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A fixed pivot constrains a point of a rigid body to remain stationary in a specified frame while permitting rotation. With an ideal pivot at $O$, the reaction force acts at $O$ and contributes no torque about that point. Angular momentum balance is therefore conveniently written about $O$.

For planar rotation about a fixed axis, $I_O\ddot{\theta }={\tau }_O$. A uniform rod of length $\ell$ pivoted at one end has $I_O=m{\ell }^2/3$; gravity supplies the restoring torque when it hangs downward. A frictional pivot can also exert a couple, so zero reaction torque is an idealization to state explicitly.

Consider an axially symmetric rigid body rotating about a fixed pivot, with transverse and axial principal moments of inertia $A$ and $C$ and orientation-dependent potential energy $Mgh\cos \theta$. Let the inclination $\alpha$ be nonvertical, $0<\alpha <\pi$, and let $n$ be the axial angular velocity of a symmetric top. Conservation of the two cyclic angular momenta makes both $n$ and the steady precession rate $\Omega$ constant. The tilt Euler-Lagrange equation, after division by $\sin \alpha$, givesWhen $\cos \alpha \ne 0$, a real precession rate exists exactly when the quadratic discriminant obeys $C^2{n}^2-4AMgh\cos \alpha \ge 0$. At horizontal inclination, the equation is linear: if $Cn\ne 0$, then $\Omega =Mgh/(Cn)$. If $Cn=0$ and $Mgh\ne 0$, there is no steady horizontal solution. Vertical inclinations require the original undivided equation instead. The moments of inertia and lever arm $h$ must all refer to the same fixed pivot, rather than mixing central and pivot values.