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

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SUMMARY

The magnitude of the angular momentum of a rigid, uniform bar rotating about its midpoint is calculated using the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. For a bar of mass m and length b, the moment of inertia is I = 1/12 * m * b². The relationship between linear speed v and angular velocity is given by v = ω(L/2), which applies to each endpoint of the bar. The final expression for angular momentum is L = 1/6 * mbv, confirming that the linear speed does not need to be doubled for both endpoints.

PREREQUISITES
  • Understanding of angular momentum and its formula L = Iω
  • Knowledge of moment of inertia for a rigid body, specifically I = 1/12 * m * L²
  • Familiarity with the relationship between linear speed and angular velocity, v = ωr
  • Basic concepts of rotational motion and rigid body dynamics
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes, including rods and disks
  • Learn about the conservation of angular momentum in closed systems
  • Explore the effects of varying mass distribution on angular momentum
  • Investigate real-world applications of angular momentum in mechanical systems
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Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators teaching concepts related to angular momentum and rigid body motion.

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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