Solving Tension in Two Wires Attached to 200 g Sphere

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The discussion revolves around calculating the tension in two wires attached to a 200 g sphere revolving in a horizontal circle at a speed of 6.80 m/s. Participants clarify that the tensions in the wires are not equal due to the sphere's weight, which affects the upper wire more than the lower one. The upper wire supports the sphere's weight while the lower wire remains slack during the initial spin-up. To solve for the tensions, it's suggested to apply Newton's laws and analyze forces in both the horizontal and vertical directions. Ultimately, the equations derived from these forces will allow for the determination of the tensions in the two wires.
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Homework Statement



Two wires are tied to the 200 g sphere shown in figure. The sphere revolves in a horiIzontal circle at a constant speed of 6.80 m/s.

http://img684.imageshack.us/img684/6537/knightfigure0761.jpg

Homework Equations


Ca=m*v^2/r
f=ma

The Attempt at a Solution


I was able to find the radius using triangles, and with that
I know the centrifical acceleration is .200*6.8^2/.866=10.679
But i have no idea where to go from here
 
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Use trig to find what tension provides a horizontal component of half the centripetal force m*v^2/r
 
Does that meen the tension of the 2 strings will be equal? that is what's throwing me off
 
Yes. The tensions must be equal because it is perfectly symmetrical.
 
Delphi51 said:
Yes. The tensions must be equal because it is perfectly symmetrical.
What about the weight of the sphere? Will it not contribute to the tension.
 
I may be wrong, but I don't see how it's possible for the tensions to be equal. In addition to the radial acceleration, there is also the vertical weight of the sphere. So, the upper string will be supporting the weight of the sphere, but the bottom string won't.

Imagine the system is just starting to spin. The sphere will hang, making the upper string tight, but the lower string will be slack. As the system spins, the ball will be forced outward. If the lower string weren't there, the top string might almost reach a horizontal angle if the rotation is fast enough. But the bottom string will kick in and keep the sphere from rising above a certain point. No matter how fast it spins, the top string will never be slack, and the way I figure it, the top string will always have greater tension than the bottom string.

Is that incorrect?
 
chudd88 said:
I may be wrong, but I don't see how it's possible for the tensions to be equal. Is that incorrect?
I think you may be correct, however, in order to find out, the OP will have to assume that they are not equal and sum forces in the x direction and y directions and use Newton's laws, and then be sure that ther lower rope does not stay slack. Juggalomike, what's the acceleration in the x direction, and in the y direction?
 
i believe acceleration in the x direction is 10.679 and the y direction is M*G
 
juggalomike said:
i believe acceleration in the x direction is 10.679
that's the net centripetal force, in Newtons
and the y direction is M*G
Is there any acceleration in the y direction? And you are confusing acceleration with force, which are related by Newton's 2nd law: F_net = ma.
 
  • #10
Let T1 and T2 are the tensions in upper and lower strings. Let 2θ be the angle between two strings. Then
T1*cosθ + T2*cosθ = m*v^2/R...(1)
T1*sinθ = T2*sinθ + mg...(2)
Solve these two equations to find T1 and T2.
 
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