Friday, 30 September 2016

classical mechanics - Normal force in a compound pendulum (physical pundulum) system?


Consider a compound pendulum pivoted about a fixed horizontal axis, illustrated by the force diagram on the right:


image was taken from hyperphysics.phy-astr.gsu.edu/hbase/pend.html#c4


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Okay, I can't figure out where the normal force on the pendlum should point (this force isn't indicated in the diagram).


On the one hand, I think it should point in the same direction as the tension force on the left-side diagram, in order to produce the rotation. But on the other hand, shouldn't the normal force point towards the pendulum rather than away from it? In any case, I know that normal face is perpendicuar to the contact surface, but how does "contact surface" fit into the context of this scenario?


Please help me understand the concepts involved.



Answer



If you think of the pendulum as swinging on a nail going through a smooth hole near the top, then the tension in the pendulum will pull in the direction of the pendulum's length. The normal force will be opposite to this (your first answer) and will be exerted by the nail on the point of the hole in contact with it. So if the nail has a circular cross section, the force will be exerted at the point of this circle at the "top end" of the diameter which is in the direction of the pendulum's length.



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