Irodov Problem 1.287

As the cylinder spins about the axis XX' it has an angular velocity Iw directed along the axis XX' . Now the axis XX' undergoes oscillations (Irodov means simple harmonic oscillations even though he does not specifically mention it in the problem).

This means that the angle of orientation of the axis XX' at some instant of time ifs given,




The angular momentum vector of the cylinder L, is thus, given by,






This change in angular momentum is provided by the torque resulting from the force acting at the bearings given by F(l/2) + F(l/2) = Fl. Hence, we have,









(see the list of moments of inertia)

Irodov Problem 1.286

Let the moment of inertia of the sphere be I. The sphere undergoes two spinning motions, i) its rotation about AB with an angular velocity w (the angular momentum vector Iw is directed along the axis AB as shown) ii) its rotation about OO' with an angular velocity w' (the vector oriented along OO').

As the axis AB rotates a result, the angular momentum vector Iw rotates with it in the plane at a rate w'. Since the angular momentum vector rotates, this can only occur because of torque acting at the bearings. The torque is directed along the tangent of the circle spanned by the tip of angular momentum vector. Since torque is given by F x r (r is directed along AB), F must be directed along the z-axis. Hence, we have,








Hence, we have,
(see the list of moments of inertia).

Irodov Problem 1.285

The easiest way to solve this problem is using Einstein's famous principle of equivalence. Since, the entire top moves with a linear acceleration w in the horizontal direction, it is as if it experiences a net force of gravity equivalent to in a direction tilted to the vertical at an angle .

Now all we have to do is to solve this problem exactly as 1.283, with the caveat that gravity is now tilted at an angle and has a different magnitude.














Thus unlike in problem 1.283, where the top precessed symmetrically about the vertical axis, in this case, the top will precess symmetrically about the tilted axis OA oriented as shown in the figure. Alternatively, one could imagine that the top itself is on an inclined plane inclined at an angle and on a planet with higher gravity of value as shown in the figure. Thus, solving the problem exactly as in 1.283 it is clear that,
and acts at an angle of tilted to the vertical as shown in the figure.