I think I 'understand' the concept of moment of inertia, but I don't see why it is proportional to r^2 (for a given point). Is the equation just empirical?
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2It comes from the definition of angular momentum when you enforce the form $L = I\omega$ – Phoenix87 Feb 22 '16 at 09:45
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It also comes from Newton's 2nd law in a rotating reference frame. – David White Feb 22 '16 at 12:00
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If we apply a force $F$ to a mass $m$ then the acceleration is given by Newton's second law:
$$ F = ma $$
and for angular motion we expect a similar equation:
$$ T = I\ddot{\theta} $$
where $T$ is the torque and $I$ is the moment of inertia.
Now suppose we are considering angular motion about a point $P$ a distance $r$ away measured perpendicular to the force:
The torque about the point $P$ is:
$$ T = Fr \tag{1} $$
The angular acceleration about $P$ is:
$$ \ddot{\theta} = \frac{a}{r} = \frac{F/m}{r} $$
and using equation (1) to substitute for $F$ gives:
$$ \ddot{\theta} = \frac{T}{mr^2} $$
Comparing this with our expected equation for angular acceleration:
$$ T = I\ddot{\theta} $$
We get:
$$ I= mr^2 $$
John Rennie
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