When a bucket full of water is rapidly whirled in a vertical circular path, water does not fall out. Why?
it is about the circular motion and gravitation. i want its answer very immediately, if it is late should not more than tommorow.complete Qs is below: when a bucket full of water is rapidly whirled in a vertical circular path, water does not fall outeven if the bucket is inverted at the maximum height. why this is so?
I believe cetrifugal force is what keeps the water in the bottom of the bucket. In a situation like the one you describe, an outward force away from the center of rotation is developed as the bucket is swung around a central axis.
In some ways, this is considered a form of gravity, such that designs for space stations include the idea of a large living space built in a circle around a central axis, spinning so that there would be an artificial gravity created along the surface of that outer ring.
You can take a bob tied to a string and and swing the bob (in this case a bucket of water) in a vertical plane. At small speed the bob oscilates. The higher the speed imparted to the bob through the string ,you get the bob raise higher and higher doing the oscillation. When the initial imparted speed crosses a limit, the bob rotates( without oscillatory motion) in a vertical plane due to the centrefugal force provided through the string . The minimumum speed V0 at the lowest postion of the vertical plane that should be provided to the bob (or a bucket of water) so that keeps on rotating in a vertical circle without string getting slack (or water in the bucket without falling down even at the topmost point of the circle when the bucket is upside down) is vo = (5gl)^(1/2). Wgre g is the acceleration due to gravitation and gl is the length of the string (or the radius of the vertical plane in which the bucket is making the circular motion).