Irodov Problem 1.331

The viscosity of the fluid causes generates a resistive torque T that tries to slow down the rotating disc. The power generated by by this torque as the disc rotates at an angular velocity of w, is given by Tw.

Now let us compute this resistive torque generated by viscosity. Consider an infinitesimally thin ring of radius r and width dr on the disc. The area of this ring is . Let the resistive viscous force acting on this thin ring be dF.

The velocity of the oil close to this ring section of the disc will be the same as that of the rotating disc equal to wr. The velocity of the oil close to walls of cavity will be 0 since the walls are stationary. Thus, the velocity gradient of the oil in the vertical region between this thin ring section of disc and walls of the cavity will be wr/h.

From the definition of coefficient of viscosity the viscous force dF generated by fluid between the upper cavity wall and the thin ring section of the disc is given by,






The same amount of resistive torque is generated by the fluid between the lower wall of the cavity and the thin ring section as well and so the net resistive viscous force is 2(dF). This resistive force 2(dF) results in a resistive torque dT = 2(dF)r given by,








The work done by the viscous forces is given by Tw equal to .

Irodov Problem 1.330

Essentially the problem asks us to find the function h = f(r), i.e. how the height of the fluid depends on the distance from the axis of rotation.

Basically what happens is that as the fluid rotates, each section of the rotating fluid exerts a centrifugal force (as seen from the frame rotating with the axis) on the next section of the fluid. Consequently the next section of the fluid increases in height to generate enough hydrostatic pressure to compensate for the centrifugal force.

Let us first start by computing the hydrostatic force exerted by an infinitesimally thin section of rotating fluid of height h and width dr.

At a depth x from the surface the hydrostatic pressure exerted by the fluid is . The force exerted over a thin section of height dx at a this depth x is given by . The net hydrostatic force exerted by the entire section of liquid is given by,



The centrifugal force exerted by this section of liquid as it rotates is given by . Here dm is the mass of the infinitesimally thin section of the liquid and is given by . The centrifugal force exerted by this section of the liquid is thus given by .

Now let us consider too adjacent infinitesimally thin sections of the liquid of height h and h+dh.
The right (lower height column) exerts a hydrostatic force of F(h) to the left while the left (higher column) exerts hydrostatic force F(h+dh) to the right. The right column also exerts a centrifugal force of towards the left. For equilibrium we have,



As dh->0 and dr->0 we can neglect second order terms in (1) such as dh squared and obtain,










Here h0 is the height of the fluid at the center of the cylinder. In practice knowing the total volume of the liquid in the container ho can be found.

b) The hydrostatic pressure at the bottom of the cylinder any radius r is simply

Irodov Problem 1.329

We assume that just before the tube in the tank the water is still i.e its velocity is 0. Let the velocity of water as it exits out of the spout be vs. The internal pressure of water just before the tube in the tank is given by the pressure due to water column and the atmospheric pressure PA as .
As the water exists out of the spout the pressure is equal to the atmospheric pressure. Applying Bernoulli's equation for just before the tube and as it exists out of the tube we can write,



Now let us consider a cross-section within the tube that has cross-section area a. Let the velocity of the water at this point be v. Since the rate of volume of the water flowing out the tube must be equal at all cross-sections within the tube we have,



Let the internal pressure at this cross-section within the tube be P. Appllying Bernoulli's equation at this cross-section within the tube and at the exit point of the tube we have,



The water exerts pressure P normal to the surface of the tube outwards and the atmosphere exerts pressure PA pushing it inwards. The net pressure exerted on the tube outwards is thus P-PA. The horizontal component of the force due to this pressure (dF) that is trying to rip this section of the tube out of the tank is given by the pressure times the vertical component of the area of the tube given by,