Physical intuition for the parallel axis theorem

Question

I am confused with the parallel axis theorem, especially with the term $M$ in $Md^2$.

I don't understand why the moment of inertia of each particle increases by a term of $(dm)(d^2)$.

Is there a physical reason for why we should use the total mass of the rotating rigid body in the extra term $Md^2$?

Take a look at the picture above. In the first picture, the axis passes through the center of mass of the disc and the dashed line is the axis about which it will rotate.

When the axis changes as shown in the second figure, the molecules in the region $A$ are farther from the axis and hence the $mr^2$ term for them increased.

For the molecules in region $B$, the distance from the axis decreased and hence the $m'r^2$ term also decreased. And in region $C$, there apparently isn't any change at all.

So it is clear that the net change is not equal to $Md^2$.

Please correct me if I am wrong somewhere. And also here $r$ is any arbitrary distance.

Answer 1

True the distance from the axis has decreased but notice that we are taking the square of the square of the distances and hence the signs go away. That is, even if the average change of a collection of quantities is zero, the sum of squared quantities would not be unless all the quantities under consideration is zero.

There is actually parallels between mathematical ideas such as standard deviation and that of moment of inertia. Understanding one may aid in understanding another. See here

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Perhaps a derivation of it may help in understanding. Here is an intuitive and simple derivation for parallel axis theorem from Resnick Halliday walker, consider the inertia along the $z$ axis:

$$I = \int r^2 dm = \int x^2 + y^2 dm$$

where $r$ is the perpendicular distance to mass elements from the $z$ axis, under a change of cordinates to $x', y'$ where $ x' = x-a $ and $ y'=y-b$ , we can write the new moment of inertia in the new system as:

$$I' = \int (x-a)^2 + (y-a)^2 dm = \int x^2 +y^2 dm - \left[ 2a \int x dm - 2b \int y dm \right] + (a^2 +b^2) \int dm$$

The first integral is just the moment of inertia along the original axis we took. If we take the original axis as the center of mass axis, the bracketed integrals are zero (why?). Hence,

$$I' = I + (a^2 + b^2 ) M$$

Now, we can find geometrically that $a^2 +b^2$ is equal to the square perpendicular distance between two axis, lets call that $L^2$

$$ I'= I + L^2 M$$

Answer 2

Consider for simplicity just two "particles" of mass m, each on a massless rod of length, say $2L$. The moment of inertia of this system wrt the COM is

\begin{equation} I = mL^2+mL^2=2mL^2. \end{equation}

Now move the axis by $x$. Then the moment of inertia becomes \begin{equation} I = m(L+x)^2+m(L-x)^2=2mL^2+2mx^2. \end{equation}

The crucial thing is that $r$ enters in the moment of inertia with power 2.

Answer 3

Think about the $M d^2$ term as follows:

• The velocity of a mass at the end of a rotating massless rod is proportional to $d$ $$v = \omega d$$
• Thus the translational momentum of said mass is proportional to $M d$ $$p = M v = M (\omega d)$$
• The angular momentum of the mass ("torque" of momentum) is proportional to $d p$ $$L = d M (\omega d) = (M d^2) \omega$$

So one of the $d$'s in mass moment of inertia comes from the fact the velocity increases with distance, and the other from the fact the moment arm of momentum increases with distance.