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A cylinder is kept on frictionless inclined plane. Why it does not roll (it slides only), although torque with respect to contact point is working on it (due to component of weight parallel to incline)?

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  • $\begingroup$ What force is present that would supply torque at the contact point? $\endgroup$ – BowlOfRed Feb 24 '16 at 8:33
  • $\begingroup$ You need friction to establish rolling. $\endgroup$ – Ankur Singh Apr 21 '18 at 17:29
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The standard translational motion derivation goes like this:
Using Newton's second law $F=ma$.
Forces perpendicular to plane
$N-mg\sin\theta = 0$
Forces parallel to plane
$mg \sin \theta = ma \Rightarrow a = g \sin \theta$

Now do it in terms of torque and angular acceleration about $X$.
Using $\tau = \dfrac {dL}{dt}$
$mg\sin \theta \cdot r = \dfrac {d(mvr)}{dt} = mr\dfrac {dv}{dt} \Rightarrow a = g \sin \theta$

What is happening is the torque exerted on the cylinder about $X$ is changing in the angular momentum of the cylinder about $X$.

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  • $\begingroup$ How does this relate to the rotation of the cylinder about its own axis? $\endgroup$ – BowlOfRed Feb 24 '16 at 9:04
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    $\begingroup$ A force acting through the centre of mass cannot impart any spin angular momentum to a body, ie the body cannot be made to rotate about its centre of mass. $\endgroup$ – Farcher Feb 24 '16 at 9:11
  • $\begingroup$ Now if you introduce sufficient friction, it starts to roll although again torque wrt contact point is provided by component of weight only (no torque due to friction force wrt contact point). What makes cylinder to roll in this case $\endgroup$ – Mohan Feb 24 '16 at 9:26
  • $\begingroup$ Since there is no friction the force from the inclined plane points to the centre of the cylinder. Hence the total angular momentum i zero. If, however, the cylinder is slightly oblate, the perpendicular force might not point at the center of mass and it would roll. $\endgroup$ – Jens Feb 24 '16 at 9:42
  • $\begingroup$ What will happen to asymmetric hollow cylinder on frictionless inclined plane. Roll or slide? $\endgroup$ – Mohan Feb 24 '16 at 10:02
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Contact point is not an inertial point of reference. So when you deal with torque equation you have to consider a ghost force acting on the center of mass. This will cancel the torque of weight around the contact point.

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