A 5.41 kg ball is attached to the top of a vertical pole with a 2.37 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.75 m/s in a horizontal circle with the string remaining taut. Calculate the angle θ, between 0∘ and 90∘, that the string makes with the pole. Use g=9.81 m/s2.

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 =

             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 = v² / g

             (1 -cos²)/ cos θ  =

             1 - cos² θ  =     cos θ

             cos² θ + 0.97044 cos θ -1   = 0

we change variable    cos  θ = x

             x² + 0.97044 x - 1 =0

             x=

           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 =

           T cos θ  = m g

resolved

           tan θ =  

           θ = tan⁻¹ ( 4.75²/ 1 9.81)

           θ = 66º

d is the correct answer

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