Finding the tension of the string (vertical circular motion)

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Homework Help Overview

The problem involves a ball attached to a string undergoing vertical circular motion, with a focus on determining the tension in the string at both the top and bottom of the circular path. The context includes concepts from dynamics and circular motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the tension in the string using radial acceleration and Newton's second law, raising questions about the correctness of their calculations and the relationships between period and frequency.

Discussion Status

Participants are engaged in verifying the calculations and clarifying the relationship between revolutions per second and period. Some corrections have been noted regarding the interpretation of the equations used, but no consensus has been reached on the overall correctness of the approach.

Contextual Notes

There are indications of confusion regarding the conversion between frequency and period, as well as the application of Newton's laws in the context of circular motion. Participants express a need to improve their understanding of these relationships.

littlebearrrr
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Homework Statement



A ball of mass 125g is attached to a string .900 meters long. It is then set into vertical circular motion with 38 RPM. What is the tension of the string at the top of the circle and at the bottom of the circle?

Homework Equations



∑Fy = may = marad

arad = v2/R = 4∏2R/T2

The Attempt at a Solution



First, I found T (period) by using the given RPM:
(38 rev/1 min)(1 min/60 s) = 0.633 rev/s --> 1.58 s per revolution = T

Used T to calculate arad (radial acceleration):
arad = 4∏2(.900 m) / (1.58 s)2 = 14.23 m/s2

Finding the tension at the top of the circle:
The only forces acting on the ball at the top are its weight and the tension force acting in the same direction (downward). With these facts, I apply Newton's second law for the vertical direction and solve for the tension -
∑Fy = may = marad = w + T
T = marad - w = (.125 kg)(14.23 m/s2) - (.125 kg)(9.81 m/s2) = 0.553 N

Finding the tension at the bottom of the circle:
At the bottom, the forces now oppose each other. The tension is directed upward, while the weight remains directed down. Again, I apply Newton's second law -
∑Fy = may = marad = T - w
T = marad + w = (.125 kg)(14.23 m/s2) + (.125 kg)(9.81 m/s2) = 3.01 N

Just checking to see if this is correct. Thank you in advance!
 
Last edited:
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Yes, it's correct, except the part where you said 0.633 rev/s = 1.58 s per revolution. That's wrong. But it didn't affect your answer.
 
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Ah, so it was just supposed to be 1.58s? Also, thanks for checking my answer. Appreciate it!
 
No, the problem is that the correct equation is 0.633 rev/s = 1/(1.58 s per revolution)
 
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Oh, got it! Thanks for pointing that out; I fixed it above. I have a bad habit of equating the wrong things when reporting answers sometimes, even when the answer is right. I definitely need to work on that.
 
littlebearrrr said:
Oh, got it! Thanks for pointing that out. I have a bad habit of equating the wrong things when reporting answers sometimes, even when the answer is right. I definitely need to work on that.

That happens because you're mentally saying "equal" when you ought to say "therefore". The mathematical symbol for therefore is an arrow \Rightarrow.
 

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