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Let $P$ be a particle of mass $m$ moving frictionlessly on the inside of an inverted cone with apex $O$ and angle of ascent $\alpha$. $P$ is subjected only to the influence of gravity and the normal force provided by the inside surface of the cone. Suppose further that that the initial velocity vector is horizontal and has magntitude $\omega_o$ and that the initial height of the particle $P$ above $O$ is $z_0$.

If we express the position of $P$ in cylindrical coordinates with $O$ at the origin and the axis of the cone being the $z$-axis, then the initial conditions are $z=z_0$, $\theta=0$ and $z'=0,\theta'=w_0$; note that the initial conditions for $r$ are determined by those for $z$ since $r=z\tan(\alpha)$.

What is the motion of the particle?

If $\omega_o=0$ then the motion will be a straight line towards $O$. Otherwise we can prove mathematically that the motion will be (periodic?) a bounded path which never reaches $O$. In otherwords, even if $\omega_o=0$ is very small, if $P$ begins at height $z_0$, it will spiral towards some minimum height $z_{min}$ (all the while increasing speed) and then spiral back up to height $z_0$.

This is highly counterintuitive for me. Why doesn't the particle just spiral towards $O$?

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  • $\begingroup$ I'm having trouble visualizing this situation, but I'm very interested. Would you be willing to provide a diagram? $\endgroup$ May 8, 2017 at 1:27

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The answer is the same reason the planets don't fall into the sun: as the particle gets closer to the center, its angular velocity goes up by angular momentum conservation. At a certain point, this costs so much energy that it's impossible to get any closer; we say the particle is 'repelled by the angular momentum barrier'. (For a real cone, friction would reduce the angular momentum over time, allowing the particle to spiral in.)

More quantitatively, the angular momentum is $$L = mv_\theta r$$ where $v_\theta$ is the tangential speed. Rearranging a little, we find that the kinetic energy due to this tangential motion is $$T_\theta = \frac{L^2}{2mr^2}.$$ That is, the energy cost of getting closer to the center diverges as $1/r^2$. In your situation, the potential energy only decreases linearly as we get closer to the center, $U \sim -r$, so eventually the angular momentum barrier will win and the particle can't get closer.

In the case of a planet orbiting the Sun, the potential energy diverges as $-1/r$, which still is beaten by the angular momentum barrier, since $1/r^2$ diverges faster. In general relativity, there is an additional term proportional to $-1/r^2$ in the potential, so things can fall into black holes.

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  • $\begingroup$ Ok. Now let's take a surface given by form $r=(-z)^n$ defined for $z<0$ with $0>n > -0.5$ Then $U=mgz=-mgr^{1/n}$, so U diverges to negative infinity faster than $-1/r^2$. Then we can have the possibility to fall into the "black hole" at negative infinity, right? $\endgroup$ May 9, 2017 at 15:43
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In real world of course it would spiral in and down because of friction and losing energy to heat and slowing down. Also if you imagine the ball rolling around at a given height in a perfect circle the speed would need to be chosen perfectly to balance forces ... Gravity plus the reaction force ( perpendicular to the cone if object free to slide ) must equate to the central force for circular motion. Since gravity and central force here are mutually at right angles we can easily relate them by vectors. The speed of rotation is readily determined from the angle and the height. If the object goes faster than this speed then it is intuitively clear that the object is going to rise up the cone and not fall down! As it spirals higher it will lose KE being converted to gravitational PE and then start coming down again and so on. Hope that is intuitive explanation which can back up with the equations if you like.

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