A physical pendulum consists of a long, thin cone suspended at its apex The height of the cone is L. What is the period of this pendulum?
the moment of inertia with respect to the center of mass of a cone is:
$$I = \frac{3}{10}MR^2$$
now lest derive the period form SHO note($sin\theta = \theta$ small angle aproximation):
$$\tau = I\alpha = mg\frac{L}{2}sin\theta$$ $$\alpha = \frac{10mgL}{6mR^2}\theta$$ $$\omega = \sqrt{\frac{5gL}{3R^2}}$$ $$T = 2\pi \sqrt{\frac{3R^2}{5gL}}$$
which is not the correct answer true answer is: $$T = 2\pi \sqrt{\frac{4L}{5g}}$$
Consider the relation ${\rm d}\tau={\rm d}I\,\ddot{\theta}$ applied to an elemental volume of the cone, consisting of a thin circular slice of radius $r$ and thickness ${\rm d}z$ at a distance $z\,$ below the vertex of the cone.
The radius of this slice is given by $r=zR/L,$ where $R$ is the radius of the cone's base and $L$ is the vertical height of the cone, and therefore the elemental mass of the circular slice is ${\rm d}m=\rho\pi r^2\,{\rm d}z,$ where $\rho$ is the constant density of the cone.
The elemental torque applied to the slice about the point of suspension at the vertex is $$ {\rm d}\tau= Fz=({\rm d}m)g\theta z=\left(\rho\pi r^2 {\rm d}z\right)g\theta z,\quad\cdots\cdots\quad(1) $$ where $\sin\theta=\theta +\mathcal{O}(\theta^3)$ is the small angle between the vertical and the line through the vertex of the cone to the center of it's base.
Also, the rate of change of elemental angular momentum of the circular slice about the vertex is $${\rm d}I\,\ddot{\theta} ={\rm d}I\,\omega^2\theta=\left({\rm d}m \times z^2\right)\omega^2\theta= \left(\rho\pi r^2{\rm d}z \times z^2\right)\omega^2\theta,\quad\cdots\cdots\quad(2) $$ where $\ddot{\theta}=\omega^2\theta $ in simple harmonic motion.
Equating $(1)$ and $(2),$ and integrating the elemental relationships over the height of the cone, using $r=zR/L,$ gives $$\tau=\int_{0}^{L}\rho\pi z^2(R/L)^2 g\theta z\, {\rm d}z = I\ddot{\theta} =\int_{0}^{L}\rho\pi z^2 (R/L)^2 z^2\omega^2 \theta \,{\rm d}z.$$ The common factor $\rho\pi (R/L)^2\theta$ cancels from both sides and on integration and rearranging, $$\omega^2=\dfrac{5}{L^5}\dfrac{L^4}{4}g . $$ Since $\omega T=2\pi$ for a simple harmonic oscillator, this gives the periodic time as $$T=2\pi \sqrt{\dfrac{4L}{5g}}\quad\cdots\cdots\quad (3).$$
In the Wikipedia article "Pendulum", the period of a pendulum is given by $$ T=2\pi \sqrt{\dfrac{I}{mgR}}, $$ where $I$ is the pendulum's moment of inertia about the suspension point and $R$ is the distance of the center of mass from the suspension point. The moment of inertia of the conical pendulum about its suspension point is $I=\dfrac{3}{5}mL^2,$ and the distance from the suspension point to the center of mass is $R=\dfrac{3}{4}L,$ and inserting these values into the Wikipedia formula gives equation $(3)$.
The slenderness of the cone is used in assuming that each small element of the disk has the same value of $\theta.$