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When a body is in circular motion, realistically, it experiences only the centripetal force, created by gravitational pull, tension etc. , which gives it acceleration towards the centre. Now, assuming we were to fill a bucket with water, and whirl it in the vertical plane. There will be a certain velocity when the water doesn't fall out even when the bucket is upside down. At that instant, if we were to write down the forces that act on the water, they would be the Normal reaction by the bucket, and it's weight. This means that the net force is downwards. Why doesn't the water fall down, since there's no force in the upward direction? Centrifugal force is a psuedo force, so it shouldn't be applicable here, as the water won't wall down even if I watch from a non inertial frame. How can one explain this phenomenon?

Also, say we were to fit a ring in a rod, and rotate the rod, the ring will eventually slip and fall off. Why does this happen if there is no outward force?

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Why wouldn't centrifugal force be applicable in a rotating system? –  Kyle Kanos Aug 9 '13 at 16:50
    
More on centrifugal vs centripetal: physics.stackexchange.com/… –  Qmechanic Aug 9 '13 at 18:38
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The water does fall down, just that the water falls down with the bucket and it does not spill. –  ja72 Aug 9 '13 at 19:19

3 Answers 3

The net force on the water at the zenith is indeed downwards.

However a net force downwards does not mean that the motion will be downwards, only the acceleration.

And the water is indeed accelerated downwards at that brief instant, which is why it continues in a circular motion, rather than flying off horizontally as it would if there were neither centripetal nor gravitational forces acting on it.

Stop the water at the zenith by stopping the bucket, and the downward acceleration will act upon it for long enough for the water to indeed move downwards - by falling out of the bucket.

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The force on an object doesn't directly determine the direction it moves in — instead it determines the direction it accelerates in. Moving in one direction but slowing down is accelerating in the opposite direction.

The net force on the water in whirling bucket, when the bucket is at the top of its whirl is indeed downward. As you say, both gravity and the bucket are pushing the water down. And as you'd expect, the water is accelerating downwards. A moment ago it was slowing moving upwards. A moment later it will be slowly moving downwards. This is true however short the moments are. That means it's accelerating downwards. The direction it's moving in is changing from sideways and slightly upwards to sideways and slightly downwards.

With the ring, you say there is no 'outward' force, but I think as you call Centrifugal a psuedo force we should call 'outward' a pseudo direction. For an observer who isn't spinning, no particular direction is outward.

The ring must in practice slip off in some particular direction. Let's suppose it's to the east. There is a force pushing the ring to the east for half the time the rod is spinning.

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What i meant to say is, the ring will eventually slip off from the rod, the outward direction being the radial ray starting from the axis of rotation. My teacher said that this happens because the normal reaction on the ring by the rod is orthogonal at any instant, and due to this, at the next instant, one of it's components sort of pushes it outward. Can that explanation be used to justify this? –  Dhananjay Gupta Aug 10 '13 at 8:57

The water does fall down!

It moves forward due to it's tangential velocity and falls down precisely the amount it is necessary to move in a circle.The argument is similar to why doesn't the moon fall "down" towards the earth even if gravity is the only force acting on the moon.

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