Suppose we have a disk (of mass $M$) connected to a rod (of mass $m$) attached to a fixed pivot. (Note that initially I thought of the rod hanging off a rope, but I realized it would be better to assume the rod is attached to a fixed pivot). The rod is horizontal along the $x$-axis, with the disk in the $+x$-direction, and the disk is spinning with angular momentum $\vec{L} = L\,\vec{e}_{x}$. Gravity imposes a force on the rod and disk downward in the $-z$-direction, but due to the way the torque is imposed, the net result is that the setup precesses about the $z$-axis.

enter image description here

We have three forces at play: the gravitation force on the disk, the gravitation force on the rod, and the pivot force keeping one end of the rod in place (which I label as $\vec{T}$).

Now the center of mass of the rod + disk is somewhere at $x > 0$ in the image depicted. Since the center of mass does not fall, we must have $T = Mg + mg$ in terms of magnitudes. Now I have two questions:

  1. Doesn't the center of mass of the rod + disk move in circular motion around the $z$-axis? If so, doesn't there have to be an additional force from somewhere causing this motion? What is this force and what am I failing to account for here?

  2. Since the center of mass itself moves circularly around the $z$-axis (due to precession), doesn't that meant there is an additional angular momentum component pointing upward? Where does this $z$-angular momentum come from? Even if it is constant, there had to have been an initial torque coming from somewhere, so where would that upward-pointing torque come from? For concreteness, imagine a demo where you hold the spinning bicycle wheel in place, and then let go. When you let go, the fact that the center of mass starts to move around the $z$-axis means there had to have been an initial vertical torque. What force is responsible for that initial torque and how can we understand it?


1 Answer 1

  1. You've argued that a centripetal force is required to maintain the motion of the center of mass (CoM) about the pivot. The most straightforward solution is that this would need to be applied by the pivot itself. If such a force were not applied by the pivot, the precession would not occur as described in the question.

  2. If the total angular momentum is initially directed along the x-axis, it will remain in the x-y plane. The component along the z-axis associated with the CoM motion about the z-axis will be cancelled by a change in the orientation of the disk. That is, the disk and rod will fall slightly from their starting position. Thus, the rotation of the disk about its axis will have an angular momentum slightly below the horizontal.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.