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I have an question about rotational motion of the following system

There is a hollow symmetric cylinder on a friction-less inclined plane (mass M, moment of inertia I, radius R, incline angle a). We place a small metallic sphere (mass m) inside the cylinder in an equilibrium state, but it can move inside without friction.

sketch

The cylinder is kept in place by external force and the small sphere is in equilibrium and not moving. The external force is removed at t=0.

Will the cylinder rotate while going down the plane? It so, how to calculate the rotational acceleration?

(I assume that a symmetric cylinder without a sphere inside will not rotate)

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  • $\begingroup$ Is the inner mass m constrained (ie fixed) to lie on the inner surface of the cylinder? Or can it leave the surface? $\endgroup$ – cms Sep 5 '19 at 16:34
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    $\begingroup$ For simplicity, it slides along the surface (constrained; like a bead along a rope) $\endgroup$ – Andriy Sep 5 '19 at 16:43
  • $\begingroup$ The weight of the sphere acts to the right of the center of support of the cylinder. This will cause a torque, giving the cylinder an angular acceleration which is independent of its motion down the incline. $\endgroup$ – R.W. Bird Sep 5 '19 at 19:16
  • $\begingroup$ That is what I would expect as well. But looks like something is missing in the assumptions, since the answers below are stating the opposite. $\endgroup$ – Andriy Sep 5 '19 at 19:54
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In order to have a rotation, you have to have some force on it that doesn't act through the center of mass.

Gravity is considered to act entirely on the center of mass and never creates a torque.

The ramp can only supply a normal force at the point of contact (since it is frictionless). This force must act through the center of mass and can impart no rotation.

Likewise the sphere inside can only supply a normal force at the point of contact, and this too must act on a line through the center of mass. So there are no forces that can apply a rotation.

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  • $\begingroup$ Does it mean that even if the sphere is placed in any point in the cylinder and the system is released, the cylinder will not rotate? (and the sphere will slide back and force inside) $\endgroup$ – Andriy Sep 5 '19 at 19:49
  • $\begingroup$ If both the inside and the outside of the cylinder are frictionless, yes. If the cylinder is not perfectly round (so normal forces don't necessarily go through center), or if there is friction, then you would have something that could supply a torque. Think of repeating this on a flat surface (no ramp). If you displace the sphere, what happens to the cylinder? It oscillates back and forth, but without rotation. $\endgroup$ – BowlOfRed Sep 5 '19 at 20:16
  • $\begingroup$ You’ve only described the instantaneous case at t = 0. The bead on the inside of the cylinder does not have an inclined normal force and cannot initially accelerate parallel to the ramp; it will move relative to the cylinder. $\endgroup$ – cms Sep 6 '19 at 18:29
  • $\begingroup$ @cms, the question doesn't ask how the ball moves relative to the cylinder. It only asks about the cylinder's rotation. The argument is the same regardless of the position of the sphere within. $\endgroup$ – BowlOfRed Sep 6 '19 at 18:54
  • $\begingroup$ I was too brief; if the bead moves it will also cause the cylinder to rotate. It may not be uniform rotation but excluding oscillator rotation sounds pedantic. $\endgroup$ – cms Sep 6 '19 at 19:23
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The normal force from the plane acting on the cylinder is thorough its center of mass, therefore not contributing to rotation. Same can be said for the metallic sphere. So nothing will rotate.

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  • $\begingroup$ But the center of mass of the system (cylinder and sphere) is shifted and normal forces are not going through it. Is it similar to the case of asymmetric cylinder? $\endgroup$ – Andriy Sep 5 '19 at 15:58
  • $\begingroup$ When I say the center of mass, I mean only the cylinder. $\endgroup$ – seilgu Sep 5 '19 at 16:00
  • $\begingroup$ If to look into the system of cylinder and sphere together. Their center of mass is shifted downward toward the sphere. Then the Normal force, that acts on the cylinder is the external force that creates torque with respect to the shifted center of mass. Isn't it? Should it result in the system rotation? $\endgroup$ – Andriy Sep 5 '19 at 16:55
  • $\begingroup$ If you want to look at the cylinder and sphere together, suit yourself. But the analysis would be harder. Since you just wanted to know about whether the cylinder will rotate, and there's no friction between the sphere and the cylinder, just analyze the cylinder separately. $\endgroup$ – seilgu Sep 5 '19 at 17:08
  • $\begingroup$ What actually will happen is that the cylinder will not rotate, but it will slide. The sphere inside would not move with the cylinder : the cylinder moves right 2cm, but the ball might just move right 1cm. which, if seen as the combined system, would be "rotating" with respect to their center of mass. $\endgroup$ – seilgu Sep 5 '19 at 17:10

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