Tension and Uniform Circular Motion (no r, ω, θ)

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SUMMARY

The discussion centers on solving for the angle θ in a physics problem involving a 0.020-kg mass attached to a 1.2-m string moving in a horizontal circle at a constant speed of 1.50 m/s. Key equations include T = mg/cosθ and Fc = Tsinθ, leading to the relationship mv²/L = mg*sin²θ/cosθ. The main challenge identified is the difficulty in simplifying sin²θ/cosθ, which prevents the direct calculation of θ. Participants emphasize the importance of recognizing trigonometric identities and the potential need for numerical methods in complex problems.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Familiarity with trigonometric identities
  • Knowledge of Newton's second law
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study trigonometric identities and their applications in physics
  • Learn about numerical methods for solving non-analytic problems
  • Explore the relationship between tension and centripetal force in circular motion
  • Practice solving problems involving angles in circular motion scenarios
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Students revisiting physics concepts, educators seeking to enhance their teaching methods, and anyone interested in mastering the application of trigonometry in physical problems.

tmanderson
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Homework Statement


A 0.020-kg mass is attached to a 1.2-m string and moves in a horizontal circle with a constant speed of 1.50 m/s, as shown in the figure. What is the measure of angle θ?

Ix591dNm.png


What is given:
  • m = 0.020kg
  • v = 1.50 m/s
  • L = 1.2 m

Homework Equations


Lengths
  • L = 1.2m
  • Lsinθ = r
Forces:
  • Tx = Tsinθ = Fc
  • Ty = Tcosθ
  • T = mg/cosθ

The Attempt at a Solution


4TEmGSHm.png


What would solve this:
  • atan(Fc/mg) – (Fc needs r)
  • atan(r/h) –(r is unknown, and h would require knowing θ or r)
One (of the many) attempted solutions:
  1. T = mg/cosθ
  2. Fc = Tsinθ
  3. Fc = mg/cosθ * sinθ
  4. Fc = mg*sinθ/cosθ
  5. Fc = mv2/r
  6. mv2/r = mg*sinθ/cosθ
  7. r is unknown, but r = Lsinθ
  8. mv2/Lsinθ = mg*sinθ/cosθ
  9. mv2/L = mg*sin2θ/cosθ
  10. mv2/mgL = sin2θ/cosθ
  11. v2/gL = sin2θ/cosθ

My problem continually being the sin2θ/cosθ, making that not tanθ. If there's something I missed, I feel like it might be some trig identity or sloppy algebra? Nothing came to mind immediately. Every solution path I went, I always ended up needing θ or 'r'. Would really appreciate some help with this one, as I've been consuming too much time with this single scenario. Thank you!
 
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tmanderson said:
My problem continually being the sin2θ/cosθ, making that not tanθ.
Why would this be a problem?
 
Orodruin said:
Why would this be a problem?

Whoops, that should have been sin^2(θ), and my problem being that it is not tanθ (nor tan^2(θ)), and not allowing me to solve for θ with known values.

EDIT: Ah, sorry - It looks like your quote was didn't copy the formatting correctly.
 
tmanderson said:
Whoops, that should have been sin^2(θ), and my problem being that it is not tanθ (nor tan^2(θ)), and not allowing me to solve for θ with known values.

What is the most basic trigonometric identity you are aware of?

(That aside, why do you think everyhing should be analytically solvable?)
 
Orodruin said:
What is the most basic trigonometric identity you are aware of?

a^2 + b^2 = c^2 ?

Orodruin said:
(That aside, why do you think everything should be analytically solvable?)

I'm sorry, I don't think I quite get what you're asking. Can you elaborate?
 
tmanderson said:
a^2 + b^2 = c^2 ?
This is not a relation between trigonometric functions, but you are close.

tmanderson said:
I'm sorry, I don't think I quite get what you're asking. Can you elaborate?
Problems do not always have an analytic solution. In many interesting problems you may have to rely on approximativr or numerical methods to obtain a numerical answer. It is not the case here, but in general you should be open to he possibility.
 
Orodruin said:
This is not a relation between trigonometric functions, but you are close.

1 - cos^2 = sin^2 ?
1 - sin^2 = cos^2 ?

Orodruin said:
Problems do not always have an analytic solution. In many interesting problems you may have to rely on approximativr or numerical methods to obtain a numerical answer. It is not the case here, but in general you should be open to he possibility.

Ah, I understand. I don't actively think that, thought it may have come off that way. I'm revisiting math/science education (sadly, not in a classroom) for the first time in 7 years (on the math side) and 10 years (on the general science side). I never had a desire to really "dig" into math until after my schooling was over. I'm as eager as ever to learn (but realize more and more every day just how much there is) and I appreciate you pointing this out because as I've been going through my mathematics (over the past year or so) I've realized that one potential detriment to online classes or teaching yourself is that you don't have a teacher giving you insights like this.

Sorry for the long winded answer to that.
 
tmanderson said:
1 - cos^2 = sin^2 ?
Yes. Use this and you will have a second order polynomial in cos theta.
 

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