Respuesta :
Answer:
θ = 66º
Explanation:
This exercise of Newton's second law must be solved in part, let's start by finding the slowing down acceleration of the ball
a = v² / r
the radius of the circle is
sin θ = r / L
r = L sin θ
we substitute
a = v² /L sin θ
now let's write Newton's second law
vertical axis
T_y -W = 0
T_y = W
radial axis
Tₓ = m a (1)
let's use trigonometry for the components of the string tension
cos θ = T_y / T
sin θ = Tₓ / T
Tₓ = T sin θ
we substitute in 1
T sin θ = [tex]\frac{m \ v^2}{L \ sin \theta}[/tex]
T L sin² θ = m v²
we write our system of equations
T cos θ = m g
T L sin ² tea = m v²
we divide the two equations
L [tex]\frac{sin^2 \theta}{cos \theta}[/tex] = v² / g
(1 -cos²)/ cos θ = [tex]\frac{v^2 }{g \ L}[/tex]
1 - cos² θ = [tex]\frac{4.75^2}{9.81 \ 2.37}[/tex] cos θ
cos² θ + 0.97044 cos θ -1 = 0
we change variable cos θ = x
x² + 0.97044 x - 1 =0
x= [tex]\frac{-0.97 \pm \sqrt{0.97^2 - 4 1} }{2}[/tex]
since the square root is imaginary there is no real solution to the problem, suppose that the radius is 1 m r = 1 m
T sin θ = [tex]\frac{m \ v^2}{ r}[/tex]
T cos θ = m g
resolved
tan θ = [tex]\frac{v^2}{ r g}[/tex]
θ = tan⁻¹ ( 4.75²/ 1 9.81)
θ = 66º
