Tension+ centripetal force

  • Thread starter klm
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  • #1
klm
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Two wires are tied to the 400 g sphere shown in figure. The sphere revolves in a horizontal circle at a constant speed of 8.30 m/s

1) What is the tension in the upper wire?
2) What is the tension in the lower wire?

28w1ipz.jpg


okay i know that m=.4 and v=8.3
i dont know r, which i thought that maybe i could find by doing the square root (1^2-.5^2) =.866
and then Force= m V^2 / R sooo.. (.4)(8.3^2)/(.866) = 31.819 . but i dont know where to go from there?
 

Answers and Replies

  • #2
klm
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okay i also think..
Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
Fnety= Tsin 26 + -T2 sin 26 -Fg= mv^2/ r
 
  • #3
klm
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so my free body diagram looks like 1 wire pulling in the 2nd quad, one in the 3 quad, and Fg going straight down on the y axis
 
  • #4
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yep. No motion in y-dir., so forces sum up to zero. In X- dir. they will contribute to centripetal force.
 
Last edited:
  • #5
klm
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but i dont understand how the tension in the 2 wires are going to be different
 
  • #6
klm
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Fnety= Tsin 26 + -T2 sin 26 -Fg= 0
T1 sin 26 - T2 sin 26 = Fg
T1 sin 26 - T2 sin 26 = 3.92

Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
-T1 cos 26 + -T2 cos 26 = 31.82
 
  • #7
klm
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^ i am not sure how to solve for T1 or T2 . can you please help me with that
 
  • #8
klm
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please can someone help me!
 
  • #9
klm
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i figured out the T2, but now i cant figure out T1. shouldnt it just be plug T2 into one of the equations and it should give you T1??
 
  • #10
Doc Al
Mentor
45,085
1,395
Fnety= Tsin 26 + -T2 sin 26 -Fg= 0
T1 sin 26 - T2 sin 26 = Fg
T1 sin 26 - T2 sin 26 = 3.92
Where does the 26 degrees come from?

Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
-T1 cos 26 + -T2 cos 26 = 31.82
Which way does the centripetal acceleration act?

Once you correct your equations, isolate the T1 term in each, like this:

[tex]T_1 \sin\theta = mg + T_2 \sin\theta[/tex]
 
  • #11
klm
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26 came from inverse tan (.5/1)
 
  • #12
klm
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does centripetal acc act on the Horizontal force?
 
  • #13
Doc Al
Mentor
45,085
1,395
26 came from inverse tan (.5/1)
But the 1 m side is the hypotenuse. That should be inverse sin (.5/1).
 
  • #14
Doc Al
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does centripetal acc act on the Horizontal force?
My question is: What's the direction of the centripetal acceleration?
 
  • #15
klm
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umm this may be wrong, but i thought centripetal acc just means that the direction is toward the center
 
  • #16
klm
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ohh wait i know what i did. it is just the angle that is messing me up! thank you for pointing that out. i dont know why, but the angle i used still let me get the tension of the first string??!
 
  • #17
Doc Al
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45,085
1,395
umm this may be wrong, but i thought centripetal acc just means that the direction is toward the center
That's exactly right. So, what direction is that? Should it be positive or negative?
 
  • #18
klm
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negative
 
  • #19
klm
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would that be right?
 
  • #20
Doc Al
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negative
Good. So fix your equation for horizontal forces.
 
  • #21
klm
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Fnetx= -T1 cos 26 + -T2 cos 26 = - mv^2/ r
 
  • #22
Doc Al
Mentor
45,085
1,395
Fnetx= -T1 cos 26 + -T2 cos 26 = - mv^2/ r
Good. Now combine with the other equation and solve.
 

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