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In Landau's Mechanics volume at section 32 solved problem no. 5, the instantaneous axis is chosen as the one which coincides with the line where the cylinder touches the plane. This is possible only if there is rolling without slipping. My question is, had it been rolling with forward slipping, the instantaneous axis should be somewhere placed outside the cylinder. Am I right?


PROBLEM 5. Find the kinetic energy of a cylinder of radius $\:R\:$ rolling on a plane, if the mass of the cylinder is so distributed that one of the principal axes of inertia is parallel to the axis of the cylinder and at a distance $\:a\:$ from it, and the moment of inertia about that principal axis is $\:I$.


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  • $\begingroup$ On the rotation plane, if you select two points on the cylinder (for example, point of contact $A$ and the furthest point from ground $B$) and draw velocity vectors $\vec{AA_1}$ (if there is no slipping $A_1\equiv A$) and $\vec{BB_1}$, then the intersection of lines $AB$ and $A_1B_1$ will give you the immediate centre of rotation $C$ (the axis will be perpendicular and passing through that point). Thus, it's straightforward to see that if $AA_1$ and $BB_1$ point in the same direction, then $C$ is outside of the cylinder and if vectors point in different directions, then $C$ is inside. $\endgroup$ Dec 4, 2021 at 18:54

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Even if there is slip between the cylinder and the ground the line of contact can be choosen as instantaneous axis of rotation. The reason is that any point of a rigid body can be selected as an origin, around which the body rotates.

It is convenient to use the line of contact in the case of rolling without slip because the instantaneous velocity of the points of the body are zero in that line. That is, it belongs momentarily to the ground frame.

It is no longer valid if there is slip.

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