# Does inertia imply that a cyclindrical mass in frictionless motion on a horizontal plane keeps moving forever ( with $\vec F = O$)?

Suppose I have a perfectly cylindrical pencil and set it in motion on a perfectly smooth and horizontal table, in such a way that the axis of the pencil is in translation.

Is the state of this object , after having been pushed by me, an example of inertial motion?

I mean can I predict that , in case the table were actually perfectly smooth, the pencil would continue to move in straight line with constant velocity ( under the hypothesis that the surface of the table has no bound)?

The problem I see is that the material points that do not belong to the axis of rotation are accelerated ( since they follow not a straight path, but a cycloidal one). But acceleration implies force, so inertia does not apply to these points, but only to those which are located on the axis.

Other problem, it seems that gravity does some work on the points that are not located on the axis. This work seems to be conpensated by the work done on each symmetric point, but is this enough to say that the sum of the forces acting in the direction of motion is equall to $$0$$?

• Is the axis of the pencil vertical or horizontal? Your question isn't quite clear. Oct 30, 2021 at 13:35
• @Mickael Seifert. - Sorry for lack of clarity. - I meant that the pencil moves horizontally ( without rotation , except w.r.t. its axis) , after having undergone a push exerted laterally. Oct 30, 2021 at 13:59
• That still doesn't quite clear it up for me. Is the pencil standing on its end (vertical axis), or is it lying flat on the surface (horizontal axis)? Oct 30, 2021 at 14:00
• Yes, it will continue to move forever. Those points are being accelerated by internal forces, but there are no external forces that affect the movement of the center of mass, which will continue to move at constant velocity.
– user65081
Oct 30, 2021 at 17:30
• I mean, there are two vertical forces, the normal and gravity, but they cancel each other
– user65081
Oct 30, 2021 at 17:31

For a system of particles, it can be shown that the center of mass of the object obeys Newton's Second Law: if $$M$$ is the total mass of all the particles in the system, and $$\vec{X}$$ is the position of the mass, then $$\vec{F}_\text{ext} = M \ddot{\vec{X}}$$ where $$\vec{F}_\text{ext}$$ is the external force on the system, i.e., the sum of all the forces exerted on the particles in the system by agents outside of the system.
In your case, the center of mass lies on the axis of the pencil. There are two external forces exerted on the particles in the system: the normal force from the table (exerted on the particles instantaneously in contact with the table) and gravity (exerted equally on every particle.) Both of these forces act equally and cancel out, since the pencil is not accelerating upwards or downwards. And since there are no horizontal components to any of the external forces, that means that the horizontal component of $$\ddot{\vec{X}}$$ is zero, which means that the center of mass (i.e., the axis) moves with constant velocity.