I came across a theorem that states that :

for a solid body (made of a collection of points $M_i$ of masses $m_i$ and speed $\overrightarrow{v_i}$ ) in rotation around a point $O$ under the action of a set of forces $\overrightarrow{F_i}$ each with its point $M_i$ of application; then we have the following equivalence : $$ \overrightarrow{\mathcal{M}}(\sum_{i} \overrightarrow{F_i})=\frac d{dt}\overrightarrow{L}=\overrightarrow{0} \Longleftrightarrow\sum_{i} \overrightarrow{F_i}=\overrightarrow{0} $$ where $$ \overrightarrow{L}=\sum_{i} \overrightarrow{L_i}=\sum_{i} \overrightarrow{OM_i}\wedge m_i \overrightarrow{v_i} $$

is the total moment. and $\overrightarrow{\mathcal{M}}(\sum_{i} \overrightarrow{F_i})$ is the total torque.

I tried to prove this theorem but I could not


3 Answers 3


The above is obviously not true when considering a force couple. Two separated equal and opposite forces have $\sum F = 0$, but a net torque, and thus $ \frac{\rm d}{{\rm d}t} L \neq 0 $.

This is because in general

$$ \sum \vec{F}_i = 0 $$ is not the same as $$ \sum \vec{r}_i \times \vec{F}_i =0 $$

I might be missing something here, but here is the development of momentum from particles to move together. Assume the body is rotating about the center of mass, and the center of mass is translating.

From the origin, the location of each particle $\vec{r}_i$ is split into the location of the center of mass $\vec{r}_C$ plus the location from the center of mass $\vec{d}_i$ such that

$$ \sum_i m_i \vec{d}_i = \vec{0} \tag{1} $$


Each particle moves with a different velocity based on the following vector field

$$\vec{v}_i = \vec{v}_C + \vec{\omega}\times \vec{d}_i \tag{2} $$

Translational Momentum

Add up the momentum of each particle

$$ \require{cancel} \begin{aligned} \vec{p} & = \sum_i m_i \vec{v}_i = \sum_i m_i ( \vec{v}_C + \vec{\omega}\times \vec{d}_i ) \\ & = \left( \sum_i m_i \right) \vec{v}_C + \vec{\omega}\times \cancel{ \left( \sum_i m_i \vec{d}_i \right)} = m \, \vec{v}_C \end{aligned} \tag{3} $$

where $m = \sum_i m_i$

Rotational Momentum

Add up all the angular momentum about the center of mass

$$ \begin{aligned} \vec{L}_C & = \sum_i \vec{d}_i \times (m_i \vec{v}_i) = \sum_i \vec{d}_i \times m_i ( \vec{v}_C + \vec{\omega}\times \vec{d}_i ) \\ & = \cancel{\left(\sum_i m_i \vec{d}_i\right)} \times \vec{v}_C + \sum_i m_i \vec{d}_i \times ( \vec{\omega} \times \vec{d}_i ) \\ & = \mathbf{I}_C\, \vec{\omega} \end{aligned} \tag{4} $$

Law of Motion

Newtons law of motion is

$$ \begin{aligned} \sum_i \vec{F}_i & = \tfrac{\rm d}{{\rm d}t} \vec{p} \\ \sum_i \vec{d}_i \times \vec{F}_i & = \tfrac{\rm d}{{\rm d}t} \vec{L}_C \end{aligned} \tag{5} $$

Now without any specific definition of the force vector field $\vec{F}_i$ unlike the velocity field the above sum of torques is not equal to zero in general even if the sum of forces is zero.

  • 1
    $\begingroup$ in the equation (4) you miss the $m_i$ ; and i dont know how you defined $\mathbf{I}_C$ $\endgroup$
    – El-Mo
    Commented Feb 26, 2021 at 15:03
  • $\begingroup$ you have to calculate vector product and you will how $I_c$ is defined. $\endgroup$
    – Numenorean
    Commented Feb 28, 2021 at 14:40
  • 1
    $\begingroup$ @El-Mo $$ \mathbf{I}_C = \sum_i m_i \pmatrix{ y_i^2+z_i^2 & -x_i y_i & -x_i z_i \\ -x_i y_i & x_i^2 + z_i^2 & -y_i z_i \\ -x_i z_i & -y_i z_i & x_i^2+y_i^2 }$$ where $d_I = \pmatrix{x_i \\ y_i \\ z_i}$ for each particle. $\endgroup$
    – JAlex
    Commented Mar 1, 2021 at 14:38

The theorem claims that torque vanishing implies forces vanishing. Your question asks for the converse. Forces sum to zero implies that torque vanishes. The first statement is true provided the torque about every point vanishes, the second is clearly false.


the theorem is just saying the opposite of what I understood


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