I came across this equation for balancing a pencil while solving some problems:
$$ml\ddot { \theta } =mg\theta $$
Where $l=$the length of the pencil, and $\theta$ is the angle it makes with vertical.
What I cannot understand is, why acceleration, $a=l\ddot { \theta }$ and not $\displaystyle \frac { l\ddot { \theta } }{ 2 } $?
Is the center of mass located at its top and not the center? Or is there something else I am missing?
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2 Answers
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What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2?
The equation you wrote doesn't mention anything about the linear acceleration.
Is the center of mass located at its top and not the center? Or is there something else I am missing?
The center of mass of the pencil is in the middle, not the top. There is likely something else you are missing. Or, rather, maybe there is something else the person who wrote what you are reading is missing.
The tipping pencil has many forces acting on it: the force of gravity acting at the center of mass; the normal force acting at the bottom; the frictional force acting at the bottom.
As far as I can tell, the easiest way to solve this problem is by applying a sum of torques analysis with the axis of rotation chosen as the bottom of the pencil. Then the sum of torques gives (the gravitation force acts at the center of mass): $$ \frac{mg l \sin(\theta)}{2} $$ and this is equal to $$ I\frac{d^2\theta}{dt^2} $$ where $I$ is the moment of inertia of a pencil about it bottom (not its middle, because the bottom was our choice for the axis of rotation) $$ I=\frac{ml^2}{3} $$ which, for small angles, gives $$ \frac{d^2\theta}{dt^2}=\frac{3}{2}\frac{g \theta}{l} $$ which is different from your equation by a factor of 3/2.
If you consider the pencil not to be a solid rod, but rather to be a point mass located at the center of mass (at l/2) then you get $I=ml^2/4$ and you recover the equation you had written.
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$\begingroup$ The answer, all over the internet, seems to be the one in the question. Links: thatsmaths.com/2014/06/26/balancing-a-pencil, youtube.com/watch?v=U3vAoJhIWms#t=29, arxiv.org/pdf/1406.1125v1.pdf $\endgroup$ Commented Feb 28, 2015 at 20:06
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$\begingroup$ The first link does exactly what I said at the bottom of my post. It models the pencil as an "inverted pendulum". I.e., a bob at the center of mass, which is ignoring the fact that it is actually a solid rod. $\endgroup$– hftCommented Feb 28, 2015 at 20:08
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$\begingroup$ Shouldn't it be $\frac{ml^2}{4}$? $\endgroup$ Commented Feb 28, 2015 at 20:09
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$\begingroup$ In those links "l" is the length of the pendulum. $\endgroup$– hftCommented Feb 28, 2015 at 20:12
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$\begingroup$ So, the L is really the length of the hypothetical pendulum you are replacing the pencil with? $\endgroup$ Commented Feb 28, 2015 at 20:17
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"Equation that is all over the internet"...
You started at http://thatsmaths.com/2014/06/26/balancing-a-pencil/
and from there, you linked to
https://arxiv.org/abs/1406.1125
which was the source for the former. In the third paragraph of that paper, it states
We model the pencil as an inverted simple pendulum with a bob of mass m at one end of a rigid massless rod of length $\ell$, the other end being fixed at a point.
...one end of a rigid massless rod....
Once you get to that line, the rest follows. "We didn't model the pencil as a pencil, so we didn't get the equation you were expecting".
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1$\begingroup$ I had seen the equation a couple more places, didn't have the link (old downloaded PDFs). They didn't contain a proper explanation and I overlooked those lines in the other links in a hurry. My mistake. $\endgroup$ Commented Mar 1, 2015 at 6:02