 Solution

The bob experiences weight $Mg$ vertically downward and a rod force towards the pivot. The tangential component of weight is $-Mg\sin \theta$, soThe rod force supplies the radial acceleration and contributes no tangential torque about the pivot.

To the same small-angle order, writing $\chi ={\omega }_{0}t+\varphi$,Thus $K+U=MgL{\theta }_0^2/2$ is constant. Put $b=\dot{\theta }(0)$, a signed angular velocity. ThenIf the printed angular speed $\beta$ is intended as the signed rate, set $b=\beta$. If it means magnitude only, $b=\pm \beta$ gives two possible phases; the amplitude is unique. At zero amplitude the phase is arbitrary.

For a finite-radius spherical pendulum bob, the PDF dimension $L$ runs from pivot to centre of mass. The parallel axis theorem gives $I_P=M{L}^2+2M{a}^2/5$. The restoring torque remains $-MgL\theta$, henceWith $a=[3M/(4\pi \rho ){]}^{1/3}=0.02605\mathrm{m}$,The finite bob increases the inertia without changing the gravitational lever arm.