Question

a tube of length L is filled completely with an incompressible liquid of mass M and closed at both ends. the tube is rotated in the horizontal plane about of its ends with a uniform angular velocity omega. what is the force exerted by liquid on wall in the middle of tube?

Answer 1

As the tube is rotating in a circular path, there is a centripetal acc inwards. The problem is to find the force on the wall in the middle of the tube. To do that, we find the force the wall exerts since they should both be equal (Newton's Third Law).
If you were in a car taking a sharp turn, you would find yourself thrown to the door on the other side away from the center of the turn. The liquid experiences something like that as well. The liquid is the first half of the tube is pushed against the liquid on the other half, or on the wall in this case which separates the liquid.
The magnitude of this "push" which is the centripetal force is given by mv^2/r. However, notice that r is not constant over here. The fluid is a continuous medium present throughout the space of the tube.
So, you can imagine the tube as consisting of successive discs of fluid. Consider once such disc present between r and r+dr. it has a mass dm=p*dr (where p is linear density of the fluid). The centripetal force due to that part dF=(pdr)*(r^2 * w^2)/r.
To find the total centripetal force, you need to integrate this expression over the limits 0 to r.

Answer 2

In this case the limits should be from 0 to L/2

Answer 3

2pi*y is perimeter. The volume of the disc is pi*y^2*dx
The integral will be that of (p*pi*y^2*dx)*(x*w^2).
<dm> <v^2/x>

Answer 4

@Aditya790 , Yep you are right :)

Answer 5

This is the right answer