A cylindrical tumbler of diameter d has smooth sides and smooth edge. A thin rod of length L is balanced on the edge of the tumbler as shown in figure. The angle $$\alpha$$ that the rod makes with horizontal for this trick to work is

A

$$\cos^{-1}\left(\left(\frac{2d}{L}\right)^{1/3}\right)$$

B

$$\sin^{-1}\left(\left(\frac{2d}{L}\right)^{1/3}\right)$$

C

$$\cos^{-1}\left(\left(\frac{d}{L}\right)^{1/3}\right)$$

D

$$\sin^{-1}\left(\left(\frac{d}{L}\right)^{1/2}\right)$$