1
$\begingroup$

Please refer to the figure below. The center of mass of the rod under the net force will accelerate. At the same time, the whole rod will also rotate around its center of mass. In his lectures, Feynman explained mathematically the reason why the center of mass translates according to Newton's law as well as why the rod also rotates around it, which I understand very well.

However, he didn't explain why the same rod (or any other object) will rotate around the fulcrum instead of its center of mass, as in the second case in the figure. My questions are:

  1. Without the fulcrum, the center of mass is the center of rotation. What is the physical law/principle/reason for this no longer being the case when the rod is pivoted at the fulcrum? In other word, what makes the fulcrum the preferred center of rotation over the rod's center of mass? Please note that I am looking for a quantitative explanation.

  2. Intuitively, I can see that if the rod is pivoted at two different points, it can neither rotate nor translate. However, again, what is the physical law for this behavior?

    enter image description here

$\endgroup$
  • 1
    $\begingroup$ Your current intuition is misleading. For a rigid body you can assume the rotation to be occurring about any point in the body. When the rod is pivoted at one and rotating with an angular velocity $\omega$ and angular acceleration $\alpha$ about the pivot, it is also rotating about the COM with the same angular velocity $\omega$ and angular acceleration $\alpha$. Just imagine the COM in the second case and the pivoted point to be in motion; you'll understand. $\endgroup$ – user139621 Jul 8 '17 at 3:07
  • $\begingroup$ @Blue Could you elaborate your explanation? I understand that for the purpose of analyzing static condition, we can equalize the sum of all the torques around any point to zero. However, when an object actually rotates, the center of rotation must be a static point and that all other points on the objects rotate around it. Hence, a static point external to the object can also act as the center of rotation. However, there is only one point on the object that can act as the center of rotation. $\endgroup$ – A Slow Learner Jul 8 '17 at 3:20
  • $\begingroup$ Imagine that in the second diagram the centre of mass is the stationary centre of rotation rather than the pivoted point. You'll realize that it is still correct to say that even in the second case the rod is rotating about COM. Similarly for any other point on the body. $\endgroup$ – user139621 Jul 8 '17 at 4:03
  • $\begingroup$ @Blue But in the second diagram, the center of mass is not stationary, and that is the reason it is not the center of rotation, as I said in the last comment. $\endgroup$ – A Slow Learner Jul 8 '17 at 4:05
  • $\begingroup$ Even in the first diagram the centre of mass is not stationary. $\endgroup$ – user139621 Jul 8 '17 at 4:09
2
$\begingroup$

There is something you forgot: there will be a reaction force at the fulcrum - this additional force ensures that the net force is just enough to accelerate the center of mass in a way that makes the angular velocity and the linear velocity consistent with the fulcrum remaining stationary (the reaction force doesn't change the torque about the fulcrum).

When you add a second pivot point, then you fix the position of two points of the rod and there will be two reaction forces to ensure that both the total net force and the net torque will be zero.

This diagram may help:

enter image description here

Assume the force is applied for a short time $\Delta t$; if the force on the pivot point (unknown) acts for the same time, we can solve for both angular and linear momentum as follows:

Linear momentum:

$$mv = (F_i - F_r)\Delta t$$

Angular momentum:

$$I\omega = F_i \Delta t L$$

Noting that the angular momentum of the rod about one end is $I = \frac{1}{12}mL^2 + m\left(\frac{L}{2}\right)^2=\frac13 m L^2$, and that $v = \frac12 \omega L$, we can solve for the reaction force, $F_r$:

$$F_r = -\frac12 F_i$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think I understand what you just explained! Could you draw a force diagram to make it clearer? I just tried to draw one and analyze the motion, but it seems like it is more complicated than I thought. $\endgroup$ – A Slow Learner Jul 8 '17 at 3:30
  • 1
    $\begingroup$ @Geophysics I have added a diagram and some math. Clearer now? $\endgroup$ – Floris Jul 8 '17 at 15:07
  • $\begingroup$ Yes I understand up to this point. Thank you for spending time making this figure. I appreciate if you could elaborate a bit more though. Could you please draw this force diagram at various positions of the rod as it rotates? I would like to see how the centripetal force that keeps every point along the rod moving in the circular path arises. $\endgroup$ – A Slow Learner Jul 8 '17 at 18:07
  • $\begingroup$ Actually, now as I am thinking about the problem, should the reaction force in your diagram be directed toward to fulcrum instead? $\endgroup$ – A Slow Learner Jul 8 '17 at 18:17
  • $\begingroup$ I suppose that immediately after the impact there will be a force along the rod keeping it rotating about the pivot (pointing initially down in my diagram) $\endgroup$ – Floris Jul 8 '17 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.