Finding the tension of the string (vertical circular motion)

In summary, the conversation discussed the calculation of tension in a string attached to a ball of mass 125g in vertical circular motion with a length of .900 meters and a speed of 38 RPM. The equation for radial acceleration was used, along with Newton's second law, to find the tension at the top and bottom of the circle. The correct value for the period was 1.58 seconds per revolution, and the final answers for tension were 0.553 N at the top and 3.01 N at the bottom.
  • #1
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!
 
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  • #2
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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  • #3
Ah, so it was just supposed to be 1.58s? Also, thanks for checking my answer. Appreciate it!
 
  • #4
No, the problem is that the correct equation is 0.633 rev/s = 1/(1.58 s per revolution)
 
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  • #5
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.
 
  • #6
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 [itex]\Rightarrow[/itex].
 

Related to Finding the tension of the string (vertical circular motion)

1. What is the equation for finding the tension of a string in vertical circular motion?

In order to find the tension of a string in vertical circular motion, you can use the equation T = mv^2/r, where T is the tension, m is the mass of the object, v is the velocity, and r is the radius of the circular motion.

2. How does the tension of the string change during vertical circular motion?

The tension of the string will change as the object moves through the circular motion. It will be highest at the bottom of the circle, when the object is moving fastest, and lowest at the top of the circle, when the object is moving slowest.

3. What factors affect the tension of the string in vertical circular motion?

The tension of the string is affected by several factors, including the mass of the object, the velocity of the object, and the radius of the circular motion. Additionally, the force of gravity and any external forces acting on the object can also affect the tension of the string.

4. How does the tension of the string affect the object's motion in vertical circular motion?

The tension of the string is what keeps the object in circular motion. If the tension is too low, the object will not be able to maintain the circular path and will fall. If the tension is too high, the object will move faster and may even break the string.

5. Can the tension of the string ever be equal to the weight of the object in vertical circular motion?

Yes, it is possible for the tension of the string to be equal to the weight of the object at certain points in the circular motion. This occurs when the object is at the top or bottom of the circle, where the tension is the lowest and highest respectively.

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