homework and exercises - Angular momentum in a rod rotating around one end?

Sorry if I can't get straight to the point, I have to give a lot of details before I actually state the question.

The formula for angular momentum is $L=I \omega$. If we look up $I$ for a thin rod pivoted around one end we get $I=\frac 13 ML^2$ so $L=\frac 13 ML^2 \omega$.

However, $L$ is also equal to $p \cdot d$, where $p$ is the linear momentum and $d$ is some distance, so $I \omega=pd$

My question is as to why in this case $d$ comes out to be what it is $($it is $\frac 23 L)$

We can work out the linear momentum of the rod:

Assume the rod, which has a mass $M$, length $L$, and a constant angular speed of $\omega$ is split into $n$ equal pieces. Then each piece has mass $\frac Mn$ and a linear momentum of $\frac Mn \cdot \omega r_i$, where $r_i$ is the distance from the pivot to the end of the $i^{th}$ segment.

The linear momentum is approximately $\sum_{i=1}^{n} \frac Mn r_i \omega$. We can make this exact by taking the limit as $n$ approaches infinity. We have $r_1=\frac Ln$, $r_2=2\frac Ln$, $\cdot \cdot \cdot \cdot \cdot r_i=i \cdot \frac Ln$, so we can substitute this in the sum:

$$ \lim_{n \to \infty} \sum_{i=1}^{n} \left(\frac Mn i \cdot \dfrac Rn \omega\right)$$ Since $M$, $\omega$, $R$, and $n$ are not relevant to the summation, we can factor them out:

$$ p=MR \omega \lim_{n \to \infty} \dfrac {1}{n^2} \sum_{i=1}^{n} i$$

$$p= MR \omega \lim_{n \to \infty} \dfrac {1}{n^2} \dfrac {n(n+1)}{2}$$

$$p=\dfrac 12 ML \omega$$

Again, we have $$L=\dfrac 13 ML^2 \omega=pd=\dfrac 12 ML \omega d$$

If we solve for $d$, we get $d=\frac 23 L$. Why is this? I was expecting $d$ to be a more relevant point, such as $L$ or $\frac 12 L$ (because that's where the center of mass is. But why $\frac 23$? It seems pretty random.

Answer

There are two parts to angular momentum that both contribute at the same time. In vector form (where × is the cross product)

$$ \vec{H}_A = I_{cm} \vec{\omega} + \vec{r}_A \times m \vec{v}_{cm} $$

For a horizontal rod rotating about end point A you have

$$ \begin{aligned} \vec{\omega} & = (0,0,\Omega) & \vec{v}_{cm} &= \vec{\omega} \times \vec{r}_A = (0,\Omega \frac{L}{2},0) \\ \vec{r}_A & = (\frac{L}{2},0,0) & I_{cm} & = \frac{1}{12} m L^2 \end{aligned} $$

So angular momentum is $H_A = \frac{m L^2 \Omega}{12} + \frac{m L^2 \Omega}{4} = \frac{m L^2 \Omega}{3} $. The above is often abbreviated by defining the moment of inertia about the end point as $I_A = \frac{1}{3} m L^2$ to get to $$H_A = I_A \Omega$$

In 3D the transformation of the inertia tensor follows the parallel axis theorem. See this answer for more details: https://physics.stackexchange.com/a/88566/392.

So you have tried to equate the two parts of angular momentum without transforming the inertia value appropriately.