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enter image description here In this diagram, assuming there is no friction between the pin joint and the rod, I understand that the rod will rotate about $O$. However, let's say that the rod is released from $ \theta = 0$ and that when $ \theta = 90$ the rod is vertical at that instant in time. If we drew the forces on the rod at this instant, there would be a weight force and a horizontal and vertical reaction force acting at $O$. What is confusing me is that surely the horizontal reaction force at this instant must be 0 since there is no other force to balance it and no acceleration occurs in the x direction (the angular acceleration at 90 degrees will be 0 and so it follows that the linear acceleration is also 0 from $ a = r \alpha$).

However, in order for the rod to keep rotating doesn't a moment need to be applied about $O$? And in the situation where $ \theta = 90$ the horizontal reaction force is the only force that can provide this moment, leading me to believe that this force is indeed non zero. Therefore, at $ \theta = 90$ will the horizontal reaction force be 0 N or non zero and if it is non zero how would you calculate it?

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Your analysis is very precise and reasonable, with respect that the horizontal forces should be zero when $\theta =90^o$.

If your question boils down to:

Do you need a horizontal force at O in order for the rod to keep rotating ?

then I think you are just forgetting that, when the rod is passing through the vertical position, at that point its center of mass already has a velocity and therefore some kinetic energy. Therefore, to keep moving it does not need any external moment.

So its because of the inertia of the rod, that it will keep moving. (it will try to mode horizontally at that point, but even at that point the reaction of the pin force it to perform a circular motion).

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  • $\begingroup$ I see my mistake now. I was thinking in terms of x-y coordinates and not normal-tangent coordinates which seems to make the analysis easier. Would I be correct to assume that in the normal direction, when theta = 90 degrees, the reaction force at the pin joint will be: $ Oy = W + mv^2/R$? $\endgroup$
    – J.Doe
    Apr 25 at 1:06
  • $\begingroup$ Yes that Would be accurate. $\endgroup$
    – NMech
    Apr 25 at 2:28
  • $\begingroup$ @J.Doe To me your thinking is deeper than most. If you still have doubts about your thinking, consider the following. If the rod was given an initial velocity and was free to rotate on the horizontal plane, then the force on the pin would be equal to $ mv^2/R$. And the rod would continue to rotate happily ad infinitum, with only the reaction force from the pin. I'd be happy to explain that further, if you need to. $\endgroup$
    – NMech
    Apr 25 at 4:13
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Since the center of mass offsets the center of support with a distance $"r"$, The rod will rotate (with point $"B"$ heads down) about the frictionless pin $(point "O")$ in a circular motion and convert the potential energy into kinetic energy. Along the path, the gravity force $(W = m*g)$ will have a radial force component and a tangential force component. The magnitudes of the component forces vary with position, but the net horizontal force remains at zero at all time, so there is no horizontal reaction at the support pin as $\sum Fx = 0$ remains valid.

Due to angular velocity and acceleration, the rod will continue to move past the vertical position $(\theta = 90^o)$ until the energy has exhausted by the drag (air friction) and the reversed kinetic motion. From here on, it swings back and forth similar to a pendulum. that will swing for a long time without interference from an external force.

Note, without externally applied force (torque, moment), the rod will not make a full circle rotation.

enter image description here

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  • $\begingroup$ Yes, conservation of mass, so it is always there and be a vertical force due to gravity. Don't confuse it as the traction force that acts along the circular path and produces torque, though in this case, the gravitational force has the same effect (produce a moment about point "O" to initiate the rotation). $\endgroup$
    – r13
    Apr 25 at 1:04
  • $\begingroup$ I see my mistake now. I was thinking in terms of x-y coordinates and not normal-tangent coordinates which seems to make the analysis easier. Would I be correct to assume that in the normal direction, when theta = 90 degrees, the reaction force at the pin joint will be: $ Oy=W+mv^2/R?$ $\endgroup$
    – J.Doe
    Apr 25 at 1:08
  • $\begingroup$ No, there is only a single mass of the rod. At 90 degrees, the tangential force component equal to zero, while the radial force equal to the weight, so Qy = -W. IMO, the energy equation you wrote involves the average (constant) velocity, which is not true for this case. You shall double-check on the velocity term using equations of motion. $\endgroup$
    – r13
    Apr 25 at 1:41
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    $\begingroup$ But the centre of mass (COM) of the rod is undergoing circular motion is it not? I understand how the tangential force component is zero but in the normal(radial direction) doesn't there need to be centripetal acceleration? Hence the reason for using $ O - W = mv^2/R$ where $R$ is the distance from the COM to the rotation axis. $\endgroup$
    – J.Doe
    Apr 25 at 2:06
  • $\begingroup$ @J Doe You are correct. I was wrong. $\endgroup$
    – r13
    Apr 25 at 6:44

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