Question

A mass m = 9.0 kg is attached to the lower end of a massless string of length L = 83.0 cm. The upper end of the string is held fixed. Suppose that the mass moves in a circle at constant speed, and that the string makes an angle theta = 21 with the vertical, as shown in the figure.

find how long it takes to make one revolution and the tension in the string

my work so far:

r=0.83tan21

Fy = Ty-G=0 so Ty=9*9.8=88.2
Fx=Tx=9a so a=88.2tan21/9

v= sqrt. a*r (since it's centripetal acceleration)=1.094786688 m/s

T= 2pi*r/v= 1.8285s but apparently that's wrong.

for tension i calculated (88.2tan21)/sin21= 94.8 which is correct

Answer 1

Assuming your tension is correct, then the inward component of tension is what is counterbalancing centreptal force.

horizontaltension=mv^2/r
tension*sinTheta=mv^2/r

94.8*sin21=m (2PIr/Period)^2/r

solving for period

Period=2PIsqrt (m* r/94.8*sin21) check that.