Solved: Conical Pendulum: Calculating Tension & Period

[Gold3nlily]

Homework Statement

Fig. 6-53 shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. (The cord sweeps out a cone as the bob rotates.) The bob has a mass of 0.012 kg, the string has length L = 0.86 m and negligible mass, and the bob follows a circular path of circumference 0.52 m. What are (a) the tension in the string and (b) the period of the motion?
Image:
http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c06/fig06_43.gif

Homework Equations

Y: Tcos@-mg=m(0)
Y: Tcos@=mg
X: Tsin@=(mv^2)/r
C=2pi*r
r=C/(2pi)
Period = (2pi*r)/v

The Attempt at a Solution

First I solved for the radius and angle:
r=C/(2pi) = 0.52/(2pi) = 0.08276
sin@=r/Length = 0.08276/0.86
@ = sin^-1 (0.08276/0.86)
@=5.52 degrees

Then I solved for tension:
Tcos@=mg
T=mg/cos@ = (0.012*9.8)/cos(5.52)
T=0.118N (this was the correct answer)

Now I cannot solve for the period.
This is what I have tried:
Period = (2pi*r)/v
Period = (0.52)/v
…need to solve for V
Tsin@=(mv^2)/r
V=((r*Tsin@)/m)^(1/2)
V=((0.08276*0.86sin5.52)/0.012)^(1/2)
V=0.755

Period = (0.52)/ 0.755
Period = 0.688 (this was incorrect)

What did I do wrong?

P.s.- my first test is the Wednesday so hopefully someone can help me really understand this problem soon. (Also, i didn't wait till the last minute this homework just came up yesterday...)

Thank you so much in advance for helping me.

 

[RTW69]

I believe the period of a conical pendulum is 2*Pi*sqrt(L*cos(theta)/g). Google period conical pendulum for explanation

 

[Gold3nlily]

I believe the period of a conical pendulum is 2*Pi*sqrt(L*cos(theta)/g). Google period conical pendulum for explanation

You're right. Thank you. <3

I found explanation here:
http://en.wikipedia.org/wiki/Conical_pendulum