Find the force at the middle point of the rod

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To find the force at the midpoint of a rotating rod clamped at one end, one must consider both the centripetal and gravitational forces acting on the rod. The centrifugal force exerted by the outer section of the rod at the midpoint is significant, while the inner section does not contribute additional force due to being supported by the pivot. The discussion emphasizes the need for integration to account for the linear mass density and the varying forces along the length of the rod. The analogy with a hanging rod illustrates how forces are distributed differently based on the rod's position and rotation. Understanding these dynamics is crucial for accurately calculating the force at the midpoint.
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Homework Statement



A rod of length L,mass m is clamped at one end to a wall.The rod is rotated with angular velocity @ (omega say).Find the force at the middle point of the rod.



Homework Equations



Force = ma
Centripetal force = mw^2r



The Attempt at a Solution



I think there is some integration in this but I can't seem to make out what all forces are applied at the mid point.
 
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When a rod is hanging from one end, the bottom half of the rod will pull down on the mid point, while the upper half will not exert any force at that point.

Here, too, the outer part is exerting a centrifugal force outward at the mid point. Assume some linear mass density of the rod, and then integrate the force from L/2 to L.

(If dx is the elementary length at a dist x, then its mass dm = rho*dx, and you know its w in terms of x.)
 
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Why is that?I mean why doesn't the inside part exert force?
 
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Refer to the hanging rod example. The weight of the lower half is supported by the tension at the mid-point, while the weight of the upper half is not. But at the topmost point, the whole weight of the rod is supported.

Here too, the centrifugal force acts outward and away from the centre. It's a very similar scenario. The part which is "outer" to the mid-point pulls outward at the mid-point. The part which is inward to the mid-point is already supported by the force at the pivot, like the topmost point of the hanging rod.
 
All right thank you.
 
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