# Spinning Mass

1. Jun 3, 2014

### gummybeargirl

1. The problem statement, all variables and given/known data
A mass of 3.50 kg is suspended from a 1.59 m long string. It revolves in a horizontal circle as shown in the figure.
The tangential speed of the mass is 2.99 m/s. Calculate the angle between the string and the vertical.

2. Relevant equations
a_c = v^2/r
F = m*a_c
tanθ=sinθ/cosθ
sin^2θ=1-cos^2θ

3. The attempt at a solution
Vertical: T*cos(θ)-m*g=0 → T=(m*g)/cos(θ)
Horizontal: T*sin(θ) = m*a_c
Combined the two: ((m*g)/cos(θ))sin(θ)=m*a_c → m*g*tan(θ)=m*a_c → g*tan(θ)=a_c
we know that a_c=v^2/r, plug that in to get g*tan(θ)=v^2/r
r=L*sin(θ) → g*tan(θ)=v^2/(L*sin(θ))

I cannot get past this point and the notes on the problem say that a quadratic equation needs to be solved at some point.

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2. Jun 3, 2014

### Nathanael

You wrote r in terms of θ and proceeded to try to solve for θ. It left to a bit of a dead end, so now try to solve for r

See if you can write tan(θ) in terms of known quantites and r (it can be done) then you'll be able to solve it

3. Jun 3, 2014

### TSny

Good work so far. Continue from your last equation. Try isolating the trig functions on one side of the equation and all the constants on the other side. Let x = cosθ and see if you can get a quadratic equation in x.

4. Jun 3, 2014

### Nathanael

I don't think that will work. The equation becomes:

$\frac{v^{2}}{gL}=\frac{sin^{2}(θ)}{cos(θ)}$

There is another way to solve for θ (involving the quadratic equation) if you solve for r and then use r to find θ (once you know r, θ is trivial)

EDIT: actually you're right you could solve it that way... (solution removed) sorry about that you are correct

Last edited: Jun 3, 2014
5. Jun 4, 2014

### gummybeargirl

I understand where this equation comes from, but I am not sure how you would go about solving that for θ.

I don't understand at all what I would use to solve for r or how this would result in a quadratic equation.
I am struggling with every aspect of this problem.

6. Jun 4, 2014

### gummybeargirl

Nevermind!!!! I figured it out, thank you all so much for you guidance in this problem.

7. Jun 4, 2014

### Nathanael

I know you said you figured it out now, but I'll still show you the other methods.

You could rewrite $sin^2(θ)$ as $1-cos^2(θ)$ (from pythagorean theorem) and then the equation becomes:
$\frac{v^2}{gL}=\frac{1-cos^2(θ)}{cos(θ)}=\frac{1}{cos(θ)}-cos(θ)$
Multiply both sides by cos(θ) and rearrange and you'll get a quadratic equation (with x=cos(θ)) which can be solved for cos(θ):
$cos^2(θ)+\frac{v^2}{gL}cos(θ)-1=0$
There will only be one root with an absolute value of less than one, so simply take the arccosine of it and you'll have θ
You had written the equation $g*tan(θ)=\frac{v^2}{r}$
Since $L^2=r^2+y^2$ (where y is the unknown side of the triangle) you could use the equation $L^2=r^2+(r*cot(θ))^2$
(... r*cot(θ) is the unknown side ...) to rewrite $tan(θ)$ as $\sqrt{\frac{r^2}{L^2-r^2}}$
Plug that expression into the "tan(θ)" part of your expression (and then square both sides and rearrange) and you'll get a quadratic equation (with $x=r^2$) which you can solve for $r^2$
$g^2r^4+v^4r^2-v^4L^2=0$