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here's a short simple answer: the Hamiltonian of the pencil can be approximated by an inverted harmonic oscillator near the equilibrium (downward parabola) . It's an easy exercise to solve.

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Instability occurring when a heavy fluid is superposed on a lighter one was first studied by Lord Rayleigh in 1883. The nature of this instability does not change when the problem is posed as a lighter fluid accelerating against a heavier one. G I Taylor first investigated it in this latter sense. The second paper in this reply will answer your question in ...

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The question is so ambiguous that it allows a resounding yes. This is because "balance" is not defined, neither are the dimensions and the material used for the pencil, nor the location of where the "balancing" is to happen. The material and the shape of the surface to balance the pencil on, are not specified, nor the length of time it should stay balanced. ...

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Let me be the firs to answer 'Yes' (more or less). As the saying goes: In theory there is no difference between theory and practice. In practice there is. What I'm getting at is that there will always be differences between theory and practice, and that it is up to the physicist to decide which assumptions/simplications are suitable and which ones are ...

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No. The weight of the pencil is roughly 1 Newton, and the area is about 500 square picometer (5 * 10-22) which means the pressure on the tip is around 2 ZettaPascal. That's quite a bit more than what graphite (or diamond) can withstand (that's masured in GigaPascal)

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No. Firstly, the point of the pencil is generally not sharp enough to have just one single atom. People attempt to make that kind of tip in STM's. Even if you somehow did manage to get it sharp enough, graphite is so soft the the weight of the pencil will crush the tip. It won't stay a single atom wide. So there is no way to balance the pencil on a single ...

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TL;DR: there are many factors that prevent a pencil remaining perfectly balanced. The most important of these is the uncertainty principle that will make the pencil fall over in less than four seconds. For details, read on... Short answer: NO. The first photon of light that hits it would disturb your perfect equilibrium. The moon's tidal forces (which are ...

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No. To balance perfectly, the pencil would have to be perfectly upright and perfectly still. The uncertainty principle limits how well you can do both at the same time. Momentum and position form a conjugate pair. $\Delta x \Delta p \geq \hbar$. Angular momentum and angular position form one too. $\Delta L \Delta \Theta \geq \hbar$ This doesn't guarantee ...

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You have ignored outside influences. Something as simple as asteroids striking the outside of the sphere would push it out of concentric balance and cause a drift toward eccentricity, eventually leading to a contact between the star and the shere (or more likely a vaporization of part of the sphere well before a contact). There would need to be active ...

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