I have started learning about rotational mechanics about a month back. I am trying to derive everything from the fundamentals.<br>

First off, I derive the relation for the moment of inertia using the Kinetic energy argument. ($I= \int_0^m r^2 dm$)
From here by analogy to linear dynamics, I state Angular momentum $\mathbf{L}= I \boldsymbol{\omega}$ Again, as an analogy to Newton's 3rd Law, I say Moment of Force (Torque) = rate of change of angular momentum= $I \boldsymbol{\alpha} $

Now, I am having trouble proving that $$I \boldsymbol{\alpha} = \mathbf{r} \times \mathbf{F}$$ ($I, \boldsymbol{\alpha}$ and $\mathbf{r}$ are w.r.t the same axis). I am able to prove it easily for a point mass system, where the force is applied on the mass directly, but unable to prove it for a general Rigid body. $I$=moment of inertia

$\boldsymbol{\alpha}$= Angular acceleration vector

$ \boldsymbol{\omega}$=Angular velocity vector

$\mathbf{r}$ = Position vector

$\mathbf{F}$ = Force vector

I tried something new but it seems contradictory...

torque= I $\alpha$ = $\alpha \int_0^m r^2 dm $ =$\alpha ( \int_0^m r \times r dm) $

now using integration by parts,
torque=$\alpha (r \int_0^m r dm - \int_0^m ((dr/dm) \int_0^m r dm)dm $

By using the relation ,
F= $ \alpha \int_0^m r dm $
torque=$ \alpha (r (F / \alpha ) - (F/ \alpha ) (\int_0^r dr)$ =$ \alpha (r (F / \alpha ) - r (F / \alpha )) =0 $

this is certainly not true as torque may not always be 0. What am I doing wrong here? Also can someone answer my original question with a similar approach?


3 Answers 3


It isn't always true that $\vec{\boldsymbol{\tau}}=I\vec{\boldsymbol{\alpha}}$. The general form of torque is $$\vec{\boldsymbol{\tau}}=\dfrac{d}{dt}\vec{\boldsymbol{L}}$$ From definition of angular momentum $$\vec{\boldsymbol{L}}=\vec{\boldsymbol{r}}\times \vec{\boldsymbol{p}}$$ gives \begin{align*} \vec{\boldsymbol{\tau}}&=\dfrac{d}{dt}(\vec{\boldsymbol{r}}\times \vec{\boldsymbol{p}})\\ \vec{\boldsymbol{\tau}}&=\vec{\boldsymbol{r}}\times \dfrac{d\vec{\boldsymbol{p}}}{dt}\\ \end{align*} and from Newton's second law of motion $$\vec{\boldsymbol{F}}=\dfrac{d}{dt}\vec{\boldsymbol{p}}$$ then $$\vec{\boldsymbol{\tau}}=\vec{\boldsymbol{r}}\times\vec{\boldsymbol{F}}\;\blacksquare$$

  • $\begingroup$ well this method raises a similar question of why L=r $\times$p (i started with the definition L=I $\omega$), Still thanks for your effort :) $\endgroup$
    – saitama
    Nov 10, 2020 at 13:15
  • 1
    $\begingroup$ Examine any intro to math physics textbook and you'll find the answer readily. Anyway, the angular momentum is defined as the cross product $L = r \times p$ because $L$ will always be perpendicular to the plane containing $r$ and $p$. Thus, the equation $L = I\omega$ is more subtle than you probably know: the inertia $I$ is actually a rank 2 tensor, also called a matrix, so only in very simple cases does $L$ point in the same direction as $\omega$ (i.e. when I is a diagonal matrix). en.wikipedia.org/wiki/Moment_of_inertia#Inertia_tensor $\endgroup$ Nov 10, 2020 at 13:58
  • $\begingroup$ Nice explanation here farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html $\endgroup$ Nov 10, 2020 at 14:00
  • $\begingroup$ @saitama because $L=I\omega$ isn't the definition of angular momentum. $\endgroup$
    – Ken
    Nov 11, 2020 at 6:34


If we were being more explicit, we define torque about some origin $\mathrm{O}$ around this origin we can rewrite an inertia element of some mass element some where as:

$$ \Delta I_{i} = |\vec{r_i}|^2 \Delta m_i$$


$$ \Delta I_i \text{ is the inertia of the mass element}$$

$$ \Delta{\vec{r_i}} \text{ is the magnitude of position vector from the defined origin O}$$

$$ \Delta{ m_i} \text{ is the mass element we are considering}$$

Though we cant directly apply the definition of inertia to the whole system, we can indeed apply the definition of it for a point mass on to a small mass element. We can think of shrinking the mass element such that the whole mass is concentrated in a small enough region that we could approximate it.

Better yet, a more intelligent way would be to write this expression using density. Assume we have an object with varying density $\rho$ which depends on $(x,y,z)$ then we can say the mass of that element in some n-volume around region of inspection is given by $ \rho \Delta V_i$ (*). This turns our equation to this form:

$$ \Delta I_i = |r_i|^2 \rho(x,y,z) \Delta V_i$$

Now, given a rigid body, we can think of as partitioning it into small n-volume elements of $ \Delta V_i$ and finding the inertia of each and summing over that. So, what we can do is sum over all the n-volume elements using an n-dimensional integral (**).

$$ I = \int |r_i|^2 \rho(x,y,z) \Delta V_i$$

The problem in your post:

In the second method, you had a step where you had done the following:

$$ |r_i|^2 = \vec{r_i} \cdot \vec{r_i}$$

And you had written as:

$$ I = \int \vec{r} \cdot \left[ \vec{r_i} \rho(x,y,z) \Delta V_i \right]$$

And integrated the above quantity by parts, however, when you do this you are integrating a vector field ($ \int \vec{r} \Delta V_i$)over a volume. This operation does not make any sense by the mathematics that I know of / can find by googling. The closest thing I found was this quora post.


For a good explanation of how to derive this in detail go see Kleppner and Kolenkow around page-245

Note: n-volume is the sort of generalization of volume



Radial force on a point:m.$\omega^2$r

Tangential force on it:mr$\alpha$ (Torque due to it :m.$r^2.\alpha$)

Sum over all particles for torque.


Think about whether or not radial force provides any torque or not.

  • $\begingroup$ this method (tang force= mr $\alpha$ ), looks obvious for a point mass m in circular motion of radius r, where force is applied on the point mass tangentially, but how is this relation valid for a continuous mass( rigid body)? (i'm assuming that by r, you mean to refer to position vector of point of application of force) $\endgroup$
    – saitama
    Nov 10, 2020 at 4:14
  • $\begingroup$ @saitama If you have understood that part, For a rigid body it is easy, We just have to sum/integrate it for every possible point on the rigid body. $\endgroup$
    – Linkin
    Nov 10, 2020 at 4:21
  • $\begingroup$ ok, so I have managed to understand why F=mr $\alpha$ for a rigid body, But in third line you say summing over all particles......gives torque. How does summing tangential force give torque? even their dimensions are also differenent $\endgroup$
    – saitama
    Nov 10, 2020 at 5:10
  • $\begingroup$ @saitama I've made an edit(I am summing torque not force), does it seem clear now? Tell me if it doesn't. $\endgroup$
    – Linkin
    Nov 10, 2020 at 5:14
  • $\begingroup$ ok now i have got it. Thanks. Also you may look into the new approach i tried (the edit in the question), and tell me what is wrong there, although I have got the answer to my main question by your method. $\endgroup$
    – saitama
    Nov 10, 2020 at 5:37

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