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

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1. Jun 22, 2015

### tmanderson

1. The problem statement, all variables and given/known data
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?

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

2. Relevant equations
Lengths
• L = 1.2m
• Lsinθ = r
Forces:
• Tx = Tsinθ = Fc
• Ty = Tcosθ
• T = mg/cosθ
3. The attempt at a solution

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!

2. Jun 22, 2015

### Orodruin

Staff Emeritus
Why would this be a problem?

3. Jun 22, 2015

### tmanderson

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.

4. Jun 22, 2015

### Orodruin

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

(That aside, why do you think everyhing should be analytically solvable?)

5. Jun 22, 2015

### tmanderson

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

I'm sorry, I don't think I quite get what you're asking. Can you elaborate?

6. Jun 22, 2015

### Orodruin

Staff Emeritus
This is not a relation between trigonometric functions, but you are close.

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.

7. Jun 22, 2015

### tmanderson

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

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.

8. Jun 23, 2015

### Orodruin

Staff Emeritus
Yes. Use this and you will have a second order polynomial in cos theta.