Finding the tension of the string (vertical circular motion)

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SUMMARY

The discussion focuses on calculating the tension in a string during vertical circular motion of a ball with a mass of 125g and a string length of 0.900 meters, rotating at 38 RPM. The tension at the top of the circle is determined to be 0.553 N, while the tension at the bottom is calculated to be 3.01 N. Key equations used include the radial acceleration formula, arad = v²/R, and the application of Newton's second law for both positions of the ball. The calculations were confirmed to be correct, with a minor clarification regarding the relationship between revolutions per second and period.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Knowledge of radial acceleration in circular motion
  • Ability to convert RPM to revolutions per second
  • Familiarity with basic physics concepts related to forces
NEXT STEPS
  • Study the derivation of the radial acceleration formula in circular motion
  • Learn about the effects of tension in various types of circular motion
  • Explore the relationship between angular velocity and linear velocity
  • Investigate the role of centripetal force in circular motion dynamics
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Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to clarify concepts related to tension and forces in rotational systems.

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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