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A rigid body is formed by two masses $m_1$ and $m_2$ ($m_1$ is not equal to $m_2$) joined by a rod of negligible mass and length $4L$. The system is initially at rest on a smooth horizontal surface. If a horizontal and perpendicular force is applied to the middle of the rod, then it follows that:

enter image description here

I think the answer is d), but I'm not sure.

This is how I reasoned: Considering that $m_1$ is left and $m_2$ is right, the center of mass (CM) is not in the middle because the masses are different, what means the CM is either left or right from the middle of the rod depending on the relationship between $m_1$ and $m_2$.

However this relationship is not stated, but we can make the following assumption: let's say that $m_1 = 2m_2$. Then, the center of mass would be left closer to $m_1$.

First, it is obvious that if the surface is frictionless, the system will accelerate from rest according to Newton's 2nd law for the CM. But... how can I determine if the system rotates or not? Will it have angular acceleration or not?

In general, how can I know exactly when a rigid body has both linear and angular acceleration?

I know that according to Newton's 2nd law for rotational dynamics, angular acceleration ($\alpha$) depends on inertia (of the system) and net torque. But since net torque is calculated respect to a rotation axis (and inertia too) how can I choose it and which is it? My professor told us there's sometimes a natural axis rotation, but I don't get it... The rod doesn't have an end fixed!

Please, I want a clear and deep explanation!

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marked as duplicate by Jon Custer, sammy gerbil, Kyle Kanos, John Rennie homework-and-exercises Sep 1 '17 at 16:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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If a system has a net torque about the center of mass then it will accelerate rotationally. More generally, the change in angular momentum about the center of mass equals the net torque about the center of mass.

The Newton-Euler laws of motion (in vector form) are:

$$\begin{align} \frac{{\rm d} }{{\rm d}t} {\bf p} & = {\bf F} \\ \frac{{\rm d} }{{\rm d}t} {\bf L}_C & = {\boldsymbol \tau}_{C} \end{align}$$

Where, $\bf p$ is linear momentum, $\bf F$ net force, ${\bf L}_C$ angular momentum about the center of mass C, and ${\boldsymbol \tau}_C$ net torque about the center of mass.

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  • $\begingroup$ To ja72. User @Floris said: "If you apply a force that is not aligned with the center of mass of the object, the result will be both rotation and translation. If the force aligns with the center of mass, the object will not change its angular momentum and therefore keeps it". In my example, it was at rest (initial angular momentum equals 0), then it will not rotate. Is that right? $\endgroup$ – Mr. Mister Sep 1 '17 at 15:50
  • $\begingroup$ Does it mean that if I apply a force at the center of mass the net torque will always be 0, although the force and position vectors aren't collineal? $\endgroup$ – Mr. Mister Sep 1 '17 at 15:58
  • $\begingroup$ Each force has a "line of action" where the force vector can slide along without changing the problem. If the line of action goes through the center of mass, then the net torque is zero. It is only the perpendicular distance from the center of mass to the line of action than matters. $\endgroup$ – ja72 Sep 1 '17 at 19:29
  • $\begingroup$ To ja72. When we calculate a torque we do it respect to a rotation axis that appears as a point in 2D. If we have a rigid body that isn't fixed at any point and calculate a torque respect to a rotation axis called O, it's like we nail an imagginary tack on O just the effects of the exerted torque, right? $\endgroup$ – Mr. Mister Sep 1 '17 at 20:25
  • $\begingroup$ The torque about the center of mass is the only thing you have to worry about. A pinned body, will have reaction forces acting and you have to include those forces in the calculation for net torque. In 3D the situation is similar. $\endgroup$ – ja72 Sep 1 '17 at 21:22

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