Question

Classical Mechanics problem:

A bucket of water is set spinning about its symmetry axis with an angular velocity of magnitude Ω. What is the shape of the water surface after it has reached equilibrium? HINT: The surface of the water is an equipotential under the combined effects of gravity and the centrifugal force. The shape is one you are familiar with and using cylindrical polar coordinates

Answer 1

Solution:

When a bucket of water is spinning about its symmetry axis with an angular velocity, $\Omega$ it is at rest with respect to the rotating frame. Let us assume that the water in the bucket has come to rest. Consider the following diagram to indicate the forces acting on the system :

fcent

Now, the equation for the system is given by :

F + cent - JP=0

where,    =The body force per unit volume

F = -pgť

Frent = - p x (RxR) = centripetal facce öp = face due to premiere gradient. = -8gê - potêx CQxR) Spa –gge + prird where, & is the distance of the volume element from the rotation axis. ů is the unit vector pointing in the direction of increasing er. Then, JP = aggê - prêx CQxRy ap = -89 and (1) I = 82² After integrating equations (1) and (2), = £ 9 -22 22 – ggz + constant. Since the system is in equilibrium, pressure, P = constant pozer-p92 - constant or 2-der + 20 29

The above equation represents a paraboloid. Thus, the surface of water in the bucket, spinning about its symmetry axis with an angular velocity, $\Omega$ is a paraboloid after it has reached equilibrium.