Thank you very much voko for such nice explanation.
I had
$$U(\theta)=mg((R+L/2)\cos\theta+R\theta\sin\theta)$$
$$\Rightarrow U''(\theta)=mg((R+L/2)(-\cos\theta)+R\cos\theta+R\cos\theta-R\theta\sin\theta)$$
For ##\theta=0##,
$$U''(0)=mg\left(R-\frac{L}{2}\right)$$
Since
$$K(\theta) = \frac{1}{2}m\left(\frac{L^2}{4}+R^2\theta^2\right)+ \frac {1}{2}I_{CM} \Rightarrow K(0)=\frac{1}{8}mL^2+\frac{1}{2}I_{CM}$$
Substituting ##I_{CM}=mL^2/6##,
$$K(0)=\frac{5}{24}mL^2$$
As you stated before, from energy conservation we have,
$$K(0)\dot{\theta}^2+U''(0)\theta^2/2=K(0)\dot{\theta}(0)$$
Differentiating wrt time,
$$K(0)(2\dot{\theta}\ddot{\theta})+U''(0)\theta\dot{\theta}=0$$
$$\Rightarrow \ddot{\theta}=-\frac{U''(0)}{2K(0)}\theta$$
Substituting the values,
$$\ddot{\theta}=-\frac{12g}{5L^2}\left(R-\frac{L}{2}\right)\theta$$
The time period can be easily calculated from the above equation.
Is the above correct?
I have checked what Kleppner has to say on small oscillations, and found that lacking. The form of the kinetic energy they have has no dependency on the coordinates, only on velocities, so there is no demonstration how to proceed in a case like you have here, even though it is taken from that same book.
Symon and Goldstein do a far better job at explaining that; Landau is also noteworthy.
I will have a look at them soon if they are available in my country. :)
