I'm learning about rotational motion and the moment of inertia. Unlike inertia that I learned before, there is a formula to calculate rotational inertia. I'm having trouble understanding why it's possible to calculate inertia for a rotating object, but not a regular moving object. After doing research, not only is there no formula for normal objects, but different interpretations of what inertia is. Is there a fundamental difference between the moment of inertia and the inertia of an object, or am I misunderstanding something?
Newtonian Mechanics – Why Can We Calculate Moment of Inertia but Not Inertia?
inertiamassmoment of inertianewtonian-mechanicsrotational-dynamics
Related Solutions
The formulae from the NASA document and Wikipedia are simply in different frames: the ones in Wikipedia assume coordinate frame whose axes pass through the center of mass whereas the NASA document assumes arbitrary cartesian coordinate frame.
Let capital Xi, Yi and Zi denote coordinates of i-th atom in an arbitrary Cartesian coordinate frame O and xi, yi and zi denote respective coordinates in the frame Ocom translated to the center of mass of the molecule.
The coordinates of the center of mass of the molecule in the O frame are:
\begin{equation} X_{com} = \frac{1}{M}\sum_{i} m_i X_i \\ Y_{com} = \frac{1}{M}\sum_{i} m_i Y_i \\ Z_{com} = \frac{1}{M}\sum_{i} m_i Z_i \end{equation}
Transformation from O to Ocom is simply translation by [-Xcom, -Ycom, -Zcom]:
\begin{equation} x_i = X_i - X_{com} = X_i - \frac{1}{M}\sum_{i} m_i X_i\\ y_i = Y_i - Y_{com} = Y_i - \frac{1}{M}\sum_{i} m_i Y_i\\ z_i = Z_i - Z_{com} = Z_i - \frac{1}{M}\sum_{i} m_i Z_i \end{equation}
Now, take for example the formula for Ixx in the O frame:
\begin{equation} I_{xx} = \sum_{i} m_i (y_i^2 + z_i^2) \end{equation}
(Note that the two-term sum in the parenthesis is simply the Euclidean distance between the i-th atom and the x axis, exactly as one would expect.)
Substituting the formulae for yi and zi above we have
\begin{equation} I_{xx} = \sum_{i} m_i [(Y_i - Y_{com})^2 + (Z_i - Z_{com})^2] \\ I_{xx} = \sum_{i} m_i (Y_i^2 + Z_i^2) + \sum_{i} m_i (Y_{com}^2 + Z_{com}^2) - 2 \sum_{i} m_i (Y_i Y_{com} + Z_i Z_{com}) \\ I_{xx} = \sum_{i} m_i (Y_i^2 + Z_i^2) + M (Y_{com}^2 + Z_{com}^2) - 2 M (Y_{com}^2 + Z_{com}^2) \\ I_{xx} = \sum_{i} m_i (Y_i^2 + Z_i^2) - \frac{1}{M}(\sum_{i} m_i Y_i)^2 - \frac{1}{M}(\sum_{i} m_i Z_i)^2 \end{equation}
which is exactly the formula for Ixx in the NASA document. Analogously for the other formulae.
Interestingly, the sum of the last two terms in the final formula is exactly what one would expect from the parallel axis theorem.
The conclusion is that you should use whichever formulae fit your coordinate frame. If the center of your coordinate frame coincides with the center of mass of the molecule, then you can use the simpler formulae from Wikipedia. Otherwise you should use those from the NASA document.
Before the collision the rod has angular momentum $I \omega_i$ where $I$ is the moment of inertia of the rod about the pivot and $\omega_i$ is the angular velocity of the rod immediately before the collision.
Whether the ball can be considered to have any moment of inertia before the collision is irrelevant, because it has no angular velocity so it has no angular momentum. If the ball did not stick to the rod after the collision, but bounced off in a straight line, then it would be even more perplexing to work out what its moment of inertia and angular velocity are, because both are constantly changing as the ball moves off in a straight line. However the ball's angular momentum is constant and easily calculated as linear momentum x perpendicular distance from the pivot.
After the inelastic collision the ball sticks to the rod, and both rotate about the pivot with the same angular velocity $\omega_f$. Both the rod and the ball now have angular momentum, the total of which is $(I+mL^2) \omega_f$. Whether you consider the ball and rod to be a single object in which there is no internal motion (ie a rigid body), or two separate objects moving such as to maintain their relative positions, does not make any difference.
It ought to be obvious from the Conservation of Angular Momentum, which you accept does apply in this case, that $\omega_i \ne \omega_f$ if $m \ne 0$.
It is not clear to me what you mean when you suggest that the mass of the ball is converted to heat. Some kinetic energy is lost in the collision, which is why it is called inelastic, and some of this loss becomes heat. But the mass of the ball (or the rod) does not change to any significant degree.
Best Answer
Classically, the inertia of something is just its mass. If you want an analogous equation, just integrate the mass density $\rho$ of the object over the volume of the object:
$$m=\iiint \text dm=\iiint\rho\,\text dV$$
Compare this to what you usually see in introductory physics as $$I=\iiint r^2\,\text dm=\iiint r^2\rho\,\text dV$$
which, for a given axis, is one element of the moment of inertia tensor.
Yes. The inertia of an object does not depend on where the mass is within the body, only on how much mass there is. The moment of inertia about a given point does depend on how that mass is distributed about the point / axis you are calculating the moment of inertia about.