Moment of Inertia and Rotational Dynamics? [closed]

I'm having problems with the intuition behind the Parallel axis theorem and the Perpendicular axis theorem.

I'm self studying Mechanics for the British Curriculum but, the book I've is missing the moment of inertia part. It's supposed to cover simple systems and doesn't contain any calculus.

I need a small explanation of the two theorems if possible. No derivation, please.

Sorry, if my question isn't appropriate for this site.

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closed as unclear what you're asking by David Z♦Nov 4 '13 at 3:08

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

I can't see what your question is; the wikipedia page pretty much explains both theorems with the formulae quite well IMO. For developing some intuition of the Parallel axis theorem, I suggest your read up angular momentum and check out the Ice Skater Angular momentum experiment (that's not the exact name, but it's somewhat like it). bsharp.org/physics/spins – shortstheory Nov 4 '13 at 3:06
Note that the wikipedia articles don't do much explaining beyond stating and proving the theorems. My understanding of @user102148's question was that he wanted an intuitive explanation of what the theorems are saying and why they are true, and the book he was using (and wikipedia) did not provide these. Is this right? – NowIGetToLearnWhatAHeadIs Nov 4 '13 at 17:53
@NowIGetToLearnWhatAHeadIs Yes, thats what I meant. Wikipedia doesn't provide a clear enough explanation for me, since English is my second language and I don't have a strong enough mathematical background (Grade 11 here). Thank you, for clearing things up. – Lenol Nov 4 '13 at 19:35

1 Answer

I will explain both theorems in turn.

Parallel Axis Theorem

I will start by explaining the parallel axis theorem. The parallel axis theorem relies on the fact that there are two kinds of angular momentum an object can have, which I will call orbital angular momentum and spin angular momentum and that these add in a simple way.

First I will explain oribital angular momentum. Let's consider a sphere of radius $r$ and mass $m$ going around the origin in a circle of radius $R$, but not changing orientation (so that the same point on the sphere always points to the right). What is the angular momentum? It is $L_{\mathrm{oribital}} = mvR = m \omega R^2$. Since this angular momentum comes from the orbital motion of the object around the origin, we have called it orbital angular momentum.

Now I will explain spin angular momentum. Let's consider sitting at the origin, but rotating about its center of mass with angular velocity $\omega$. What is the angular momentum? It is $L_{\mathrm{spin}} = I_\mathrm{cm} \omega = \frac{2}{5} m r^2 \omega$, where $r$ is the radius of the sphere itself. Since this angular momentum comes from the object spinning about its center of mass, we have called it spin angular momentum.

Now the content of the parallel axis theorem is just that if an object is orbiting the origin and spinning about its center of mass at the same time, then the total angular momentum is just the sum of the spin and orbital angular momentum: $L_{\mathrm{total}} = L_{\mathrm{orbital}} + L_{\mathrm{spin}}$ (as opposed to a more complicated sum law like $L_{\mathrm{total}} = \sqrt{L_{\mathrm{orbital}}^2 + L_{\mathrm{spin}}^2}$).

Now actually the typical statement of the parallel axis theorem is a special case where you consider an object rotating at angular velocity $\omega$ about an origin which is not its center of mass. Now this is equivalent to the object orbiting the origin with velocity $\omega R$ (where R is the distance from the origin to the center of mass), and spinning with angular velocity $\omega$. Then the total angular momentum is $L_{\mathrm{total}} = L_{\mathrm{orbital}} + L_{\mathrm{spin}} = m R^2 \omega + I_{\mathrm{cm}} \omega = (m R^2 + I_{\mathrm{cm}})\omega$. Thus we find that the moment of inertia about an axis not going through the center of mass is $m R^2 + I_{\mathrm{cm}}$, where $m$ is the mass of the object, $R$ is the distance from the rotation axis to the center of mass, and $I_{\mathrm{cm}}$ is the moment of inertia about the center of mass. Thus the total moment of inertia is an orbital moment of inertia ($mR^2$) plus a spin moment of inertia ($I_\mathrm{cm}$). This is the real statement of the parallel axis theorem.

Perpendicular Axis Theorem

Now I will explain the perpendicular axis theorem. It says that if you have a planar object, and you pick two perpendicular lines in the plane of the object (call these the $x$ and $y$ axies), then the sum of the moments of inertia $I_x$ and $I_y$ for rotations about the axes is equal to the moment of inertia $I_z$ for rotations about the $z$ axis: the line normal to the object intersecting the two other lines.

It's basically saying that the distance from the origin, which determines $I_z$ can be decomposed into a distance from the $x$ axis, which determines $I_x$ and a distance from the $y$ axis, which determines $I_y$.

The proof is trivial. Suppose we have an object of mass $m$ at a position $(x,y)$. Then its $I_x$ is $my^2$ and its $I_y$ is $mx^2$. Meanwhile its $I_z$ is $mr^2 = m(x^2 + y^2) = mx^2 + my^2 = I_x + I_y$. The general case follows by superposition.

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How did you get that orbital angular momentum is $mvR$? Isn't momentum mass * velocity? Where did the R come from? – 1110101001 Jul 28 '15 at 21:48
@1110101001 angular momentum is different from $mv$, which is usually called momentum, but can be called linear momentum to distinguish it from angular momentum. Describes motion going around a point, and the further away from the point you are, the more angular momentum you have. This is why there is an $R$ in the formula. – NowIGetToLearnWhatAHeadIs Jul 29 '15 at 3:01