Why do we take pseudo-force to pass through center of mass of a rigid body while calculating torque on the body using a non-inertial frame of reference?

I was just told this fact in my class and never understood intuitively as to why only center of mass is the point at which the resultant pseudo force is applied in the calculation of the resultant torque on the rigid body calculated using the non-inertial frame of reference.

  • $\begingroup$ Do you have a specific example of this? $\endgroup$ Commented Mar 8, 2020 at 21:10
  • $\begingroup$ Since our intuition is based on our experiences with mechanics observed using the ground reference frame, which is approximately Newtonian, inertial forces are counter-intuitive. Especially when combined with attitudinal quantities such as angular momentum, it is difficult to answer questions such as the one you have posed, using empirical examples. I have added a detailed answer here. More simplistically, one can explain this (sans the why) by considering the rigid body and its angular momentum when modeled as a point mass. $\endgroup$
    – kbakshi314
    Commented Mar 10, 2020 at 4:15

2 Answers 2


Pseudo forces act on the center of mass due to Newton's second law. The law ties forces to the motion of the center of mass. This is because momentum is defined as mass times the velocity of the center of mass. And forces affect the motion of the center of mass due to Newton's laws

$$ \vec{F} = \frac{\rm d}{{\rm d}t} \vec{p} = \frac{\rm d}{{\rm d}t} (m \,\vec{v}_{\rm CM}) = m \,\vec{a}_{\rm CM} $$

In reverse, a fictions force is one to adjust for the apparent motion of the center of mass in a non-inertial frame, and thus it has to be considered to be acting along the center of mass in order to modify only the motion of the center of mass, and not the rotation about the center of mass.

  • 1
    $\begingroup$ I mean that would all pseudo forces act on center of mass. Even forces like coriolis force which are a result of observing from rotating frame of reference? $\endgroup$ Commented Mar 9, 2020 at 3:28
  • $\begingroup$ @AbsoluteZero - yes, unless the rotating reference frame was accelerating rotationally. Then a pseudo-torque would need to be added to the body to account for the change in angular momentum. $\endgroup$ Commented Mar 9, 2020 at 12:44
  • $\begingroup$ Related: physics.stackexchange.com/a/228467/392 $\endgroup$ Commented Mar 9, 2020 at 15:42
  • $\begingroup$ Just because a force affects COM motion doesn't mean it acts through the COM though $\endgroup$ Commented Mar 9, 2020 at 20:48
  • $\begingroup$ @AaronStevens yes it does because a pseudo force only affects the COM motion and has no effect on the net torque on the body. $\endgroup$ Commented Mar 9, 2020 at 21:56

Let us analyze a system of particles to understand how to use the idea of d'Alembert, fictitious, inertial or pseudo forces in order to apply Newton's second law of motion when using a non-inertial reference frame.

Consider a set of material particles or point masses $\{P_i\}_{1 \leq i \leq n}$ comprising a body. If we are using a inertial or Newtonian reference frame with the non-inertial acceleration of $\vec{a}$ w.r.t. a Newtonian reference frame, to obtain the acceleration of the center of mass of this body, then Newton's law may be written as $$m_i \vec{a}_i = \vec{F}_i - m_i \vec{a},$$ for every particle indexed $1 \leq i \leq n$, where $\vec{a}_i$ is the acceleration of particle $P_i$ of mass $0 < m_i$ w.r.t. the non-inertial reference frame and $\vec{F}_i$ is the resultant force acting on the particle $P_i$. The proof of this statement is evident on applying Newton's second law using the inertial reference frame as $$m_i (\vec{a} + \vec{a}_i) = \vec{F}_i,$$ since $\vec{a} + \vec{a}_i$ is the corresponding acceleration of the particle $P_i$ w.r.t. the Newtonian reference frame. Further, using the equation $m_i \vec{a}_i = \vec{F}_i - m_i \vec{a}$, if we sum over all the particles on the left and right hand sides, we obtain $$m \sum_{i = 1}^n \vec{a}_i = \sum \vec{F_i} - m \vec{a},$$ where $m = \sum_{i = 1}^n m_i$ is the mass of the body. We notice from the definition of the center of mass denoted $C$ that $\vec{a_C} := \sum_{i = 1}^n \vec{a_i}$ is the acceleration of the center of mass of the body w.r.t. the non-Newtonian reference frame and that $\vec{F} := \sum_{i=1}^n \vec{F_i}$ is the resultant force acting on the center of mass. Notice that we did not make an assumption on the whether the acceleration $\vec{a}$ arises due to the translational or attitudinal motion of the non-inertial reference frame w.r.t. the inertial reference frame. More specifically, in the general case, the pseudo-force $m \vec{a}$ is comprised by terms including the translational pseudo-force, attitudinal (non-inertial frame angular acceleration induced pseudo-force), Coriolis and centripetal accelerations of the non-inertial reference frame w.r.t. the inertial reference frame.

Finally, note that we have not applied the assumption of the body being rigid. This assumption is useful when applying the Newton's second law to deduce the angular momentum conservation (torque) equation when using an inertial or non-inertial reference frame. Let us calculate the torque acting on the system of particles considered to see how the center of mass appears to be the point of application of the pseudo-forces in the context of the torque acting on the body. In particular, the forces $\vec{F}_i$ acting on every particle $1 \leq i \leq n$ may be decomposed as the sum $\vec{F}_i = \vec{F}_i^{ext} + \sum_{i \neq j} F_{ij}$, where $\vec{F}_i^{ext}$ is the resultant force acting on the particle $P_i$ due to material objects or entities external to the body and $F_{ij}$ is the force acted on the particle $P_i$ by the particle $P_j$ belonging to, or internal to the body. Newton's third law may then be applied in this case to calculate the torque at a point, say O, using the non-inertial reference frame as $$\tau_O = \sum_{i = 1}^n \vec{r}_{iO} \times \left( \vec{F}_i^{ext} + \sum_{i \neq j} F_{ij} - m \vec{a} \right) = \vec{r}_{iO} \times \vec{F} - \vec{r}_{CO} \times m \vec{a},$$ since $\sum_{i = 1}^n r_{iO} = \vec{r}_{CO}$, and $\vec{F}_{ij} = - \vec{F}_{ji}$ for all $1 \leq i, j \leq n$ and $i \neq j$ implies that $\sum_{i}^n \vec{F}_i^{ext} = \vec{F}$ and that the expression $\sum_{i = 1}^n \vec{r}_{iO} \times \sum_{i \neq j} F_{ij}$ vanishes. The final part of the previous statement can be verified by decomposing the relevant displacement vectors as $$\sum_{i = 1}^n \vec{r}_{iO} \times \sum_{i \neq j} F_{ij} = \sum_{i = 1}^n \vec{r}_{iC} \times \sum_{i \neq j} F_{ij} + \sum_{i = 1}^n \vec{r}_{CO} \times \sum_{i \neq j} F_{ij},$$ where $\sum_{i = 1}^n \sum_{i \neq j} F_{ij}$ vanishes because $\vec{F}_{ij} = - \vec{F}_{ji}$, implying that $\sum_{i = 1}^n \vec{r}_{CO} \times \sum_{i \neq j} F_{ij} = 0$, and noting that $\sum_{i = 1}^n \vec{r}_{iC} \times \sum_{i \neq j} F_{ij} = \sum_{i \neq j} \vec{r}_{ij} \times \vec{F}_{ij}$ since $\vec{F}_{ij} = - \vec{F}_{ji}$, vanishes because of the implicit assumption that $\vec{F}_{ij} \propto\vec{r}_{ij}$. If we assume that the body is rigid, that is $\vec{r}_{ij} = const.$ for all $1 \leq i, j \leq n$, then the implicit assumption that the internal forces are central forces, that is $\vec{F}_{ij} \propto\vec{r}_{ij}$, becomes explicit. This is because centrality is a property of forces that material particles in the Newtonian mechanical systems can exert on each other when stationary w.r.t. each other (in contrast to the non-central forces such as friction, where the particles may move with respect to each other). The rigid body assumption, that is relative stationarity of particles of the body, is also applied to obtain the moment of inertia representation of the rate of change of the angular momentum of the body. We do not obtain such a representation here, but we now have an expression for the right hand side of the rate of change of the angular momentum calculated using the non-inertial reference frame $\frac{d}{dt} \left( \sum_{i = 1}^n \vec{r}_{iO} \times m_i \frac{d}{dt} \vec{r}_{iO} \right) = \sum_{i = 1}^n \vec{r}_{iO} \times m_i \frac{d^2}{dt^2} \vec{r}_{iO} = \tau_O$, so that we note the subtle point that the angular momentum conservation principle (the rate of change of angular momentum is zero if there are no external forces, including the non-inertial forces, acting on a system of particles) holds true if and only if $\vec{F}_{ij} \propto\vec{r}_{ij}$ for all $1 \leq i, j \leq n$ such that $i \neq j$. A system of particles which consitute a rigid body satisfies this assumption but there may be non-rigid systems of particles which satisfy this assumption.


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