I think my answer should be right, but it isn't energy problem

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Homework Help Overview

The problem involves a simple pendulum consisting of a mass attached to a rigid rod, which is initially inverted and then released. The participants are tasked with determining the speed of the mass at its lowest point and the tension in the suspension at that instant.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply conservation of energy to find the speed at the lowest point, but questions the discrepancy between their result and the book's answer. Some participants question the interpretation of the term "inverted" and its implications for the height of the mass. Others discuss the nature of tension in the suspension and whether it can be simply equated to the weight of the mass.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and clarifying concepts related to energy conservation and forces acting on the pendulum. Some guidance has been offered regarding the nature of tension and centripetal acceleration, but no consensus has been reached.

Contextual Notes

There is uncertainty regarding the definition of "inverted" and its effect on the calculations. Additionally, the participants are grappling with the implications of tension in the context of centripetal motion.

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Homework Statement



A light rigid rod of length L has a mass m attached to its end, forming a simple pendulum. It is inverted and then released (I don't know what that's supposed to mean). What are:

a) the speed v at its lowest point
b) the tension T in the suspension at that instant

Homework Equations





The Attempt at a Solution



at the beginning, when it is dropped down, we should have K = 0 and U = mgL

at the bottom when it is finished dropping, we should have K = 1/2mv^2 U = 0

1/2 mv^2 = mgl

v = sqrt(2gL)

but the book says 2sqrt(gl)...

what did I do wrong?

thanks
 
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I'm guessing "inverted" means the mass is a height L above the pivot point, so it's height above its lowest point is actually 2L rather than L
 
WOW. Thanks so much, I would never have known...

What about a "suspension"?

b) the tension T in the suspension at that instant..

Shouldn't it just be the negative weight? what answer book says 5mg.. :S

Thanks
 
holezch said:
WOW. Thanks so much, I would never have known...

What about a "suspension"?

b) the tension T in the suspension at that instant..

Shouldn't it just be the negative weight? what answer book says 5mg.. :S

Thanks
the 'tension T in the suspension at that instant' means 'what is the tension in the rod when the mass is suspended by the rod at the low point of its motion (that is, upon returning to its original position before it was inverted, but this time with a speed associated with it). This becomes a centripetal acceleration question, the tension is not the negative of the weight, because it is accelerating inwardly. The sum of forces in the y direction is not 0.
 

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