Another Angular Momentum

  • Thread starter Jacob87411
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  • #1
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A light, rigid rod 1 meter in length roatets about an axis perpendicular to its length and its center. Two particles of masses 4 KG and 3 KG are connected to the ends of the rod. What is the angular momentum of the system if the speeds of earch particle is 5 m /s

Curious if this is right

L=Iw
Both are point masses so I=MR^2, so I
=(4KG)(.5M^2) + (3KG)(.5^2)

w=r/v
w=.5/5 = 1/10

So plug this all back into L=IW

[(4KG)(.5M^2)+(3KG)(.5M^2)]x.1
 

Answers and Replies

  • #2
BobG
Science Advisor
Homework Helper
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One slight change.

[tex]v=r\omega[/tex]

so:

[tex]\frac{v}{r}=\omega[/tex]
 
  • #3
171
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Ah so its 5/.5 so 10, thanks!
 

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