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

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* 9.81`

T cos θ = 58.17

`T sin \theta =(mv^2)/(r)`

`T sin \theta =(5.93* 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 - ((58.17)/(T))^2) =56.93`

T² - 56.93T - 3383.75 = 0

T = 93.22 N

`cos \theta = (58.17)/(93.22)`

θ = 51.39°