Answer:
The expression would be ω = \(\sqrt{\frac{g}{L sin 0 } }\)
Explanation:
Given that ω is the angular velocity
g is the acceleration due to gravity
L is the length
θ is the angle of downward tilt
For an object we compare the horizontal and vertical component of the forces acting on the body;
For vertical component
T sinθ = mg............1
For the horizontal component
T cos θ = \(\frac{mv^{2} }{R}\) .............2
R is our radius and is = L cos θ
v = ωR
substituting into equation 2 we have
T cos θ = m(ωR\()^{2}\) /R
T cos θ=m(ω\()^{2}\)R ..................3
Now comparing the vertical and the horizontal component we have;
equation 1 divided by equation 3 we have
T sin θ /T cos θ = mg / m(ω\()^{2}\)R
Tan θ = g / (ω\()^{2}\)R............4
Making ω the subject formula we have;
(ω\()^{2}\) = g/ R Tan θ
But R = L cos θ and Tan θ = sin θ/ cosθ
putting into equation 4 we have;
(ω\()^{2}\) = g /[( L cos θ) x( sin θ/ cosθ)]
(ω\()^{2}\) = g/ L sinθ
ω = \(\sqrt{\frac{g}{L sin 0 } }\)
Therefor the expression for the angular velocity ω in terms of g, L and angle θ would be ω = \(\sqrt{\frac{g}{L sin 0 } }\)
