# Apply force to one end of a rod in space: why not purely translational? [duplicate]

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

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.

## marked as duplicate by Hritik Narayan, HDE 226868, Kyle Kanos, ja72, John Rennie newtonian-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 28 '15 at 5:16

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.