Calculating Tensions in a Vertical Circular Motion Problem

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SUMMARY

The discussion focuses on calculating the tensions in a rope during a vertical circular motion problem involving a stuntman with a mass of 70 kg swinging from a 4.0-meter rope. The tension at the beginning of the motion is zero due to the initial velocity being zero. At a height of 1.5 meters, the velocity is calculated to be 7 m/s, leading to a tension of 857 N using the formula F = m * v^2 / r. The conversation emphasizes the importance of using conservation of energy for solving such problems, as it simplifies calculations and provides a more intuitive understanding of the forces involved.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Familiarity with the principles of conservation of energy
  • Knowledge of circular motion dynamics
  • Ability to perform basic trigonometric calculations (e.g., arccos)
NEXT STEPS
  • Study the application of conservation of energy in circular motion problems
  • Learn how to derive tension in different positions during circular motion
  • Explore the effects of varying mass and radius on tension in circular motion
  • Investigate the relationship between velocity and tension at different heights in vertical circular motion
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Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for effective teaching methods for these concepts.

cy19861126
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Homework Statement


A stuntman whose mass is 70 kg swings from the end of a 4.0-m-long rope along the arc of a vertical circle. Assuming he starts from rest when the rope is horizontal, find the tensions on the rope that are required to make him follow his circular path,(a) at the beginning of his motion, (b) at a height of 1.5 m above the bottom of the circular arc, and (c) at the bottom of his arc.


Homework Equations


F = ma
F = m * v^2/r


The Attempt at a Solution


a) Using the equation F = m*v^2/r, F is 0 because v is 0
b) First, I used the energy approach attempting to solve the problem:
mgh = 0.5mv^2 + mgh,
9.8 * 4 = 0.5 * v^2 + 9.8 * 1.5
v = 7m/s.

As I get velocity, I plugged the value into
F = m * v^2 / r
F = 70kg * 7^2 / 4
F = 857N

I do not think this is right, since the value is too large. Also, I found out that I can calculate the angle of the distance he travelled: arccos(2.5/4) = 0.89rad --> pi/2 - 0.89 rad = 0.68 rad, this is the angle which he travelled. But I do not know where to plug this in anywhere.
 
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Why do you think this value is too large? Compare your tension with the weight of the stuntman; does the value seem more realistic now?
 
Last edited:
Hootenanny said:
Why do you think this value is too large? Compare your tension with the weight of the stuntman; does the value seem more realistic now?
Are you saying I did this right? Yeah, now you've said it, I think it is realistic... Wow, I thought I didn't get this chapter at all
 
cy19861126 said:
Are you saying I did this right? Yeah, now you've said it, I think it is realistic... Wow, I thought I didn't get this chapter at all
Yep it looks spot on to me :smile:. Conservation of energy is by far the easiest approach to solving circular motion, many students overlook this method initially as they want to use all the new equations (toys) they've been given :wink:. They forget their basic principles, well that's what I find anyway...
 

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