0
$\begingroup$

So basically my question is when a force acts on a rigid body I know that the part of the force perpendicular to the rotation axis of the object causes a torque and therefore a rotation (If it is the only force on the object), but does the whole magnitude of the force then also contribute to the objects liner motion or only the magnitude left over by the torque?

if its the whole magnitude why/how is this so?

$\endgroup$
1
  • $\begingroup$ The real question is where is the axis of rotation in this situation. $\endgroup$ Oct 10, 2021 at 23:38

2 Answers 2

2
$\begingroup$

but does the whole magnitude of the force then also contribute to the objects liner motion or only the magnitude left over by the torque?

The force does not "divide itself" up between linear motion and rotation. Linear motion and rotation are the results of the same magnitude of force. Rotation occurs when the line of action of the net force is not through the center of mass (COM) which results in torque. Linear motion of the COM is the same regardless of the line of action of the net force.

I should add that the agent supplying the force does more work when it causes rotation in addition to linear motion because it is providing both rotational kinetic energy to the object as well as translational kinetic energy. So in that sense it is the work done by the force that is divided up between linear and rotational motion.

Hope this helps.

$\endgroup$
1
  • $\begingroup$ When applied as an impulse, an off-center force will travel a greater distance and do more work, supplying the extra energy for rotation. $\endgroup$
    – R.W. Bird
    Oct 10, 2021 at 18:41
0
$\begingroup$

Newton's second law states that for a set of forces acting on a system, there is a corresponding acceleration of the centre of mass that is proportional to the net force. The law does not state that the force is "spent" to produce the acceleration; a force is not a fuel (and in particular a force is not a "conserved quantity" so it is not like energy).

In fact, it is possible to derive the rotational law from the second law, so it is merely a restatement of the same principle. That is, for a net torque, there is a corresponding net angular acceleration. The torque is not "spent" to produce the angular acceleration.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.