In any rigid body if I apply a force at a distance $r$ from the centre of mass (CM), it will cause rotation about CM but will it also cause translation of CM. Will the velocity of CM change why or why not? And if it will change to what measure?

Suppose this for example: an astronaut floating freely in space wanting to go to his spaceship uses air jets attached to his hand, if he keeps his hand perpendicular to his body will it just cause him to rotate about his stomach or will he also move in the direction away from the escaping air from the jets along with his rotation.

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    $\begingroup$ Think about it more. Try pushing objects near you! $\endgroup$ Jun 26, 2017 at 10:40
  • $\begingroup$ Duplicate? physics.stackexchange.com/q/295731/104696 $\endgroup$
    – Farcher
    Jun 26, 2017 at 10:43
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    $\begingroup$ @HritikNarayan its not that simple with friction and other things involved. $\endgroup$
    – user45838
    Jun 26, 2017 at 10:44
  • $\begingroup$ Hit a ball or something, then? $\endgroup$ Jun 26, 2017 at 10:49
  • $\begingroup$ @HritikNarayan i want to know to what measure also, mathematically, in paper. $\endgroup$
    – user45838
    Jun 26, 2017 at 10:56

1 Answer 1


In fact, in a an environment dominated by friction, the answer might not be obvious. To find the answer analytically, recall that a rigid body is particular collection of particles. Let us decompose the force acting on particle $i$ into $\vec F_i^{ext}$ which is due to particles external to the system and $\sum_j\vec F_{ij}$, where $\vec F_{ij}$ is the force on $i$ due to $j$. Hence $$m_i\ddot{\vec r}_i=\vec F_i=\vec F_i^{ext}+\sum_j{\vec F_{ij}}.$$ Sum over all particles, $$\sum_im_i\ddot{\vec r}_i=\sum_i\vec F_i^{ext}+\sum_{ij}{\vec F_{ij}},$$ use the fact that internal forces cancel up (third Newton's law), $\sum_j\vec F_{ij}=0$, and use the definition of the centre of mass vector, $\vec R_{cm}$, $$\sum_im_i\ddot{\vec r}_i=\frac{d^2}{dt^2}\sum_im_i\vec r_i=M\ddot{\vec R}_{cm},$$ where $M$ is the total mass of the system. Therefore $$M\ddot{\vec R}_{cm}=\vec F^{ext},$$ where $\vec F^{ext}$ is the total external force acting on the system. As you can see, the centre of mass has acceleration in the same direction as the (total) external force. This video shows clearly how force and centre of mass translation are in the same direction even though the force was not applied into the centre of mass.


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