Calculate Tension in a String: Ball at Top & Bottom

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SUMMARY

The discussion focuses on calculating the tension in a string when a ball is revolved in a vertical circle. The ball has a mass of 0.300 kg and moves at a speed of 4 m/s with a radius of 72 cm. The correct equations for tension at the top and bottom of the path are established as FsubT + mg = m(v^2/r) for the top and FsubT - mg = m(v^2/r) for the bottom. The correct tension values are determined to be 3.72 N at the top and 9.61 N at the bottom of the path.

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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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