# One force applied to one point of a rigid body: centre of mass and torque [duplicate]

Let us suppose that one force is applied to a point of a rigid body that is not acted upon by any other force. I think an example can approximatively be a rock in deep space, far from any relevant gravity force, pushed by an astronaut.

I would intuitively say that the rock would in general be subject to a non-null torque, unless the line passing through the centre of mass of the body and the point where the force acts is parallel to the direction of the force... Nevertheless, I am not able to prove it to myself: I have tried to reason by using a non-inertial frame accelerating together with the centre of mass of the body, but I have reached no result.

I am a beginner in the study of torque and rotation mechanics, but I am very curious about the issue. I have searched the Internet for an explanation of what happens in such situations, but I have found nothing and. How can it be proved, by using rigourous mathematical tools, that the torque is null if the line passing through the centre of mass of the rigid body and the point where the force acts is parallel to the direction of the only one force acting on the rigid body?

EDIT: I suspect that this question's answers may solve my question, but, although my thirst for knowledge is great, I have not got the needed tools to understand whether and how it relates to my question: for ex. my textbook has not explained concepts like the moment of inertia yet at the point where my studies are. I will return to it to understand it better once I masterize those tools.

## marked as duplicate by ja72, ACuriousMind♦, Kyle Kanos, Chris Mueller, Ryan UngerApr 17 '15 at 22:54

Torque $\tau = \vec{r} \times \vec{f} = I \vec{\alpha}$
$r$ here is moment of arm, ie. the distance(& perpendicular distance) from the axis of rotation of the body. You can only have and define torque if you have an axis about which the subject will rotate. It will be an absolute force driven motion, with NO torques compelled to induce.
• The correct equation is $$\vec{\tau} = \frac{{\rm d}I \vec{\omega}}{{\rm d}t} = I \vec{\alpha} + \vec{\omega} \times I \vec{\omega}$$ – ja72 Apr 17 '15 at 14:11