Question

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.

Answer 1

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 = $5.93\times 9.81$

T cos θ = 58.17

$T sin \theta =\dfrac{mv^2}{r}$

$T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}$

$T sin^2 \theta =56.93$

$sin^2 \theta = 1 - cos^2 \theta$

$T (1 - cos^2 \theta) =56.93$

$T (1 - (\dfrac{58.17}{T})^2) =56.93$

T² - 56.93T - 3383.75 = 0

T =  93.22 N

$cos \theta = \dfrac{58.17}{93.22}$

θ = 51.39°