Prof. A. N. Maheshwari

Absolute space, inertial and non-inertial frames

Newton's Bucket Experiment

Let the water in the bucket rotate with a constant angular velocity w about the central vertical axis of the cylindrical bucket.

The liquid surface takes the shape as shown in the figure. What has been shown is a vertical cross-section of the fluid.

The surface of the fluid is curved. In fact it is paraboloidal.

We fix the coordinate system by measuring vertical distance from the lowest point in the curved surface of the fluid; it will be by symmetry on the central axis.

The fluid below the level y=0 is rotating uniformly with angular velocity w.

Let us consider a fluid element at a distance r from the axis. Let the area of its cross-section be DA   and its width be Dr.

Let r be the density of the fluid. The mass of the fluid element will be rDADr.

This condition on variation of pressure in the radial direction can be obtained provided at a given level pressure varies with distance from the central axis.

The centripetal force is provided by the pressure difference at the two opposite faces of the fluid element.

p (r + Dr) DA - p (r) DA = DA Dr r w² r

or

Let p = p_c be the pressure at the axis of rotation (r = 0) at a given depth. Then integrating the above equation
we get

p_c will be varying with vertical depth by the usual relation
p_c(h) = rgh,

as the fluid develops pressure because of the weight of the fluid above.

But, how does the pressure ½rw²r²       arise?

It is due to the weight of the fluid at a given r between the level y=0 and the surface of the liquid.

If the height of the fluid surface at a given r is y as measured from the level y=0, the additional pressure will be rgy.

Equating rgy and ½rw²r², we obtain the equation of the surface. It is

It is the equation of a parabola. Hence the liquid surface develops a paraboloidal shape.

Prof. A. N. Maheshwari

Inertial and Non-inertial Frames

Prof. A. N. Maheshwari

Prof. A. N. Maheshwari