# Free Body Diagram of Precessing Top

Consider a heavy, symmetric top spinning on a table, initially inclined at some angle from the vertical. Assuming this angle is fixed, it follows that the weight $$\textbf W$$ and normal force $$\textbf F_N$$ from the table must act as an external couple on the top, providing the net external torque $$\boldsymbol{\tau}^{\text{ext}}$$ that causes the direction (but not magnitude) of its angular momentum $$\textbf L$$ to change, specifically to rotate in a uniform circular motion. As a result, the top's center of mass also undergoes uniform circular motion. However, assuming the table is frictionless, the only external forces acting on the top are $$\textbf F^{\text{ext}}=\textbf W+\textbf F_N$$ and neither of these has a centripetal component, so how can this be consistent with the precession of the center of mass in a circle?

• A top would move differently on ice than a table. The tip would not stay fixed. There is a a horizontal friction force. Commented Jan 14 at 14:27
• Who says that the center of mass precesses in a circle? IIRC the motion on a frictionless table involves the center of mass remaining at rest (in some frame) and the contact point "orbiting" that point in a circle. Commented Jan 14 at 14:38
• Thank you for your clarifying comments, it was my mistake to assume that the tip must be fixed in the inertial frame of the lab if there is no static friction. So in theory, it seems that the tip will remain fixed iff the angular precession velocity $\Omega$ is sufficiently slow, specifically iff $\Omega\leq\sqrt{\frac{\mu_sg}{R}}$ where $\mu_s$ is the coefficient of static friction and $R$ is the CoM orbital radius. Commented Jan 14 at 16:23
• Or actually, there are $2$ factors: the slower the CoM's angular precession velocity $\Omega$ AND the smaller its tilt angle $\beta$ from the vertical, the "easier" it is for the tip to be fixed (the condition becomes $\sin(\beta)\Omega^2\leq\mu_sg/d$ where $d$ is the distance between the tip and the CoM). Although iirc there are only certain values of $\beta$ that admit $\dot{\beta}=0$ so including nutation effects might make it a bit more complicated. Commented Jan 14 at 16:40