The statement

I attempted to prove the following statement:


for a rigid body which is only rotating about its axis and where our reference is a point on the axis of rotaion. For simplicity let the axis of rotation point along $\hat k$.

What I have done

My first step was:

$$\vec\tau_{net}=\sum_{i}m_i(\vec r_i \times \vec a_{i})$$

where $m_{i}$ are the constituent masses of the rigid body.

Notice that:

$$\vec a=r_{\perp}\alpha\hat {\theta}-r_{\perp}\omega^2 \hat r$$

where $\hat r= \cos (\theta) \hat i+\sin (\theta) \hat j $ and $\hat {\theta}= -\sin (\theta) \hat i+\cos (\theta) \hat j $.

Using this identity we find that:

$$\vec r \times \vec a = r_{\perp}^2 \alpha \hat k - r_{\parallel}r_{\perp}\alpha \hat r - r_{\parallel}r_{\perp}\omega \hat {\theta}$$

where $r_{\perp}$ is the perpendicular distance of the point (that we are considering) from the axis while $r_{\parallel} \hat k=\vec r - r_{\perp} \hat r $

Plugging this into the sum yields:

$$\vec\tau_{net}=I\vec \alpha - \sum_{i} m_{i} (r_{\parallel,i}r_{\perp,i}\alpha \hat r_{i} + r_{\parallel,i}r_{\perp,i}\omega \hat {\theta _{i}})$$

The second term of the right hand side seems complicated and I do not see a way to show that it amounts to $0$.

The questions

  1. Are my deductions wrong? If so, could you please highlight in which step?

  2. If they are not wrong, could you explain how the last expression simplifies to the desired result?

  3. Also, in case the last expression does not simplify to the desired result, could you provide a more general formula for torque?

If you could show some math in your response or provide some useful resources, it would be appreciated. Please point out any step in my working which is unclear.


2 Answers 2


I attempted to prove the following statement:


You will not be able to prove that because it is not true. The correct form is

$$\vec\tau_{\text{net}} = I\vec\alpha + \vec\omega\times(I\vec\omega)$$

You are well on your way to proving that.

The second term, $\vec\omega\times(I\vec\omega)$, can be viewed as an analog of a fictitious force. However, because angular acceleration is the same in a body-fixed frame and an inertial frame, it is not quite fictitious. Another way to look at this is that from the perspective of an inertial frame, $\vec\tau_{\text{net}} = \frac{d}{dt}\left(I_{\text{inertial}}\,\omega\right)$. The problem here is that the inertia tensor is not constant from the perspective of an inertial frame.

  1. Kinematics

    Take a body that is purely rotating about its center of mass, with rotational velocity $\boldsymbol{\omega}$.

    The translational velocity of a particle i located at $\boldsymbol{r}_i$ from the center of mass is

    $$ \boldsymbol{v}_i = \boldsymbol{\omega} \times \boldsymbol{r}_i \tag{1}$$

    Now if the rotational velocity changes with rotational acceleration $\boldsymbol{\alpha}$ then the translational acceleration of particle i is

    $$ \boldsymbol{a}_i = \boldsymbol{\alpha} \times \boldsymbol{r}_i + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \boldsymbol{r}_i) \tag{2}$$

  2. Translational Momentum

    Show that the net momentum of the body is zero

    $$\begin{aligned}\boldsymbol{p} & =\sum_{i}m_{i}\boldsymbol{v}_{i}\\ & =\sum_{i}m_{i}\left(\boldsymbol{\omega}\times\boldsymbol{r}_{i}\right)\\ & =\boldsymbol{\omega}\times\sum_{i}m_{i}\boldsymbol{r}_{i}\\ & =0 \end{aligned} \tag{3}$$

    since by definition of the center of mass $\sum_i m_i \boldsymbol{r}_i = 0$

  3. Rotational Momentum

    This is used in the definition of the mass moment of inertia tensor ${\rm I}$

    $$\begin{aligned}\boldsymbol{L}_{{\rm net}} & =\sum_{i}\boldsymbol{r}_{i}\times m_{i}\left(\boldsymbol{\omega}\times\boldsymbol{r}_{i}\right)\\ & =\sum_{i}m_{i}\left(\boldsymbol{\omega}\left(\boldsymbol{r}_{i}\cdot\boldsymbol{r}_{i}\right)-\boldsymbol{r}_{i}\left(\boldsymbol{r}_{i}\cdot\boldsymbol{\omega}\right)\right)\\ & =\left(\sum_{i}m_{i}\left(\boldsymbol{r}_{i}^{\top}\boldsymbol{r}_{i}-\boldsymbol{r}_{i}\boldsymbol{r}_{i}^{\top}\right)\right)\boldsymbol{\omega} \\ & = {\rm I}\; \boldsymbol{\omega} \end{aligned} \tag{4}$$

  4. Forces

    Each particle must have a force acting on it of $$\boldsymbol{F}_i = m_i \boldsymbol{a}_i \tag{5} $$

    Show that the net force on the body is zero

    $$ \begin{aligned}\boldsymbol{F}_{{\rm net}} & =\sum_{i}m_{i}\left(\boldsymbol{\alpha}\times\boldsymbol{r}_{i}+\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{r}_{i})\right)\\ & =\boldsymbol{\alpha}\times\left(\sum_{i}m_{i}\boldsymbol{r}_{i}\right)+\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\left(\sum_{i}m_{i}\boldsymbol{r}_{i}\right))\\ & =0 \end{aligned} \tag{6} $$

  5. Torques

    The net torque about the center of mass is $$ \begin{aligned}\boldsymbol{\tau}_{{\rm net}} & = \sum_{i}\boldsymbol{r}_{i}\times \boldsymbol{F}_i \\ & =\sum_{i}\boldsymbol{r}_{i}\times m_{i}\left(\boldsymbol{\alpha}\times\boldsymbol{r}_{i}+\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{r}_{i})\right)\\ & =\sum_{i}m_{i}\boldsymbol{r}_{i}\times\left(\boldsymbol{\alpha}\times\boldsymbol{r}_{i}\right)+\sum_{i}m_{i}\boldsymbol{r}_{i}\times\left(\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\boldsymbol{r}_{i})\right)\\ & =\left(\sum_{i}m_{i}\left(\boldsymbol{r}_{i}^{\top}\boldsymbol{r}_{i}-\boldsymbol{r}_{i}\boldsymbol{r}_{i}^{\top}\right)\right)\boldsymbol{\alpha}+\boldsymbol{\omega}\times\left(\sum_{i}m_{i}\,\left(\boldsymbol{r}_{i}\times(\boldsymbol{\omega}\times\boldsymbol{r}_{i})\right)\right)\\ & ={\rm I}\;\boldsymbol{\alpha}+\boldsymbol{\omega}\times\boldsymbol{L}_{{\rm net}} \end{aligned} \tag{7} $$

So the rotational dynamics for a rigid body in vector form are

$$ \boxed{ \boldsymbol{\tau}_{\rm net} = {\rm I}\;\boldsymbol{\alpha}+\boldsymbol{\omega}\times {\rm I}\;\boldsymbol{\omega} } \tag{8} $$

where $\boldsymbol{\tau}_{\rm net}$ is the net torque about the center of mass, $\boldsymbol{\omega}$ and $\boldsymbol{\alpha}$ are the rotational velocity and acceleration vectors of the body, ${\rm I}$ is the mass moment of inertia tensor (along the inertial basis-vectors) summed up at the center of mass.

For clarity I often designate point C as the center of mass and write the above as $$\boldsymbol{\tau}_C = {\rm I}_C\;\boldsymbol{\alpha}+\boldsymbol{\omega}\times {\rm I}_C\boldsymbol{\omega}$$

  • $\begingroup$ Thanks for your response! I had some additional doubts on the topic which I posted here: physics.stackexchange.com/questions/746072/… . If you could take a look at it, it would be appreciated. $\endgroup$ Jan 18, 2023 at 16:16
  • $\begingroup$ @physicslearner789 - just added an answer to the follow-up question. $\endgroup$ Jan 18, 2023 at 18:12

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