Physics Question Help?

Question

A 21g ball is fastened to one end of a string 68cm long and the other end is fixed to a point on a ceiling that makes an angle of 28 degrees with the vertical. This angle remains constant as the ball rotates in a horizontal circle. The angle theta would remain constant only for a particular speed of the ball in a circular path. The accleration og gravity is 9.8m/s2. FInd this speed. Answer in m/s

Answer 1

Equations we need are


Fc=ma so a=Fc/m
a= V^2/R
V=sqrt(Fc R/m)
R=L sin(A)
Fc=mg sin(A)cos(A) finally

V=sqrt(mg sin(A)cos(A) L sin(A)/m)
V=sqrt(g sin(A)cos(A) L sin(A))
V=sin(A) sqrt(g L cos(A) )
V=sin(28) sqrt( 9.8 x .68 cos (28))
V=1.14 m/s