Calculate Tension in a String: Ball at Top & Bottom

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To calculate the tension in a string with a ball revolving in a vertical circle, the equations of motion are applied. At the top of the path, the tension is calculated using the equation mg + FsubT = m(v^2/r), resulting in a tension of 3.72N. At the bottom, the correct equation is FsubT - mg = m(v^2/r), which resolves the tension to 9.61N. The importance of correctly defining the positive and negative directions in the equations is emphasized. Proper application of these principles is crucial for accurate tension calculations in circular motion scenarios.
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Homework Statement



A ball at the end of a string is revolved at a uniform rate in a vertical circle of radius 72cm. If its speed is 4m/s and its mass is 0.300kg, calculate the tentsion in the string when the ball is (a) at the top of its path and (b) at the bottom of its path.

Homework Equations



F=ma, and variations thereof to include the sum of all forces

The Attempt at a Solution



I got (a) by using mg+FsubT=m(v^2/r) and I came up with 6.66N-2.94N=3.72N, but for some reason I can't get (b). I'm doing this for practice for an exam, so I'm doing the problems that have the answers in the back so that I can check myself, and the answer for (b) is supposed to be 9.61N, but I can't figure out how to get there.

The equation that I'm using for (b) is mg-FsubT=m(v^2/r), because that looked right from my free body diagram (with x reference frame point down), but I keep ending up with 2.94N-6.66N=-3.72N ...

This is obviously not right... the answer is positive and 3x the magnitude on top of it, so... I don't know what I'm doing wrong. Any help is greatly appreciated.
 
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"mg-FsubT=m(v^2/r)"

your signs/directions are wrong here.

If you take upwards positive... downwards negative...

FsubT - mg = mv^2/r

now solve for FsubT
 
Excellent. Many thanks.
 

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