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The following diagram shows the situation described in the question:

enter image description here

I learned that at the moment the object loses contact with the sphere, there won't be a normal force acting on the object. However, I don't quite understand how this is achieved, as the displacement of the object isn't in the same direction as the force of gravity at that moment. Thus, shouldn't the component of gravity that is perpendicular to the object keep the object on the sphere, which in return resulting the surface to exert a normal force on the object. Furthermore, after the object becomes airborne, does it travel in a circular path until hitting the floor (assuming no air resistance)?

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3 Answers 3

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The component of gravity in the normal direction does pull the object toward the sphere, but in order to stay on the sphere, the object requires a centripetal force of $F_\mathrm{c} = \frac{m v^2}{r}$. Applying Newton's Second Law in the normal direction gives us $$F_{\mathrm{g,N}} - N = F_\mathrm{c}$$ where $F_{\mathrm{g,N}}$ is the normal component of gravity and $N$ is the normal force. Since $N$ is necessarily nonnegative, the object will not be able to stay on the circular trajectory if $F_\mathrm{c}>F_{\mathrm{g,N}}$, and since $F_{\mathrm{g,N}}$ decreases while $F_\mathrm{c}$ increases as the object goes down the sphere, the object leaves the surface at the point where $F_\mathrm{c}=F_{\mathrm{g,N}}$.

Once the object is no longer in contact with the sphere, gravity is the only force acting on it, so it follows the tradition projectile trajectory of a parabola.

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  1. The component of gravity that is perpendicular to the surface of the sphere acted purely as the centripetal force at the point it loses contact
  2. No, it will travel in a parabolic curve as only gravity is acting on it
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  • $\begingroup$ For #2, won't it just fall straight down, since it loses contact when the sphere goes vertical? $\endgroup$
    – Dave
    Commented Apr 30, 2020 at 1:43
  • $\begingroup$ the point it loses contact is not the point the slope of the sphere is perpendicular to the ground; it will seperate before this point due to the inability for the component of gravity (that points to the center of the sphere) to provide enough centripetal force to maintain circular motion $\endgroup$ Commented Apr 30, 2020 at 1:46
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It will still be normal. That is, it will be pointing exactly to the left. But it’s magnitude will have fallen to 0. So at that point it doesn’t really matter which way it points.

0 is a sneaky number.

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