Answer:
[tex]\alpha =20.564^{o}[/tex]
Explanation:
Given data
Radius R=6.80 m
Velocity v=18 km/h =5 m/s
First we need to set up for Force in vertical (y) and horizontal (x)
∑Fy=0=TCosα-W
∑Fx=ma=Fc=TSinα
Solve Vertical force equation for T,substituting mg for W
[tex]TCos\alpha -W=0\\TCos\alpha=W\\TCos\alpha=mg\\T=(mg/Cos\alpha)[/tex]
Substitute expression for Fc and T into horizontal force and simplify it we get
[tex]Fc=TSin\alpha\\ where\\Fc=(mv^{2}/R )\\So\\(mv^{2}/R )=(mg/Cos\alpha )Sin\alpha\\(mv^{2}/R )=mg*tan\alpha\\ tan\alpha =\frac{v^{2} }{Rg}\\ \alpha =tan^{-1}(\frac{(5m/s)^{2} }{(6.80m)(9,8m/s^{2} )} )\\ \alpha =20.564^{o}[/tex]
