What Causes Tension Changes in a Vertical Circle Motion?

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In a vertical circular motion problem, a 0.22-kg ball experiences varying tension in the stick at different positions. At the three o'clock position, the tension is 19 N. The tension at the twelve o'clock position is calculated as 12.532 N, while at the six o'clock position, it is 25.468 N. The value of 6.468 N comes from calculating the weight of the ball (mg) multiplied by three, which is necessary for determining the tension changes at different points in the circle. Understanding these calculations is crucial for analyzing forces in circular motion.
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Homework Statement


A 0.22-kg ball on a stick is whirled on a vertical circle at a constant speed. When the ball is at the three o'clock position, the stick's tension is 19 N.


Homework Equations


mass m = 0.22 kg
tension in 3'o clock position T = 19 N
tension in 12 'o clock position T ' = T-3mg
= 19 - 6.468
= 12.532 N
tension in 6 'o clock position T " = T + 3mg
= 19 + 6.468
= 25.468 N



The Attempt at a Solution



Can someone please help me understand where the [6.468] came from?
 
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matthayzon89 said:

Homework Statement


A 0.22-kg ball on a stick is whirled on a vertical circle at a constant speed. When the ball is at the three o'clock position, the stick's tension is 19 N.


Homework Equations


mass m = 0.22 kg
tension in 3'o clock position T = 19 N
tension in 12 'o clock position T ' = T-3mg
= 19 - 6.468
= 12.532 N
tension in 6 'o clock position T " = T + 3mg
= 19 + 6.468
= 25.468 N



The Attempt at a Solution



Can someone please help me understand where the [6.468] came from?


You said yourself T' = T-3mg

mg = 2.156 so 3mg is 3 times that which gives you 6.468
 
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