Physical pendulum

A uniform rigid rod of length $l$ and mass $M_r$, pivoted at one end with a point mass $m$ at the other, has pivot moment of inertia $I=l^2(m+M_r/3)$. Its gravitational potential energy is $-gl(m+M_r/2)\cos \theta$. The planar Euler-Lagrange equation is $I\ddot{\theta }+gl(m+M_r/2)\sin \theta =0$, giving the displayed small-amplitude oscillation period. The limit $M_r=0$ gives a simple pendulum, whereas $m=0$ gives the uniform-rod physical pendulum. Thinness does not specify a rod's mass, so both contributions must be stated or explicitly neglected.

For a plane lamina suspended at O, treat removed pieces as negative masses and attached pieces as positive ones. Add their moments of inertia about O using the parallel axis theorem, and add their first moments to locate the centre of mass. If the first moments lie along one axis, the gravitational restoring coefficient is $g|{\sum }_i{m}_i{x}_i|$. Dividing it by the total inertia gives the small-oscillation angular frequency squared. Both the restoring moment and inertia change when material is moved.

For a uniform cube of mass $M$ and side $a$ hinged along an edge, the centre is $a/\sqrt{2}$ from the hinge and $I_h=2M{a}^2/3$. Removing a concentric sphere of radius $a/2$ removes mass $\pi M/6$ and hinge inertia $\pi M{a}^2/10$, without shifting the centre. Combining the reduced weight and reduced inertia in ${\omega }^2=Mgd/I_h$ proves the displayed frequency ratio. Updating only the inertia would miss the change of gravitational restoring torque.

With a pivot at distance a from the centre of mass, the parallel axis theorem and small-angle restoring torque give $T^2=(4{\pi }^2/g)[a+I_{\mathrm{CM}}/(Ma)]$. Differentiation, or the arithmetic–geometric mean inequality, minimizes the period at the radius of gyration $\sqrt{I_{\mathrm{CM}}/M}$. The minimum is $T_{\min }=2\pi \sqrt{2a_{\min }/g}$. The allowed pivot locations must include this point; a constraint keeping the pivot outside the body can move the physical optimum to a boundary.

A homogeneous spherical bob rigidly fixed to a massless support has centre-of-mass distance $L$ from its pivot and intrinsic moment of inertia $2M{a}^2/5$. The parallel axis theorem and linear gravitational restoring torque give the title period for small oscillations. Relative to a point mass at the same centre, the leading fractional period increase is $a^2/(5L^2)$. If $L$ is instead measured to the sphere's surface, the centre distance must first be changed to $L+a$.

A uniform circular disc of radius $a$ and mass $m$, pivoted on a perpendicular axis through its rim, has moment of inertia $I_A=3m{a}^2/2$ by the parallel axis theorem. Its hanging small oscillations have period $2\pi \sqrt{3a/(2g)}$. Measured from the upright position, the falling equation is $\ddot{\theta }=2g\sin \theta /(3a)$. Exact upright rest remains an unstable equilibrium; the energy level of that state describes a limiting separatrix after perturbation, with ${\dot{\theta }}^2=4g(1-\cos \theta )/(3a)$. The center acceleration and pivot reaction follow by resolving radial and tangential components.

In nondimensional time the quadratic damping removes energy from the pendulum at rate $-c|\omega {|}^3$. Downward resting positions are nonhyperbolic asymptotically stable spirals, even though the linearized equation has a centre. Upward resting positions remain saddle equilibria. Each fixed-sign branch of $\omega$ has a linear equation for ${\omega }^2$ as a function of angle.

For equal hoop and bead masses, with the hoop suspended from its circumference, the quadratic mass and stiffness matrices are proportional toThe dimensionless squared frequencies are the roots of $2c^2-5c+2$, namely $c=1/2$ and $c=2$.