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Newton's bucket problem

  1. Mar 16, 2008 #1
    How does the water in a spinning bucket arrange itself, given that its surface must be perpendicular to the force that holds the water? Solve for the Lagrangian forces of constraint.

    As a convention, I'm writing position, velocity, and acceleration as such:
    position: a velocity: a. acceleration: a..

    So here's what I've got so far. We know that the lagrangian is:

    L = T - U

    I chose cylindrical coordinates, so I got (a is the angle):
    T = (1/2) * m * r. + (1/2) * m^2 * a^2 + (1/2) * m * z.^2
    U = mgz
    Constraint1: m = (pi) * r^2 * h * density

    dL/dr = 0, dL/dr. = 0, D/DT * dL/dr. = 0
    dL/da = 0, dL/a. = m^2 * a. D/DT * dL/a. = m^2 * a..
    dL/dz = mg, dL/z. = m * z. D/DT * dL/z. = m * z..

    So for my first equation of motion, solving for motion in the a direction:
    0 = -m^2 * a..

    Which isn't exactly allowing me to solve for much of anything... where did I go wrong?
     
  2. jcsd
  3. Mar 17, 2008 #2
    I dont really know what type of answer the question is asking for, but here is my attempt on a solution that gives the height y at distance x from the buckets edge along the radius. Not sure if its entirely correct tho.

    w=anglespeed(?)
    R=radius
    a=centripetal acceleration=v^2/(R-x)=W^2*(R-x)
    the rake angle b of the water is given by tan b=a/g=(W^2/g)(R-x)
    if h is the distance along the y-axis upwards down, for small dx, dh=(W^2/g)(R-x)dx
    Integration gives that y=H-(W^2/g)(Rx-(x^2)/2)
    where H is the height if the water at the edge of the bucket, ie the maximum height. Its also possible to calculate H in terms of the height of the water in rest, by integrating again and use that the total volyme of the water is constant, but this is sufficient if you only want the shape of the surface.
     
    Last edited: Mar 17, 2008
  4. Feb 22, 2010 #3
    I am working on this problem as well, as part of my physics class this semester.
    I haven't finished it, but I have most of the problem completed. I noticed some mistakes in your work.

    First, for finding the langrangian forces of constraint, you want to apply your constraint as late as possible. Also, the equation you'll use to find the forces of constraint is as follows:
    dL/dq - D/Dt*dL/dq. + lambda * dQ/dq = 0,
    where lambda * dQ/dq = Fq, the force in the q direction. q represents a generalize coordinate.

    Your equation for the kinetic energy in cylindrical coordinates is wrong, it is as follows:
    T = (1/2) * m * r.^2 + (1/2) * m * (r*a.)^2 + (1/2) * M * z.^2
    and U = m * g * z

    So then we have
    L = T - U = (1/2) * m * r.^2 + (1/2) * m * (r*a.)^2 + (1/2) * M * z.^2 - m * g * z

    dL/dr = m * r * a.^2; dL/dr. = m * r.; D/Dt * dL/dr. = m * r..
    dL/dr - D/Dt * dL/dr. + lambda * dQ/dr = (m * r * a.^2) - (m * r..) + lambda * dQ/dr = 0

    dL/da = 0; dL/da. = m * r^2 * a.; D/Dt*dL/da. = m * (r^2 * a.. + 2 * a. * r.)
    dL/da - D/Dt * dL/da. + lambda * dQ/da = -m * (r^2 * a.. + 2 * a. * r.) + lambda * dQ/da = 0

    dL/dz = -m * g; dL/dz. = m * z.; D/Dt * dL/dz. = m * z..
    dl/dz - D/Dt * dL/dz. + lambda * dQ/dz = -m * g - m* z.. + lambda * dQ/dz = 0

    lambda * dQ/dq = Fq, so we solve for the force in each direction:

    Fr = m * r.. - m * r * a.^2
    Fa = m * (r^2 * a.. + 2 * a. * r.)
    Fz = m * z.. + m * g

    This is where I am in the problem. These forces are the lagrangian forces of constraint - at this point, you need to factor in the spinning, and then somehow use that to find the shape of the water. I'm basically stuck, at this point.
     
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