Circular Motion at an angle, finding angles and tension

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SUMMARY

The discussion focuses on calculating the speed, angle, and tension of a circular motion system involving a ride for children. The seats weigh 10.0 kg, and the chains make an angle of 28 degrees to the vertical when empty. The calculations reveal that the speed of each seat is 5.19 m/s, and the angle with a child in the seat is approximately 53.68 degrees. The tension in the chain during the ride, based on the calculations provided, is estimated to be 827.3 N, although this value may require further verification.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Familiarity with Newton's laws of motion
  • Knowledge of trigonometric functions in physics
  • Ability to apply equations of motion, specifically Fc = mv²/r
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  • Review the principles of circular motion and tension in strings
  • Learn about the effects of mass on tension and angle in circular systems
  • Explore the use of trigonometric identities in physics problems
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Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for practical examples of tension and angle calculations in real-world applications.

etern1ty
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Homework Statement


At the royal show there is a ride for younger children where they sit in a seat and it spins in a circle. The seats weigh 10.0kg and by comparison the mass of the chains holding the seats is negligible. When the system rotates the chains make an angle of 28 degrees to the vertical if there is nobody sitting in them. The radius of the system supporting the chains is 4m.The length of each chain supporting a chair is 2.50m long. a) what is the speed of each seat? c) what angle would the chain make with the vertical with the child in a seat? d) what is the tension in the chain during the ride with the child in the seat?

Homework Equations


Fc = mv^2/r
W=mg
T=ma+mg

The Attempt at a Solution


Part a)
W=mg=10*9.8 = 98N
I found the horizontal component of the weight at the end of the chain = 98tan28=52.11N
As Fc=Horizontal Weight to keep it in position, i figured that 52.11=mv^2/r
R=2.5sin28 = 1.17m+the 4m off the base of the support = 5.17m
52.11=mv^2/5.17 = 10v^2/5.17. V=*square root*(52.11*5.17)/10=5.19ms-1

Part c)
In this i wasnt sure how to find the two unknown variables, as i don't know both V and theta so i assumed V was the same as part a), 5.19ms-1
Fc=mv^2/r = (10+40)*5.19^2/(2.5sin*theta)+4
As Fc=Horizontal component of Weight, Fc=mgtan theta.
= [(10+40)*5.19^2/(2.5sin*theta)+4]=490tan*theta
And so rearranging gave me [(50*5.19^2)/(2.5*490+4)]=sin*theta*tan*theta.
from this i solved in solver of my calculator and got an angle of 53.68 degrees. I am pretty certain to say this is wrong as the mass has increased, and so the angle should decrease no? how would you solve this question for both V and theta?

Part D)
I only know the equation that i think is right for tension, i have no idea what to do for this part of the question. I've tried using the weight down of 490N (50*9.8) and using the angle gathered in part C and using trig to solve, in which i got a tension of 827.3N but I am not certain about that at all.

If a picture is needed for the question, please ask and ill try upload one
 
Last edited:
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Hi etern1ty, welcome to PF.

If the velocity remains constant, θ is independent of mass in the chair. But tension will increase.
 
Last edited:

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