0
$\begingroup$

The parallel axis theorem states that you can relate the moments of inertia defined with the center of mass as the origin to the moments of inertia defined with respect to some other origin. It is summarized in this equation:

$I=I_c+Mh^2$

For a cone, the center of mass is $1/4$ of the height from the base so we can define $I_{base}$ using the parallel axis theorem. However, since the parallel axis theorem does not care for the sign of $h$, the moments of inertia defined at $h/4$ above the center of mass are equivalent to the base.

cone with moments of inertia defined with respect to different origins

The moment of inertia can be thought of as the opposition that a body exhibits to having its speed of rotation about an axis altered by the application of a torque. This means that turning the cone about $I_1$ in the base will require the same force as it takes to rotate it about $I_1$ which is at the center of the cone. Intuitively, I would say that is harder to rotate the cone about its base than it is to spin it around its center so the fact that they have the same moments of inertia is confusing to me. Can someone please point out the flaw in my logic? Also, as a bonus question, is the parallel axis theorem still valid when you extend the origin outside of the shape?

$\endgroup$

3 Answers 3

0
$\begingroup$

Consider the following conceptual steps

  1. The MMOI of a point particle of mass $m$ that is orbiting at a distance $d$ is $$I = m\, d^2$$
  2. The motion of a rigid body is decomposed into the motion of the center of mass, and the motion about the center of mass.
  3. The MMOI of a rigid body about the center of mass $I_c$ is derived from the angular momentum of the body whilst rotating about the center of mass only.
  4. The MMOI of a rigid body about any other point $I$ is the superposition of the MMOI due to rotation $I_c$ and due to the motion of the center of mass as if the body was a concentrated point mass $m d^2$ $$ I = I_c + m d^2$$
$\endgroup$
0
$\begingroup$

The difficulty you are having is that you are looking at the problem from the wrong perspective.

I cannot draw a three-dimensional picture on this screen so here is a section of the cone through its centre of mass, $CoM$, and its apex, $C$ with axes $X$ and $Y$ perpendicular to the plane.

enter image description here

From your diagram it looks as thought your orange axis $I_1$ which I have labelled $Y$ has larger moment of inertia that your green axis $I_1 (Y)$.

The coloured circles in my diagram show the motion of particles at positions $A, \, B,\, C$ and $Y$ about my two axes.
Note that the asymmetry between the motion of the particles which make up the cone about those two axes are much less pronounced than you visualised when looking at your diagram and according the the parallel axes theorem the sum of all the contributions of the particles which make up the cone is the same irrespective if you use axis $X$ or axis $Y$.

I have not proved that they are equal but tried to show you, by hand waving, that they might be equal.

The axes can be outside the object as explained in the Wikipedia article Parallel axis theorem.

$\endgroup$
0
$\begingroup$

I mean is it acceptable to just say, “your intuition is wrong”...?

Trying to foster intuition

It gets harder to rotate something the further the masses in it are distributed from the axis of rotation. The axes closest to the mass-in-aggregate all pass through the center of mass, roughly speaking that is why we calculated a center of mass in the first place. The effect is linear in mass but quadratic in distance—what you see as a minimum at the center of mass is a delicate balance between these two, a motion towards more mass (moving the axis to the base of the cone) unfortunately shifts the minority of mass (the tip) further away from the rotation axis and the quadratic distance makes up for the linear mass improvement. On the other hand moving the other way, the mass of the base is greater and overwhelms bringing the tip closer. That's why it's the minimum.

We can formalize this idea that there has to be a minimal axis inside the figure, without the exact derivation that it goes through the CoM, simply by looking at the extreme case: an axis on the boundary, any boundary, such that the entire figure is to one side of a plane on the axis. The question is, if we move the axis normal to the plane, can we prove that it’s always easier to move the axis into the figure? If so, there must be a easiest axis somewhere in the figure. But the rest is just basic calculus, the kinetic energy expression must be differentiable with respect to this axis motion, it just involves some additions and subtractions and multiplications and not much else. And we know that if we move the axis away from the figure it gets harder to rotate it, so moving it into the figure must make it easier.

The actual derivation is super easy

Every physicist should just know this derivation, it is a very quick argument about vectors.

That it's the center of mass, well, that requires the full derivation. Approximate with a set of point masses, WLOG the axis is the $\hat z$-axis, use an angular velocity $\omega$ and cylindrical coordinates... the velocity is $v_i=\sqrt{x_i^2 + y_i^2} ~\omega$... Total kinetic energy $$K=\frac12 \sum_i m_i \omega^2 (x_i^2 +y_i^2),$$ and the question is, if we add some infinitesimal value $(\mathrm dx, \mathrm d y)$ to all of these masses uniformly how does it impact the kinetic energy, which is just $$K+\mathrm d K= \frac12 \omega^2 \sum_i m_i \big((x_i+\mathrm d x)^2 +(y_i+\mathrm d y)^2),\\ \mathrm d K/\omega^2=\mathrm d x~\sum_i m_i x_i +\mathrm d y~\sum_i m_i y_i.$$ The minimum kinetic energy is therefore found when both $\sum_i m_i x_i = \sum_i m_i y_i = 0$ which is precisely the equation for saying that the axis intersects the center of mass. So you can kind of just “see it” from the kinetic energy expression.

A slight elaboration of this proves the parallel axis theorem, assume $x_i, y_i, z_i$ are relative to a center of mass at a macroscopic position $\delta x, \delta y, \delta z$... Now you cannot neglect the $\delta x^2$ term but of course you don't want to—you have chosen $x_i, y_i$ to eliminate the cross term above and so you just have, $$ K +\delta K = \frac12 \sum_i m_i \omega^2 (x_i^2 +y_i^2) + \frac12 M \omega^2(\delta x^2+\delta y^2),$$ dividing by $\frac12 \omega^2$ we find easily $\delta I = M(\delta x^2+\delta y^2).$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.