I have drawn a slender rod released from rest. According to Newton's 2nd law, the horizontal displacement of the center of mass, which is located at the centroid, must remains constant as there are no forces acting on it horizontally. So, why if I skecht the rod at different times it's very clear that the horizontal displacement of com is changing. I'm very confused, what is my mistake?

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  • $\begingroup$ A drawing is not a physical object! You can draw successive positions of an object moving upwards, but this does not mean that it defies the law of gravity. $\endgroup$ – sammy gerbil Jun 22 '18 at 22:16

Do a free body diagram of the falling rod. There are three cases to consider, and as the body falls it will transition from one to another.

There are 5 parameters to consider. The position of point A along the horizontal $x_A$ and its derivatives, the position of point A away from the ground $y_A$ and its derivatives, the angle of the rod $\theta$ from vertical, the horizontal reaction force $A_x$ at A and the vertical reaction $A_y$ also at A. For each scenario, there should be 3 unknowns to be solved from the 3 equations of motion.


  1. Fixed End

    Friction on A causes that point to remain fixed in space and the center of mass to move to the right. This ends when $ A_x \gt \mu A_y$.

    $$ \begin{array}{r|l} \mbox{variable} & \mbox{state} \\ \hline \theta & \mbox{unknown} \\ y_A & \mbox{fixed at 0} \\ x_A & \mbox{fixed at 0} \\ A_y & \mbox{unknown} \\ A_x & \mbox{unknown} \\ \end{array}$$

  2. Sliding End

    Friction at A is overcome, and the end slides along the horizontal axis. Sliding friction causes the center of mass to continue moving to the right, but the location of point A is no longer known.

    $$ \begin{array}{r|l} \mbox{variable} & \mbox{state} \\ \hline \theta & \mbox{unknown} \\ y_A & \mbox{fixed at 0} \\ x_A & \mbox{unknown}\\ A_y & \mbox{unknown} \\ A_x & \mbox{dependend, } A_x = \mu A_y \\ \end{array}$$

  3. Flying End

    The rotation of the rod is high enough to lift the end of the rod. The center of mass moves in a projectile motion at this point, under the influence of gravity only.

    $$ \begin{array}{r|l} \mbox{variable} & \mbox{state} \\ \hline \theta & \mbox{unknown} \\ y_A & \mbox{unknown} \\ x_A & \mbox{unknown}\\ A_y & \mbox{fixed at 0} \\ A_x & \mbox{fixed at 0} \\ \end{array}$$

If the plane is frictionless then the rod transitions from Case 1 to Case 2 immediately and $A_x=0$.

  • $\begingroup$ Hi! I've learned that the position of the end of the rod in contact with a frictionless surface is not fixed. In your $\endgroup$ – Sama Jun 23 '18 at 13:22
  • $\begingroup$ Hi! I've learned that the position of the end of the rod in contact with a frictionless surface is not fixed. In case 2 you say that the center of mass is moving to the right while the end of the rod moves to the left, so, as far as I understand this motion one to the rigth and the other to the left will cause the position of the center of mass to remain constant with time, am I rigth? One more thing, how do you know in the third case it will follow a projectil motion? $\endgroup$ – Sama Jun 23 '18 at 13:49
  • $\begingroup$ In the second case if there is no friction then there isn't a horizontal force accelerating the center of mass. It will move with constant velocity along the horizontal direction. So it will maintain whatever initial velocity it had to begin with. If the rod was at rest before let go w/out friction, then the center of mass will not move horizontally ever. $\endgroup$ – ja72 Jun 23 '18 at 15:16
  • $\begingroup$ For case three, it is the same as seen in this video where an object rotating under the influence of gravity appears to have complex motion, but the center of mass moves in simple projectile fashion. $\endgroup$ – ja72 Jun 23 '18 at 15:19

You are assuming that the end of the rod at the origin remains in place, if the floor is friction-less that wont be the case. As you have it right now, there's a reaction force in X direction at that point. I wish this was a comment and not an aswer but the poor system of this site wont allow me....

  • $\begingroup$ why is there a reaction force in my case? $\endgroup$ – Sama Jun 22 '18 at 19:09
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    $\begingroup$ @Sama Because that end of the rod would be trying to slide to the left. The fact that it is not doing so means there must be some force on it to the right keeping it in place. $\endgroup$ – BowlOfRed Jun 22 '18 at 19:19

Simply put, the rod is not in freefall - it's experiencing forces from both the wall and the floor.

It appears the falling rod is wedged in a corner. The vertical wall imparts a force to the rod which causes the center of mass to move horizontally. You will also notice that the horizontal floor imparts a vertical force, causing the left end of the rod to remain motionless while the right end falls.

If you dropped this rod so that it could fall freely (not near any walls or floor), the center of mass would fall straight downward, and the rod would not rotate about its axis.


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