Conservation of Energy for a rotating Rod

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Homework Help Overview

The discussion revolves around the conservation of energy in the context of a rotating rod, specifically focusing on the potential energy associated with its center of mass and the implications for torque due to gravity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the reasoning behind the expression for potential energy as MgL/2, questioning the location of the center of mass and its relevance to torque. There are attempts to clarify the relationship between height and potential energy as the rod rotates.

Discussion Status

Several participants have provided insights regarding the center of mass and its effect on potential energy calculations. There is an ongoing exploration of the implications of these concepts on torque, with some participants acknowledging previous answers while still engaging with the topic.

Contextual Notes

Participants note that the mass of the rod is uniform and that the height used in potential energy calculations is based on the center of mass being at L/2. There is a mention of neglecting other masses in the problem setup.

AROD
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Homework Statement


rod%20problem.jpg

Homework Equations



Rotational Kinetic Energy = 1/2*I*w^2

The Attempt at a Solution



I was just wondering if someone explain to me why the potential energy is MgL/2 ?

Is this not the same for the torque from gravity?
 
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As the hint in a) states, you can neglect the mass m. And since the mass M is uniform, you can presume all its weight is in the center, hence MgL/2
 
Because the rod's center of mass is placed at L/2 over the ground.

Edit: oops, already answered :)
 
AROD said:
I was just wondering if someone explain to me why the potential energy is MgL/2 ?
The rod is uniform. Where is its center of mass?
Is this not the same for the torque from gravity?
The torque from gravity about the pivot point depends upon the angle of the rod.

Edit: Already answered twice!
 
ok so then its just mgh with h = L/2 . that makes sense

then as it tips the hypotoneuse would be L/2, and the height would be this times the cos of theta.

thanks
 

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