Circular motion (Centripetal force)

Question

A U-tube with the distance between the limbs 25.0cm. The U-tube is filled with a liquid. Calculate the difference in height of the liquid level when

 

a) The tube moves along the horizontal with constant velocity  0.50 ms^-1.

 

b)The tube moves along the horizontal with uniform acceleration 1.00 ms^-2.

 

c)The tube is rotated above the vertical-axis, with the speed of 30 revolution per minute.

*Neglect effect of viscosity and surface tension.

Answer 1

If there is a difference in height between the sides of the U-tube, then that means that indicates a difference in pressure.  Since pressure arises from the weight (force due to gravity) of the fluid, then a pressure difference means there is a non-zero net force.

 

a) For a constant velocity, that means that there is no acceleration.  According to Newton's second law, F=m*a, so if a=0, then F=0, and therefore this is no height difference.

 

b) For a constant acceleration along the horizontal x-axis, that means that the net force is acting along that horizontal axis.  If the u-tube is moving to the right, then the left side of the tube will have a higher liquid level than the right side of the tube.  The pressure of a fluid is given by P = ρ*g*h, where ρ is the density, g is gravitational acceleration, and h is the height difference.  (This is derived from Pressure = weight/Area, where weight = m*g.)  We also know that P is also equal to Force/Area.  So we can set these two equations equal to each other and solve for h. 

Remember that the force here (again, Newton's second) is equal to mass*acceleration, where the acceleration is given in the problem, and the mass is the mass of liquid in the horizontal part of the u-tube (since that is the location of the net force we are considering).  This can be determined by the volume of liquid in the tube time the density: m = ρ*V = ρ*l*A (since the length times the cross sectional area is the volume).

The initial equation is thus,

 

                            P = F/A = ρ*g*h  ---now you can use what I've just described to plug in the other

 

                                                         values and solve for h.  You'll notice that some terms will cancel out.

 

 

c) Here is where we consider centripetal force as the tube rotates about the y-axis.  Remember that centripetal force is always towards the center, so again we will take the force to be along the horizontal axis due to the height difference in the two columns.  This time, instead of using F = m*a, we will use the angular version of this equation because we are given the angular velocity.  F(centripetal) = m*α (mass*alpha where alpha is angular acceleration).  We can write alpha in terms of omega, which is angular velocity, ω.  

        

           α = r*(ω²) ----> F = m*r*(ω²)    -----> this comes from the fact that alpha = (v²)/r where 

                                                                                v = ω*r

 

          Since we are still dealing with a pressure difference, we use the same pressure equation from the previous section.  P = ρ*g*h, we can set this equation equal to Force/Area as we did before.  

 

P = ρ*g*h = (m*r*ω²)/A   Again, we have to use the mass of the liquid in the horizontal part of the tube

 

                                        (this does not change, because we are assuming the liquid is incompressible) 

 

The value of r is not the entire length of the horizontal part of the tube, but rather half of that.  This is because the centripetal force is acting on the center of mass of the tube section we are considering.  Thus, r= l/2.  To find ω, we have to convert revolution per minute into radians per second.  One complete revolution is 2*π, and 1 minute = 60 seconds.  ---> (2*π)/60s = 1 rev/min

 

Now you have the information to solve for the height difference again.  Plug in the relevant variables to solve for h algebraically.