How to calculate the period of this pendulum

Question

I have a pendulum with the following properties:

1. The diameter of the sphere is 1 meter.
2. The sphere of the pendulum is made of pure titanium.
3. The pendulum is connected to the roof with a rigid string 12 meters long. It does not have mass. The union between the string and the sphere is rigid, so it cannot rotate.
4. The oscillation angle is 60 degrees from the vertical.

We suppose there's no air, so there's no friction.

With this data, I need to know the period of this pendulum. I have not worked with pendulums so I do not know how to start. Any hint/guide?

I know that with 1. and 2. we can obtain the mass of the pendulum, using the density of the Titanium, $d_{titanium}=4506\;kg/m^3$. The volume of the sphere is $V=\frac{4}{3}\pi$, so the mass is $m=d\cdot V=4506\cdot\frac{4}{3}\pi=6803\pi$.

Answer 1

Let's assume that the gravitational acceleration $g$ is known. You did not specify where the string is connected to the sphere (center or a point on the surface). I will use the center. If it's at the surface, all you need to do is change the distance to the center by adding half the diameter. The moment of inertia of the sphere with respect to the center of the mass is $$I_{CM}=\frac 25 MR^2$$ With respect to the point on the roof around which it oscillates it is $$I=I_{CM}+ML^2=M\left(\frac25 R^2+L^2\right)$$ If you want, you can say that $R^2\ll L^2$. In that case the problem will not depend on the radius of the sphere.

Then using the zero of potential energy at $\theta=0$, where $\theta$ is measured from the vertical axis, you have $$MgL(1-\cos\theta)+\frac12I\dot\theta^2=MgL(1-\cos\theta)+\frac12M\left(\frac25 R^2+L^2\right)\dot\theta^2=MgL(1-\cos\theta_0)$$ $\theta_0$ is the initial angle of $60^\circ$. Calculating $\dot\theta$ from here, you get $$\dot\theta=\sqrt{\frac{2gL}{\frac25 R^2+L^2}(\cos\theta-\cos\theta_0)}$$ Using $\dot\theta=\frac{d\theta}{dt}$, you get $dt=\frac{d\theta}{\dot\theta}$. To get the period$$T=2\int_{-\theta_0}^{\theta_0}\frac{d\theta}{\dot\theta}=2\sqrt{\frac{L^2+\frac25R^2}{2Lg}}\int_{-\theta_0}^{\theta_0}\frac{d\theta}{\sqrt{\cos\theta-\cos\theta_0}}$$