Bounty Ended with 50 reputation awarded by HyperLuminal occurred Jul 25 '15 at 2:32 8 deleted 1 character in body edited Apr 19 '15 at 0:30 Floris 108k1111 gold badges195195 silver badges331331 bronze badges Modeling the pencil as a uniform rod with mass $$m$$, length $$\ell$$, moment of inertia $$I=\frac{1}{12}m\ell^2$$$$I=\frac{1}{3}m\ell^2$$, the torque $$\Gamma$$ when it is at an angle $$\theta$$ to the vertical is Modeling the pencil as a uniform rod with mass $$m$$, length $$\ell$$, moment of inertia $$I=\frac{1}{12}m\ell^2$$, the torque $$\Gamma$$ when it is at an angle $$\theta$$ to the vertical is Modeling the pencil as a uniform rod with mass $$m$$, length $$\ell$$, moment of inertia $$I=\frac{1}{3}m\ell^2$$, the torque $$\Gamma$$ when it is at an angle $$\theta$$ to the vertical is 7 deleted 5 characters in body edited Apr 17 '15 at 22:12 Floris 108k1111 gold badges195195 silver badges331331 bronze badges AFTERWORD my daughter just pointed me to an interesting post that calculates the same thing - and comes up with a very similar answer. The author signs him/her selfhimself as "Alemi". There is a contributor on this site with the same handle. I think I recognize the style of the thought process, so I am going to give a belated tip of the hat. Incidentally, that post comes up with a value of about 3.6 seconds. Which is astonishingly similar to the value I got. AFTERWORD my daughter just pointed me to an interesting post that calculates the same thing - and comes up with a very similar answer. The author signs him/her self as "Alemi". There is a contributor on this site with the same handle. I think I recognize the style of the thought process, so I am going to give a belated tip of the hat. Incidentally, that post comes up with a value of about 3.6 seconds. Which is astonishingly similar to the value I got. AFTERWORD my daughter just pointed me to an interesting post that calculates the same thing - and comes up with a very similar answer. The author signs himself as "Alemi". There is a contributor on this site with the same handle. I think I recognize the style of the thought process, so I am going to give a belated tip of the hat. Incidentally, that post comes up with a value of about 3.6 seconds. Which is astonishingly similar to the value I got. 6 Fixed the calculation of I (I=\frac13 m \ell^2 not 1/12... since pivoting at the end.) edited Apr 17 '15 at 21:51 Floris 108k1111 gold badges195195 silver badges331331 bronze badges 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 30four seconds. For details, read on...$$I\ddot\theta = \frac12 m g \ell \theta\\ \ddot\theta = \frac{6 m g}{\ell}\theta$$$$I\ddot\theta = \frac12 m g \ell \theta\\ \frac13 m \ell^2 \ddot\theta = \frac12 m g \ell \theta \ddot\theta = \frac{3 g}{2\ell}\theta$$ Putting $$\frac{6 m g}{\ell} = \alpha^2$$$$\frac{3 g}{2 \ell} = \alpha^2$$, we can integrate this twice to get a $$C_1 = \frac{v_0}{\ell \alpha} = \frac{v_0}{\ell\sqrt{6 m g / \ell}} = \frac{v_0}{\sqrt{6 m g \ell}}$$$$C_1 = \frac{v_0}{\ell \alpha} = \frac{v_0}{\ell\sqrt{3 g /2 \ell}} = \frac{v_0}{\sqrt{\frac32 g \ell}}$$$$= \sqrt{\frac{\ell}{6 m g}}\sinh^{-1}\left(\frac{\theta \sqrt{6 m g \ell}}{v_0}\right)$$$$= \sqrt{\frac{2\ell}{3 g}}\sinh^{-1}\left(\frac{\theta \sqrt{\frac32 g \ell}}{v_0}\right)$$ Let's assuming that "definitely falling" corresponds to an angle of 0.5 degrees, or about 0.01 rad. We can put the values in the above equation, and find t $$\approx$$ 426 s. One photon. Forty-twoSix seconds. 42 That is indeed the answer to the ultimate questiona shockingly short time. And pencils cannot be balanced on their tip.. but it's about to get worse:$$\Delta v = 1.6\cdot 10^{-16}m/s$$$$\Delta v = 4\cdot 10^{-16}m/s$$$$t = \frac{1}{\alpha}\sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)\approx 26 s$$$$t = \frac{1}{\alpha}\sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)\approx 3.7 s$$ So a perfectly balanced, "theoretical" pencil that stands on its mono-atomic tip, will fall on average in under half a minutehandful of seconds because of the uncertainty principle.AFTERWORD my daughter just pointed me to an interesting post that calculates the same thing - and comes up with a very similar answer. The author signs him/her self as "Alemi". There is a contributor on this site with the same handle. I think I recognize the style of the thought process, so I am going to give a belated tip of the hat. Incidentally, that post comes up with a value of about 3.6 seconds. Which is astonishingly similar to the value I got. 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 30 seconds. For details, read on...$$I\ddot\theta = \frac12 m g \ell \theta\\ \ddot\theta = \frac{6 m g}{\ell}\theta$$ Putting $$\frac{6 m g}{\ell} = \alpha^2$$, we can integrate this twice to get a $$C_1 = \frac{v_0}{\ell \alpha} = \frac{v_0}{\ell\sqrt{6 m g / \ell}} = \frac{v_0}{\sqrt{6 m g \ell}}$$$$= \sqrt{\frac{\ell}{6 m g}}\sinh^{-1}\left(\frac{\theta \sqrt{6 m g \ell}}{v_0}\right)$$ Let's assuming that "definitely falling" corresponds to an angle of 0.5 degrees, or about 0.01 rad. We can put the values in the above equation, and find t $$\approx$$ 42 s. One photon. Forty-two seconds. 42 is indeed the answer to the ultimate question. And pencils cannot be balanced on their tip.$$\Delta v = 1.6\cdot 10^{-16}m/s$$$$t = \frac{1}{\alpha}\sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)\approx 26 s$$ So a perfectly balanced, "theoretical" pencil that stands on its mono-atomic tip, will fall on average in under half a minute because of the uncertainty principle. 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...$$I\ddot\theta = \frac12 m g \ell \theta\\ \frac13 m \ell^2 \ddot\theta = \frac12 m g \ell \theta \ddot\theta = \frac{3 g}{2\ell}\theta$$ Putting $$\frac{3 g}{2 \ell} = \alpha^2$$, we can integrate this twice to get a $$C_1 = \frac{v_0}{\ell \alpha} = \frac{v_0}{\ell\sqrt{3 g /2 \ell}} = \frac{v_0}{\sqrt{\frac32 g \ell}}$$$$= \sqrt{\frac{2\ell}{3 g}}\sinh^{-1}\left(\frac{\theta \sqrt{\frac32 g \ell}}{v_0}\right)$$ Let's assuming that "definitely falling" corresponds to an angle of 0.5 degrees, or about 0.01 rad. We can put the values in the above equation, and find t $$\approx$$ 6 s. One photon. Six seconds. That is a shockingly short time... but it's about to get worse:$$\Delta v = 4\cdot 10^{-16}m/s$$$$t = \frac{1}{\alpha}\sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)\approx 3.7 s$$ So a perfectly balanced, "theoretical" pencil that stands on its mono-atomic tip, will fall on average in a handful of seconds because of the uncertainty principle.AFTERWORD my daughter just pointed me to an interesting post that calculates the same thing - and comes up with a very similar answer. The author signs him/her self as "Alemi". There is a contributor on this site with the same handle. I think I recognize the style of the thought process, so I am going to give a belated tip of the hat. Incidentally, that post comes up with a value of about 3.6 seconds. Which is astonishingly similar to the value I got. 5 added 2731 characters in body edited Apr 16 '15 at 16:00 Floris 108k1111 gold badges195195 silver badges331331 bronze badges 4 fixed factor 2 in the conversion from exponential notation to sinh edited Apr 15 '15 at 12:44 Floris 108k1111 gold badges195195 silver badges331331 bronze badges 3 added information about graphite edited Apr 14 '15 at 16:10 Floris 108k1111 gold badges195195 silver badges331331 bronze badges 2 added detailed calculation edited Apr 14 '15 at 12:55 Floris 108k1111 gold badges195195 silver badges331331 bronze badges 1 answered Apr 14 '15 at 2:22 Floris 108k1111 gold badges195195 silver badges331331 bronze badges