What Causes Tension Changes in a Vertical Circle Motion?

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SUMMARY

The discussion centers on the tension changes experienced by a 0.22-kg ball on a stick whirled in a vertical circle at a constant speed. 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, derived from the equation T' = T - 3mg, where 3mg equals 6.468 N. Conversely, at the six o'clock position, the tension increases to 25.468 N, calculated using T" = T + 3mg.

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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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