Sliding Down a Sliding Half-Cylinder

Is this not extremely overrated?

At the point when the ramp has moved $X$ to the left, and the radial line from the box to the centre of the ramp makes an angle $\theta$ with the vertical, the box will have moved a distance $-X + \sin\theta$ to the right. Thus the horizontal coordinate of the centre of mass of the system (relative to the initial position of the box) will be $\tfrac{1}{3M}\big[-2MX + M(-X + \sin\theta)\big] \; = \; -X + \tfrac13\sin\theta$ Since the only forces acting on the block and ramp are either internal or vertical, the horizontal coordinate of the centre of mass remains constant, and hence $X = \tfrac13\sin\theta$ . When the block hits the ground, $\theta = \tfrac12\pi$ , and so $X = \boxed{\tfrac13}$ .

In the horizontal direction the center of mass does not move (no external forces in this direction). Let us compare the initial and final states, selecting a system of reference so that initially the center of mass is at $x=0$ . In the final state the sphere moves to the left by $-x_0$ and the mass moves to the right by $R-x_0$ , so that it stays on the surface of the sphere. The center of mass is at $[-2Mx_0+M(R-x_0)]/3M =0$ . This yields $x=0.333m$ .