Tension in a string holding a spinning object

[Tibriel]

Homework Statement

You spin a 1.20 kg mass at the end of a 1.10 m string. The tangential velocity is 13 m/s. What is the tension in the string?

Homework Equations

x = angle string makes to the vertical
L = string length
r = L*sin(x)
T*cos(x) = mg (because the sum of the vertical forces has to be zero)
T*sin(x) = Fc = mv^2/r

The Attempt at a Solution

Below is the solution I've come up with. My real question is if there is an easier way of solving this. or something I'm missing. In class we ended up putting the 2 equations below into desmos.com and picking the first intersection before 90 degrees to see what the solution should be.

After setting up the free body diagram I get the equations above. Substituting L*sin(x) in for r and I get
T = mg/cos(x)
T = mv^2/[L*sin^2(x)]

we can set these equal to each other and eliminate the mass and I get
sin^2(x)/cos(x) = v^2/(Lg)

substitute: sin^2(x) = 1-cos^2(x)
since the whole right side is made up of known values let's just call all that 'b'
and while we're at it let's set cos(x) = a to make things look simpler
that leaves us with
(1 - a^2)/a = b
1 - a^2 = ab
a^2 + ab = 1
a^2 + ab + (b^2)/4 = 1 + (b^2)/4 <--- complete the square
(a + b/2)^2 = 1 + (b^2)/4
a + b/2 = root [1 + (b^2)/4]
a = -b/2 + root [1 + (b^2)/4]
so now we can say the angle, x = Cos^-1 (a)
and then substitute that into here: T = mg/cos(x), to get the Tension.
T = mg/a

 

[kuruman]

The problem does not specify if the plane of the circle is vertical or horizontal. If vertical, then the problem has to specify at what point of the motion you have to find the tension. It does not do that so you have to assume that the circle is horizontal. What is "x" in your equations? Perhaps you may wish to post your free body diagram. Is this a spherical pendulum?

On edit: Yes, it looks like you are solving the spherical pendulum case. Your solution has gone around in a circle (no pun intended). If, in your final answer, you put in your definition a = cos(x), you get T = mg/cos(x) which you knew all along. I suggest the following: Find expressions for sin(x) and cos(x) using the horizontal and vertical equations, then put these in sin²(x) + cos²(x) = 1 and solve the quadratic in T.

 

[Tibriel]

Yes, you are spinning the object in a horizontal circle, on Earth. So I would imagine that g definitely plays a role in the solution.
x is described in the original post as the angle the string makes to the vertical. Just imagine spinning something in a horizontal circle over your head. I pulled the following picture from the web. It's the same FBD with the exception that I'm using the compliment of the angle shown.

 

[haruspex]

now we can say the angle, x = Cos^-1 (a)
and then substitute that into here: T = mg/cos(x), to get the Tension.

Yes, but of course you do not need to find x. Just use T=mg/a.