A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 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, between 0° and 90°, that the string makes with the pole. Take ????=9.81 m/s2.

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Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = [tex]5.93\times 9.81[/tex]

T cos θ = 58.17

[tex]T sin \theta =\dfrac{mv^2}{r}[/tex]

[tex]T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}[/tex]

[tex]T sin^2 \theta =56.93[/tex]

[tex]sin^2 \theta = 1 - cos^2 \theta[/tex]

[tex]T (1 - cos^2 \theta) =56.93[/tex]

[tex]T (1 - (\dfrac{58.17}{T})^2) =56.93[/tex]

T² - 56.93T - 3383.75 = 0

T =  93.22 N

[tex]cos \theta = \dfrac{58.17}{93.22}[/tex]

θ = 51.39°