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Why does applying force to a rod in space (or any other isolated systems) on one end leads to both translational and rotational motion?

Diagram of rod and force in space

Here's my logic: suppose there is a rod in space. We magically apply force on one end, perpendicular to the rod. From Newton's second law of motion: $$F_y=ma_{cm,y}.$$ The only force acting in the $y$-direction is $F$, so $a_{cm}$ must equal to $F/m$. Consequently, the rod purely translates as a whole and it doesn't rotates. There is no net external torque because the end of the rod is accelerating linearly along with the center of mass.

The answers to this related Phys.SE question assumes that there is torque, which I can't see why there should be.

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marked as duplicate by Hritik Narayan, HDE 226868, Kyle Kanos, ja72, John Rennie newtonian-mechanics Sep 28 '15 at 5:16

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Your object has a center of mass, and for most practical purposes you can treat all of the mass being centered at that point. Mass has inertia and resists force by Newton's 2nd law as you have written. But If you apply the force not through the center of mass, but rather away from it you now have to treat the mass in a distributed manner. The mass at the center has an inertia, and resists the applied force at a greater magnitude than the mass closer in proximity to the applied force. This leads to a larger translational acceleration than at the center. But because the center mass and mass near the applied force are rigidly coupled, the rod turns.

In other words if you apply the force through the center of mass you accelerate all the mass at the same rate. But if you apply force away from the center of mass you will accelerate mass in the proximity of the applied force at a greater rate than what's applied through the center. And as levers work, it will take allot less force to accelerate that local mass, but also you are "leaving mass behind" away from the proximity.

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The problem is when you say "There is no net external torque because the end of the rod is accelerating linearly along with the center of mass." That's not how torque works; torque doesn't care about the net acceleration. The torque is only a function of where you apply the force and in what direction. In such a case as you have described, the $\vec r\times \vec F$ is non-zero, so there is a torque.

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