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A skier rolling down a frictionless hill where she should fly off tangentially and lose contact with ground.

In answer of this question, solution states that $mg\cos x = mv^2 / R$ at that point (to make normal reaction zero).

But I think $mg\cos x$ and centripetal force both act inwards towards the circle. So normal reaction is sum of those two forces and not zero.

Where am I missing something important?

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    $\begingroup$ Centripetal force is what opposes the centrifugal pseudoforce. The component of gravity is your centripetal force in this case. $\endgroup$ – JMac Apr 11 '19 at 13:02
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If there is an outward normal force of $N$ on the skier then the net (outward) force on the skier is $N-mg\cos \alpha$. If the skier is moving in a circle with radius $R$ and speed $v$ then

$N-mg\cos \alpha = -\frac{mv^2}{R} \\ \Rightarrow N = mg \cos \alpha - \frac{mv^2}{R}$

But we must have $N \ge 0$ (the skier does not have magnetic boots !). So if the skier is in contact with the slope then $mg \cos \alpha \ge \frac{mv^2}{R}$.

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  • $\begingroup$ That's it.. Thank you so much! $\endgroup$ – user68153 Apr 11 '19 at 14:28
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The centripetal force required to keep the skier on her circular path is, and to prevent her from flying off, is:

$$F_{centripetal} = m \frac{v^2}{R}$$

The only force that can act as centripetal force is gravity. The gravitational force that the skier actually feels in the direction of the center is:

$$F_{gravitational} = m g \cos \alpha$$

While that force is greater than the required centripetal force, the skier will stick to the ground. Then the ground will generate an opposite force equal to the difference between them to prevent the skier from sinking into the ground. The resulting normal force is zero.

Once the $F_{gravitational}$ becomes equal to or less than the required $F_{centripetal}$, the skier will fly off.

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$mgcos(\alpha) = mv^2/R$ does not mean that the forces are opposite to each other and balance themselves out.

You got the wrong idea here.

Consider the situation from scratch. The skier is going to slip because it says so and if he or she is falling from there, by analysis you know that $mgcos(\alpha)$ exists in the direction pointing to the centre. So, that should be the centripetal force. That is what they have meant.

And as for the normal force $N$, it will surely be zero if the skier is going to leave the surface. We assume that to mean that the skier is very close to getting out of there, which is literally like half out of the surface. But the analysis is all done on the basis of the skier being considered as a point mass, so that clearly meant $N=0$ anyway.

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  • $\begingroup$ Sorry but I didn't get the second part. If mgcosx acts as the centripetal force, how will N be equal to zero? $\endgroup$ – user68153 Apr 11 '19 at 14:13
  • $\begingroup$ Well, if the skier is going to leave the surface, then the normal reaction has to be zero shouldn't it be because he is practically in thin air? $\endgroup$ – Karthik Apr 11 '19 at 14:21
  • $\begingroup$ Well, I mean why exactly at that point? $\endgroup$ – user68153 Apr 11 '19 at 14:27
  • $\begingroup$ Normal force is the contact force between the skier and the surface! If there is no contact, then there cannot be any normal force. $\endgroup$ – Karthik Apr 11 '19 at 14:41

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