How Do Tensions in Upper and Lower Strings Differ in Uniform Circular Motion?

AI Thread Summary
In the discussion about tensions in upper and lower strings during uniform circular motion, participants analyze the forces acting on a 4.00 kg object rotating at a constant speed of 6.00 m/s. It is established that the tensions in the upper and lower strings are not equal due to the differing roles they play in supporting the object's weight and providing centripetal force. A suggestion is made to draw a free body diagram (FBD) to clarify the forces involved. Participants emphasize the importance of maintaining consistent sign conventions for the forces in their calculations. Overall, understanding the distinct contributions of each string's tension is crucial for solving the problem accurately.
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Homework Statement


A 4.00 kg object is attached to a vertical rod by two strings. The object rotates in a horizontal circle at constant speed 6.00m/s. Find the tension in the upper string and the lower string.
View attachment phy2.bmp

Homework Equations


C=2\pir
T=(M)V^2)/r


3. The Attempt at a Solution [
The tension should be easy. Since you have the speed and r. and the mass. Will the tension be the same for both strings though?
 
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student 1 said:

Homework Statement


A 4.00 kg object is attached to a vertical rod by two strings. The object rotates in a horizontal circle at constant speed 6.00m/s. Find the tension in the upper string and the lower string.
View attachment 14466

Homework Equations


C=2\pir
T=(M)V^2)/r


3. The Attempt at a Solution [
The tension should be easy. Since you have the speed and r. and the mass. Will the tension be the same for both strings though?

Although I can't see your attachment I can say that the tension in the two strings, will not, in general be equal. I hope the attachment is a FBD, if it isn't, I suggest that you draw one.
 
Anyone help me?? I know the top string has force of Tcos@-Mg in the y direction and Tsin@ in the x direction. I need help on the bottom string would it be pulling the ball down as well so you would have T+MG for y direction and x direction would be TSin@?
 
student 1 said:
Anyone help me?? I know the top string has force of Tcos@-Mg in the y direction and Tsin@ in the x direction.
Good :approve:
student 1 said:
I need help on the bottom string would it be pulling the ball down as well so you would have T+MG for y direction and x direction would be TSin@?
You need to be careful here. Firstly with the direction, for the top string you have taken up as positive (i.e. the ball's weight is negative), however for the bottom string you have taken down to be positive (i.e. the ball's weight and the tension in the bottom string is positive). You need to chose one convention and stick to it throughout.

Secondly, as I said previously, the tensions in the two strings needn't be equal. Therefore, you need to use different symbols for the tension in the upper and lower string, perhaps T1 and T2, as opposed to simply T.

In addition, instead of considering the string separately, you should consider the forces acting on the ball, which are the tensions of the two strings and it's weight.
 
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