Question

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.

Answer 1

θ = 66º

Step-by-step 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 =
`(m \ v^2)/(L \ sin \theta)`

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
`(sin^2 \theta)/(cos \theta)` = v² / g

(1 -cos²)/ cos θ =
`(v^2 )/(g \ L)`

1 - cos² θ =
`(4.75^2)/(9.81 \ 2.37)` cos θ

cos² θ + 0.97044 cos θ -1 = 0

we change variable cos θ = x

x² + 0.97044 x - 1 =0

x=
`(-0.97 \pm √(0.97^2 - 4 1) )/(2)`

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 =
`(m \ v^2)/( r)`

T cos θ = m g

resolved

tan θ =
`(v^2)/( r g)`

θ = tan⁻¹ ( 4.75²/ 1 9.81)

θ = 66º