Solving Tension in Two Wires Attached to 200 g Sphere

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Homework Help Overview

The problem involves analyzing the forces acting on a 200 g sphere that revolves in a horizontal circle at a constant speed of 6.80 m/s, with two wires attached to it. The discussion centers around the tension in the wires and the effects of centripetal force and gravitational force on the system.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the tensions in the two strings, questioning whether they must be equal due to the symmetry of the setup. Some participants suggest using trigonometric relationships to analyze the forces involved.

Discussion Status

There is an active exploration of the forces acting on the sphere, with some participants providing guidance on using Newton's laws to analyze the system. Multiple interpretations regarding the equality of tensions are being discussed, particularly in relation to the gravitational force acting on the sphere.

Contextual Notes

Participants note the importance of considering both the radial acceleration and the weight of the sphere, as well as the potential for one of the strings to be slack during the motion. There is an emphasis on ensuring that the lower string does not remain slack as the system spins.

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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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