Question

A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.654.65 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 ????=9.81g=9.81 m/s2. angle: °

Answer 1

Answer:
`\theta =45.73^(\circ)`

Step-by-step explanation:

Given

Length of string =2.15 m

mass of ball =5.49 kg

speed of ball=4.65 m/s

Here

Tension provides centripetal acceleration

`T\cos\theta =mg`-----1

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

Divide 2 & 1

`tan\theta =(v^2)/(rg)`

`tan\theta =(4.65^2)/(2.15* 9.8)`

`tan\theta =1.026`

`\theta =45.73^(\circ)`