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A particle of mass $m$ is moving on the inner side of smooth circular cylinder of radius $R$ whose $Oz$-axis is vertical and directed downwards. The particle started its motion from the $x$-axis with velocity $V$ which was parallel to $y$-axis.

Which forces are acting on the particle during its motion?
How can I show that the Energy conservation law holds and the Angular momentum conservation law holds with respect to the $Oz$-axis?

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    $\begingroup$ 1)The normal directed towards the center of the cylinder. 2)Gravity.3) the normal from the ground, which probably cancels with gravity if the surface is horizontal. $\endgroup$
    – udiboy1209
    Jul 28 '13 at 15:53
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    $\begingroup$ So none of these three forces do work, there's your energy conservation. The total torque results to zero, angular momentum conserved. $\endgroup$
    – udiboy1209
    Jul 28 '13 at 15:55
  • $\begingroup$ in similar problem (everything the same, just particle is moving on the smooth inner side of a circular cone) they don't mention normal force from the ground in the solution. Is there that force in my problem for sure? What if it doesn't exist? Then gravitational force does work? $\endgroup$
    – gov
    Jul 28 '13 at 17:28
  • $\begingroup$ If there's no normal from the ground, there will not be ground underneath that particle. So it will definately fall $\endgroup$
    – udiboy1209
    Jul 28 '13 at 17:44

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