- #36

Yoonique

- 105

- 0

Isn't the 2 factor because the radius of the circle is L/2?TSny said:I don't get the 2 in the numerator of the first factor.

To simplify further, think about the condition that holds at equilbrium.

At equilibrium, GMmL/(2r

_{0}

^{2}) = κθ

_{0}. So I sub it in. I'll get -δθ(κ - κθ

_{0}/r

_{0}) = Iα

So ω

^{2}= 2κ(r

_{0}- θ

_{0})/(mL

^{2}r

_{0}).

Therefore T = 2π√(mL

^{2}r

_{0})/2κ(r

_{0}- Lθ

_{0})

But my given answer is T = 2π√(mL

^{2}r

_{0})/2κ(r

_{0}- Lθ

_{0}cosθ

_{0}). Why is there an extra cosθ

_{0}...