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- 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 [itex] \theta [/itex].

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 [itex] \theta [/itex] with the vertical direction. The answer to the question can be obtained by setting [itex] \theta =0 [/itex] 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 [itex] v^2 = g r \cos \theta [/itex].

Conservation of total mechanical energy requires that [itex]2 g h_0 =2 g r \cos \theta + v^2 [/itex], where [itex] h_0 [/itex] is the height of the leftmost hill wrt the center of the leftmost hill. Substituting the first equation into the second one we get

[tex] h_0 = \frac{3}{2} r \cos \theta [/tex]

Since the height of point A is [itex] r \cos \theta[/itex], the starting point should be

[tex] h = \frac{r}{2} \cos \theta [/tex]

above A for the skier to lose contact at A.

If [itex]\theta=0 [/itex] one gets [itex] h = r/2 = 18\, {\rm m} [/itex], which is the solution given by the book.

The problem in my opinion is that [itex] h [/itex] is a decreasing function of [itex] \theta [/itex]. This means that the height for losing contact at an angle [itex] \theta>0 [/itex] is less than [itex] 18\, {\rm m} [/itex]. This in turn means that if the first hill is [itex] 18\, {\rm m} [/itex] 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.

The figure is below. I added point A and angle [itex] \theta [/itex].

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 [itex] \theta [/itex] with the vertical direction. The answer to the question can be obtained by setting [itex] \theta =0 [/itex] 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 [itex] v^2 = g r \cos \theta [/itex].

Conservation of total mechanical energy requires that [itex]2 g h_0 =2 g r \cos \theta + v^2 [/itex], where [itex] h_0 [/itex] is the height of the leftmost hill wrt the center of the leftmost hill. Substituting the first equation into the second one we get

[tex] h_0 = \frac{3}{2} r \cos \theta [/tex]

Since the height of point A is [itex] r \cos \theta[/itex], the starting point should be

[tex] h = \frac{r}{2} \cos \theta [/tex]

above A for the skier to lose contact at A.

If [itex]\theta=0 [/itex] one gets [itex] h = r/2 = 18\, {\rm m} [/itex], which is the solution given by the book.

The problem in my opinion is that [itex] h [/itex] is a decreasing function of [itex] \theta [/itex]. This means that the height for losing contact at an angle [itex] \theta>0 [/itex] is less than [itex] 18\, {\rm m} [/itex]. This in turn means that if the first hill is [itex] 18\, {\rm m} [/itex] 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.

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