Rotational Dynamics Homework: Find Initial Angular Mom & Ang Vel When Bar Vert

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Homework Help Overview

The problem involves a uniform bar hinged at one end and released from a horizontal position, focusing on finding the initial angular momentum and the angular velocity when the bar is vertical. The subject area pertains to rotational dynamics and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve for the initial angular momentum and angular velocity using energy conservation and rotational dynamics equations. Some participants question the treatment of kinetic energy components in the context of the problem.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem, including the moment of inertia and its dependence on the axis of rotation. Some guidance has been offered regarding the definitions of kinetic energy in rotational motion.

Contextual Notes

There are indications of confusion regarding the treatment of kinetic energy terms and the appropriate moment of inertia to use for the given setup. The original poster expresses uncertainty about neglecting linear kinetic energy in their approach.

Clara Chung
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Homework Statement


A uniform bar AB of length L is freely hinged at one end A and released from a horizontal position.
a) Find the initial angular momentum. Ans: 3g/2L
b)Find the angular velocity when the bar is vertical. Ans: (3g/L)^(1/2)

Homework Equations


momentum of inertial of a rod through end =(ML^2)/3

The Attempt at a Solution


I have no problem with part a, I did it like this
Let N be the moment, I be the inertia of moment, a be the angular acceleration.
N = Ia
mg(L/2) = ML^2 /3 a
a= 3g/2L

But I have problem with part b, I tried to do it with energy conservation
Kr + Kt = PE
Iw^2 /2 + mw^2(L^2)/2 = mgL/2
1/3 L w^2 + w^2L = g

I would be correct if I neglect the linear KE, but I don't think it should be neglected, is my method of doing part b wrong?
 
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For rotation about a fixed axis, A, the total KE of the object is (1/2)IAω2.

It is possible to express this as KE = (1/2)Mvcm2 + (1/2)Icmω2. But this would not be helpful for this problem.
 
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TSny said:
For rotation about a fixed axis, A, the total KE of the object is (1/2)IAω2.

It is possible to express this as KE = (1/2)Mvcm2 + (1/2)Icmω2. But this would not be helpful for this problem.

Thanks. what is Icm?
 
cm is center of mass
 
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Moment of inertia, I, depends on the origin. The moment of inertia about the center of the rod is different than the moment of inertia about one end of the rod.
 
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