A 22 g ball is fastened to one end of a string

Question

86 cm long and the other end is held fixed at
point O so that the string makes an angle of
35
◦
with the vertical, as in the figure. This
angle remains constant as the ball rotates in
a horizontal circle. The angle
θ
would remain
constant only for a particular speed of the ball
in its circular path.
The acceleration of gravity is 9
Find the velocity

Answer 1

F=Fx+Fy

Two equations:

1.)

Fx=F*Sin35=m*v^2/r

2.)

Fy=F*Cos35-Fg

Fy=F*Cos35-m*g

Solve equation 2 for F:

m*g=F*Cos35

Solving for F:

mg/Cos35=F

Plug 2 into 1 and simplify:

(mg/Cos35)*Sin35=m*v^2/r

mg*Tan35=m*v^2/r

Masses cancel:

g*Tan35=v^2/r

Solve for v:

sqrt[r*(g*Tan35)]=v

Use the last equation, I think.

And plug in the following:

g=9
r=0.86m
Tan35=0.7 *** I think