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The parallel axis theorem states that $I_{\rm new} = I_{\rm original} + Mh^2$ where $h$ is the displacement from the original axis of rotation. However, I'd like to describe a situation which proves that this doesn't always work.

Say we have an equilateral triangle with one side pointed straight upwards and the vertex to the right:

enter image description here

Say that we take this vertical length as the axis of rotation. $I = \frac{1}{8}Ma^2$ then where $a = \rm side\ length$. Say we move the axis of rotation $a$ length $h$ to the left (situation 1) and then again $a$ length $h$ to the right (situation 2) with $h$ being equal to the height of the triangle. In both cases, according to the formula, we end up with the same final moment of inertia.

enter image description here

Situation 1: The longest side (with most mass) is $h$ away from the axis of rotation, and the vertex is $2h$ away. Situation 2: The longest side (with most mass) is $h$ away from the axis of rotation, and the vertex is $0h$ away.

The moment of inertia must be smaller in situation 2. Clearly the formula can't always be applied. Could someone please describe the situations in which it can and can't?

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  • $\begingroup$ Hi Edward. I've added diagrams to illustrate what I think you mean. If the diagrams are wrong please feel free to roll back my edit. $\endgroup$ Jan 23, 2018 at 9:56
  • $\begingroup$ Hey @Qmechanic. Thanks so much, they’re perfect. I would have had I known how. $\endgroup$ Jan 23, 2018 at 11:59
  • $\begingroup$ Hi Edward Garemo: That was @John Rennie who did the diagrams, not me! $\endgroup$
    – Qmechanic
    Jan 23, 2018 at 12:10

1 Answer 1

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The parallel axis theorem states:

Suppose a body of mass $m$ is made to rotate about an axis $z$ passing through the body's centre of gravity. The body has a moment of inertia $I_\text{cm}$ with respect to this axis. The parallel axis theorem states that if the body is made to rotate instead about a new axis $z′$ which is parallel to the first axis and displaced from it by a distance $d$, then the moment of inertia $I$ with respect to the new axis is related to $I_\text{cm}$ by

$$ I = I_\text{cm} + md^2 $$

Note the text I've emboldened. You cannot apply the parallel axis theorem to your initial situation because the axis does not pass through the centre of gravity.

Let's denote the distance from your original rotation axis to the centre of mass by $d$:

enter image description here

So that $I_\text{cm} = I - md^2$. Then the moment of inertia for your situation $1$ is:

$$ I_1 = I - md^2 + m(h+d)^2 $$

And the moment of inertia for your situation $2$ is:

$$ I_2 = I - md^2 + m(h-d)^2 $$

And as you rightly say $I_1 \ne I_2$.

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