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.

  • $\begingroup$ Why not write those limits of sums as integrals instead? $\endgroup$
    – Ayesha
    Commented Feb 17, 2014 at 2:57
  • $\begingroup$ @Ayesha I am not too comfortable going straight to integration. I know how to do it, but I know the way I understand it now is fundamentally wrong (thinking of the differentials as very small quantities). My original plan was to arrive at a Riemann sum which I could convert to an integral but I arrived at this. $\endgroup$
    – Ovi
    Commented Feb 17, 2014 at 3:31
  • $\begingroup$ Hi Ovi. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Commented Feb 18, 2014 at 14:03
  • $\begingroup$ @Qmechanic Oh ok thanks I didn't know the policy, I retagged it now. This site is a lot better than math stackexcahnge because if you tag a question as hw there they mostly give you hints and I tought that is what would happen here (although this was inspired by hw but not a hw question) $\endgroup$
    – Ovi
    Commented Feb 21, 2014 at 3:09

3 Answers 3


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.


Good work and a good idea. d = L/2 would correspond to the moment of inertia for a point mass M at distance L/2. Momentum p = M $\omega$/2 L/2.

What would you get if the mass of the rod was concentrated at the two end points, each 1/2 M ? One point zero, the other M/2 $\omega$ L. In other words d = 1.

So the mass distribution along the rod plays a role.

For one rod pivoted around one end you have 1/3 ML$^2$. Half the length of that rod around one end: 1/3 M/2 (L/2)$^2$. Two of those would give you one rod pivoting about it center. I = 1/12 ML$^2$ Exactly the difference between d = 1/2 and d = 2/3 ! What you found here is the rule for moment of inertia around an off-center axis: I = I$_c$ + M d$^2$ where I$_c$ is moment of inertia around axis through center of mass and d is the distance between that axis and the (parallel) off-center axis.


What your result is telling you is that if a point particle and a rod of length $L$ have the same linear momentum $p$, then the angular momentum of the rod about one end will be the same as the angular momentum of the point particle about an axis at distance $d$ from it provided that $d=\frac23 L$. There is nothing remarkable about this result.

Your summation is telling you that the linear momentum of the rod is equal to the mass of the rod times the velocity of the centre of mass, which is not surprising.


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