Answer:
a)[tex]\omega =\sqrt{\dfrac{g}{L}}[/tex]
b)N= 21.29 rpm
Explanation:
Given that
Mass of the ball =m
Length of string = L
Lets take angular speed = ω
The centripetal force on the ball
F = m ω² L
To complete the circle ,at the top condition the force due to gravity should be equal to the centripetal force
Gravity force = mg
F= mg
m ω² L = m g
ω² L = g
[tex]\omega =\sqrt{\dfrac{g}{L}}[/tex]
When L= 2 m
Lets take g =10 m/s²
[tex]\omega =\sqrt{\dfrac{g}{L}}[/tex]
[tex]\omega =\sqrt{\dfrac{10}{2}}[/tex]
ω = 2.23 rad/s
To convert in rpm
[tex]\omega =\dfrac{2\pi N}{60}[/tex]
N=Speed in rpm
[tex]2.23 =\dfrac{2\pi N}{60}[/tex]
N= 21.29 rpm