Force on different points on a body not passing through the centre of mass [duplicate]

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I was studying about centre of mass and I found that if the line of action of force passes through centre of mass then it will execute pure translation. Moreover acceleration of centre of mass is net force applied on the body divided by the total mass of the body and my textbook says that its valid for all points on the body where the force is applied. I can't get this as when the line of action of force will not pass through centre of mass then body must also rotate and i think the acceleration(translational) of centre of mass must change (due to rotation of body). Assume that the body is rigid. Any help is appreciated.

marked as duplicate by sammy gerbil, Jon Custer, John Rennie, ZeroTheHero, Kyle KanosAug 24 '17 at 10:10

• You also want the answer when the force is not through the centre plus there's no rotation – Shashaank Mar 31 '17 at 19:42
• @Shashaank there is rotation i mean the body does rotate. – DarkSideofPhy Mar 31 '17 at 19:43
• Why do you think " translational " acceleration will change due to rotation? – Utkarsh futous Mar 31 '17 at 19:54
• That is what i don't understand "why it will not change ?" – DarkSideofPhy Mar 31 '17 at 19:59
• @Pinku Hint: Newton's law of equality of action -Reaction – Shashaank Mar 31 '17 at 20:40

What is valid is that the acceleration of the center of mass is described only by the net forces on the body regardless of the point of application. Only when concerned with rotation (or the motion away from the center of mass) you need to consider the position of line of action of the forces.

Read this answer on the derivation of the equations of motion. Specifically the motion of a rigid body when a force is applied not at the center of mass and the acceleration of that point.

\begin{align} \sum \vec{F} &= m \vec{a}_A - m \vec{c}\times \vec{\alpha} + m \vec{\omega}\times\vec{\omega}\times\vec{c} \\ \sum \vec{M}_A &= I_C \vec{\alpha} + m \vec{c} \times \vec{a}_A - m \vec{c} \times \vec{c} \times \vec{\alpha} +\vec{\omega} \times I_C \vec{\omega} + m \vec{c} \times \left( \vec{\omega} \times \vec{\omega} \times \vec{c} \right) \end{align}

Point A is a point away from the center of mass C, and the vector $\vec{c}$ is the location of the COM relative to A.

The case described on the book is where the point of interest is the center of mass (or A = C, and $\vec{c}=0$) which makes the equations the familiar ones

\begin{align} \sum \vec{F} & = m \vec{a}_C \\ \sum \vec{M}_C & = I_C \vec{\alpha} + \vec{\omega} \times I_C \vec{\omega} \end{align}

If the line of action of the force is not through the center of mass, there is going to be a net moment about the center of mass $\sum \vec{M}_C \ne 0$ which is going to cause rotational acceleration $\vec{\alpha}$. But the linear acceleration of the center of mass $\vec{a}_C$ only depends on the net forces $\sum \vec{F}$.

But for any other point (AC) the acceleration of that point $\vec{a}_A$ depends on the rotational acceleration $\vec{\alpha}$ and hence the line of action of the applied forces.

A corollary to the statement on the book is that when a pure torque is applied (i.e. net force is zero), the body will rotate about the center of mass.

As you already know when applied force is passing through the com there is no rotational effect. The total mass of the body can be assumed to be in the com and hence F=ma is directly applicable. But when the case is not that, the force applied produces a torque on the body hence producing an angular and linear acceleration. Now the linear acceleration of every point on the body depends on its distance from the origin i.e the com. And here center of mass of a body does not rotate about its own axis. It translates during the motion. Hope this clarifies somewhat