that portion it can be arbitrarily short.

Ok, in a way the subtext of the original problem is "just focus on the very top of the hill, without bothering about what might happen elsewhere".
The fact that I'm not able to picture in my mind the shape of the hill such that nothing actually happens elsewhere is really nagging me :)

Consider the boundary case, where it just remains in contact over some section. If you compare it with a certain other scenario you can see the answer immediately.

You make it sound pretty easy. There must be something I'm not seeing.

I'm assuming that the function giving the outline of the hill has two inflections, where the sign of $y''$ changes. Specifically, it is positive before the first inflection point, negative between the first and the second inflection points and then again positive.
Of course, as long as $y''>0$ there is no problem with contact, because the centrifugal force is actually helping with it.
As soon as $y''<0$ the centrifugal force might cause the loss of contact. The problem is a bit messy, though, since different effects should be taken into account. One of them is of course the change in the radius of curvature. But there is also the fact that the skier is climbing a slope, and hence his speed is decreasing with height. Finally, the fraction of the skier's weight actually helping him with staying in contact with the snow depends on the slope of the hill.

I thought of tackling the problem through a differential equation, but at first sight it looks a bit complex, and right now I have not the time to look into it.

As I mention, your last comment seems to suggest that the answer is easy to find, but so far I'm not seeing it.At this point I'm just curious of what the shape of the hill would be.

 

[jbriggs444]

At this point I'm just curious of what the shape of the hill would be.

As I understand it, you are searching for the shape of a curve where the supporting force from the hill is arbitrarily small. Ballistic, actually.

 

[haruspex]

If there is no contact force the hill might as well not be there.

 

[FranzDiCoccio]

Ok so, the "other scenario" was a hill shape on which the contact force is absent.
Actually, I should have thought of that!

So if the central part of the hill is a parabolic arc, then there is a critical velocity for the skier at the inflection point connecting such arc with the concave part of the ski run.
If the skier is faster than that, he never touches the hill until he reaches the next inflection point.
If the skier is slower, he won't lose contact.
So I need to check that it's possible for the skier to be slower than the critical value at the inflection point, and yet faster than the critical value at the top of the hill.

(I edited below this point, because the previous version contained a few silly mistakes)

I did some quick calculations assuming that the top of hill is described by the function
$$y=-\frac{x^2}{2R}+R$$
for $-x_0\leq x\leq x_0$, where $R$ is the radius of curvature of such shape at $x=0$. If my calculations are right and that was a ballistic trajectory, the velocity at $\pm x_0$ would be such that
$$v_0^2=2Rg \left(1+\frac{x_0^2}{R^2}\right)$$
On the other hand, for the velocity at $x=0$ on the hill to be $v^2=g R$, it should be $v_0^2=g R+g\frac{x_0^2}{R}$ at $-x_0$.
Thus the condition to be met appears to be
$$g R+g\frac{x_0^2}{R}< 2Rg \left(1+\frac{x_0^2}{R^2}\right)=2 \left(gR+g\frac{x_0^2}{R}\right)$$
which is always true.
After all this is also apparent from the expression of $v_0$.

At this point (even later in the night) it appears that with a parabolic hill, either the skier is always in contact with the snow (including at the top of the hill), or he is never in contact with it.
So i still cannot figure out the shape of the hill...
Better go to bed.
Many thanks to you all, I'll think again about this when I'm fresher.

 

[Lnewqban]

 

[FranzDiCoccio]

ok... the dirtbike jumps before the top... That is also my intuition. But I think the hills are not shaped for avoiding that effect.
I am wondering whether it's (theoretically) possible to build a hill where the motorbike would lose contact at the topmost point only, like the skier in the problem.

One possibility is perhaps to add a $x^4$ term in the function above. This should not change the curvature radius at the top of the hill, but changes its shape.

 

[Lnewqban]

ok... the dirtbike jumps before the top... That is also my intuition. But I think the hills are not shaped for avoiding that effect.
I am wondering whether it's (theoretically) possible to build a hill where the motorbike would lose contact at the topmost point only, like the skier in the problem.

One possibility is perhaps to add a $x^4$ term in the function above. This should not change the curvature radius at the top of the hill, but changes its shape.

For a body without self-propulsion (like our skier), the profile of that uphill could be very close to the shape of the flight path of a projectile which launch angle is 90°-θ and launch velocity equals the skier's natural velocity at point A.

The shape around the topmost point would need to have a smaller height than the vertex of the parabola (projectile flight path) in order to produce some "air time".

For a self-propelled vehicle, the energy produced by the engine or motor between point A and the topmost point would need to be considered.

 

[FranzDiCoccio]

For a body without self-propulsion (like our skier), the profile of that uphill could be very close to the shape of the flight path of a projectile which launch angle is 90°-θ and launch velocity equals the skier's natural velocity at point A.

The shape around the topmost point would need to have a smaller height than the vertex of the parabola (projectile flight path) in order to produce some "air time".

Yes, I understand this. Only, we cannot lower the topmost point alone, so the "lowering" (and the loss of contact) would start below that point. Of course we can argue that the "jumping point" is arbitrarily close to the topmost point.
I was just trying to imagine a reasonably simple smooth function with the require property. But perhaps it does not exist.

For a self-propelled vehicle, the energy produced by the engine or motor between point A and the topmost point would need to be considered.

Ok, yes. In fact I think that with a self propelled vehicle we can safely assume that the hill can be climbed at the constant velocity that ensures the loss of contact exactly at the topmost point.