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Suppose I have a rod of uniform mass as follows. I'm confused how to pick the moment of inertia for this rod.

One possibility would be to use the formula corresponding to this picture :

enter image description here

$$I = \frac{1}{12} m \ell^2$$

Alternatively, I could use the moment of inertia corresponding to this picture :

enter image description here

$$I = \frac{1}{3} m \ell^2$$

The kinetic energy is :

$$\begin{align} T &= \frac{1}{2} (m \lVert v_{COM}\rVert^2 + I\dot{\theta}^2) \end{align} $$

Notice, if I plug in the different formulas for $I$, they yield different results. So they are not equivalent.

It also seems clear to me that the second formula for $I$ seems more appropriate because it matches the path of the rod.

In the Physics SE Chat, someone said you can calculate the moment of inertia about any point, it doesn't have to be the point of rotation. They said choosing that point doesn't necessarily mean the object will rotate around that point.

Can someone explain why both formulas could be used to model this system? In particular, could you elaborate on the differences between the two formulas for $I$ in terms of what information they would tell me about the system?

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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Aug 6, 2020 at 0:50

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The short answer is that when you write

$$T = \frac{1}{2}M v_{CM}^2 + \frac{1}{2}I\omega^2$$

you are assuming that the moment of inertia is being calculated about the center of mass. It's true that the moment of inertia can be calculated about whatever axis you'd like; however, if you don't choose the center of mass, then the expression for the total kinetic energy does not neatly separate into a rotational part and a translational part. I'll show you why that is.


Consider an object which is made up of a bunch of small pieces. The $i^{th}$ piece is located at position $\mathbf r_i$ and has mass $m_i$.

The total kinetic energy of the system is

$$T = \sum_i \frac{1}{2}m_i \dot {\mathbf{r}}_i^2$$ but this expression is generally not so useful. We want to decompose this expression into a translational piece and a rotational piece. We need to pick a position about which to calculate the moment of inertia, so let's call that position $\mathbf R$.

If we define the position of the $i^{th}$ mass relative to $\mathbf R$ to be $\mathbf u_i = \mathbf r_i-\mathbf R$, then we can write

$$T = \sum_i \frac{1}{2}m_i (\dot{\mathbf u_i} + \dot{\mathbf R})^2= \sum_i \frac{1}{2}m_i \dot{\mathbf u}_i^2 + \sum_i \frac{1}{2}m_i \dot{\mathbf R}^2 + \sum_i m_i \dot{\mathbf u_i}\cdot \dot{\mathbf R}$$

If the system in question is a rigid object so that the distances between all of the constitutent pieces are constant, then we can write $\dot {\mathbf u}_i = \boldsymbol \omega \times \mathbf u_i$ where $\boldsymbol\omega = \mathbf u_i\times\dot{\mathbf{u}}_i/u_i^2$ is the angular velocity of the object. One can show that $\boldsymbol\omega$ is the same for every part of a rigid object. As a result, defining $M = \sum_i m_i$ to be the total mass,

$$T = \frac{1}{2}M\dot{\mathbf R}^2 + \sum_i \frac{1}{2}m_i (\boldsymbol \omega \times \mathbf u_i)^2 + \sum_i m_i(\boldsymbol\omega \times \mathbf u_i) \cdot \dot{\mathbf R}$$


More algebra can be done to make this expression more recognizable, but this is enough to show what we want. The first term is the translational kinetic energy we'd get if we imagined the entire mass concentrated at $\mathbf R$. The second term is the rotational kinetic energy calculated about the point $\mathbf R$.

The third term is generally messy and complicated. However, note that it can be rearranged: $$\sum_i m_i(\boldsymbol\omega \times \mathbf u_i) \cdot \dot{\mathbf R} = M\dot{\mathbf{R}}\cdot \left(\boldsymbol\omega \times \left[\mathbf R_{CM} - \mathbf R\right]\right)$$

where $\mathbf R_{CM} = \sum_i m_i \mathbf r_i/M$ is the center of mass position. Therefore, if we simply choose $\mathbf R=\mathbf R_{CM}$, then this term is equal to zero and the total kinetic energy can be written as the simple sum of a translational part and a rotational part.

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