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Center of Rotationby SS2006
Tags: rotation 
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#1
Dec805, 02:52 PM

P: 84

If w = 1.0 rad/s
0000 lenght 1 = 2 m lenght 2 = 3 m lenght 3 = 4 m each circl is a object weighing 1 kg. find center of rotation and tension in the massless rods getting center of rotation was easy but tension.. well i know if theres 3 objects, to get tension, i used m*wsquare*r from each object to the center of rotation giving me 2 tensions, that worked, but with 4 objects, what do we do here. ( iknow we should get 3 tensions) 


#2
Dec805, 04:28 PM

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P: 1,772

It's much easier than you might think. After solving for the outer tensions. Look at one of the "inside masses." The ouside tension is pulling outward, and the inside tension pulls inward. The net force here must be centripetal force. If you did things correctly, it doesn't matter which inner mass you choose, because the tension must be the same on each (tension pulls equqally at both ends).



#3
Dec805, 04:51 PM

P: 84

is this right then?
t1 = 4 N t3 = 5N t2 = 1 N (difference between 2 ends?) letmeknow thanks 


#4
Dec805, 06:27 PM

Sci Advisor
HW Helper
P: 1,772

Center of Rotation
YOu have 1 and 3 correct, but the middle tension is not.
call the four particles a, b, c, d. You can see that the centripetal force on the c must be 1 N right? Well, in perfect uniform circular motion, the centripetal force must be the net force. Tension 3 is pulling on c to the right with 5N (since 5 N is pulling on d at the other end), but the net force on c must be 1 N. How much must tension 2 be to make it so? You find the same answer if you use the same longic on b. 


#5
Dec805, 07:10 PM

P: 84

i get 4 N when using C
and 6N when using B shudnt they be teh same for C fnet + 5 =1 fnet = 4 for b 4 + fnet = 2 fnet = 6 


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