What is the magnitude of the angular momentum of the bar?

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Homework Help Overview

The problem involves a rigid, uniform bar of mass m and length b rotating about an axis at its midpoint, with a focus on determining the magnitude of its angular momentum given the linear speed at the endpoints of the bar.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between linear speed and angular velocity, particularly questioning why the linear speed at the endpoints does not require doubling in calculations. There is exploration of the angular momentum equation and its application to different points on the bar.

Discussion Status

Participants are actively engaging with the concepts of angular momentum and linear speed, with some guidance provided regarding the uniformity of angular velocity across the rigid body. There is a recognition of the need to clarify misunderstandings about the application of the equations involved.

Contextual Notes

There is some uncertainty regarding the interpretation of linear speed in relation to the endpoints of the bar and how it affects the calculations for angular momentum. Participants are reflecting on previous errors in reasoning to avoid similar mistakes in the future.

Toranc3
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Homework Statement



A rigid, uniform bar with mass m and length b rotates about the axis passing through the midpoint of the bar perpendicular to the bar. The linear speed of the end points of the bar is v . What is the magnitude of the angular momentum of the bar?

Homework Equations



L=Iw
I=1/12(ML^(2) This L is the length.
v=wr


The Attempt at a Solution



L=1/12*m*B^(2)*w

v=wr
it says that the linear speed of the bar at the end points is v
When I first read that I automatically thought that I would have to double v=wr because it has two endpoints. However, that is wrong. Why is it that I am only supposed to use a single v=wr?
 
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Hi Toranc3! :smile:
Toranc3 said:
v=wr
it says that the linear speed of the bar at the end points is v
When I first read that I automatically thought that I would have to double v=wr because it has two endpoints. However, that is wrong. Why is it that I am only supposed to use a single v=wr?

v = ωr applies to the speed v of any point on the object at distance r from the centre of rotation

so v = ω(L/2) for one end, and v = ω(L/2) for the other end :wink:
 
tiny-tim said:
Hi Toranc3! :smile:v = ωr applies to the speed v of any point on the object at distance r from the centre of rotation

so v = ω(L/2) for one end, and v = ω(L/2) for the other end :wink:

Hi tiny tim, thanks for replying.
So going back to angular momentum equation L=Iw, the omega applies only to a certain point of the rotation. Therefore when I substitute v=wr I am only replacing v/r=w at that certain point. So that is why I do not double it. Would it be correct to say that? The final answer is L=1/6mbv, forgot to add that in.
 
(try using the Quick Symbols box next to the Reply box :wink:)
Toranc3 said:
So going back to angular momentum equation L=Iw, the omega applies only to a certain point of the rotation. Therefore when I substitute v=wr I am only replacing v/r=w at that certain point.

no, ω is the same for every point of a rigid body

v is what varies :wink:

(ie the speed, v, is proportional to the distance, r, from the fixed point)
 
Well I am still not so sure why but I am going to put this error that I did in my head so I will not make the same error again. Thanks though
 

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