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DanielSank
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A small query regarding the equation for balancing Balancing a pencil

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Qmechanic
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I came across this equation for balancing a pencil while solving some problems:
  $ml\ddot { \theta } =mg\theta $
$$ml\ddot { \theta } =mg\theta $$ Where $l$=the$l=$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 }$$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?

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?

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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A small query regarding the equation for balancing a pencil

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?