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Take for example the case of a rod rotating about an axis passing through its centre of mass and perpendicular to it. It has a ring hung from one of its sides. The rotation of the rod causes the ring to move out wards and ultimately fall off the rod. Wrt ground the ring should experience a centripetal force inwards (right?) But still it moves outwards.Why? What is the force acting outwards wrt ground?

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marked as duplicate by sammy gerbil, Jon Custer, stafusa, Daniel Griscom, Kyle Kanos Nov 7 '17 at 11:04

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When the rod rotates then due to inertia it has a tendency of slipping away from the axis of rotation. If there is not sufficient friction between the ring and the rod then it slips away. Wrt to ground no force acts in outward direction. If there will be sufficient friction between the ring and the rod then force of friction will provide necessary centripetal force and the ring will rotate not skid.

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Centripetal force is responsble to make the ring move in circular motion with the rod.

with respect to the ground .. the ring should move towards the tangential direction if no centripetal force...(Like a stone tied to a string after being given a circular motion and is let go )

Here the the ring has the tendency to fly off the tangent.

But is made to move around due to the centripetal force of friction.

but at every instant of time it wants to move tangentially ... now the tangential direction is changing all the time....

so the next instant the inetia causes it to move in the past tangential direction (which now at the next instant has a radial component) enter image description here

hence we get a radial direction motion as well... hence the ring falls off

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