Describing the motion of a bubble through water

[ttja]

Homework Statement

A bubble of air which is 1 mm in diameter is released without initial velocity in the
volume of glycerol at room temperature. Describe the motion of the bubble as a
function of time assuming that its diameter remains unchanged. Note that friction
force acting on the bubble is given by

Drag Force = -4[tex]\pi[/tex][tex]\eta[/tex]Rv

where h is the viscosity of the liquid, R is the radius of the bubble, and v is its
velocity. Note that the numerical coefficient in this formula is different from that for a
solid sphere moving in the liquid (which is 6π). Explain qualitatively why the
numerical coefficients in tbubble are different.

Homework Equations

F=ma
F=[tex]\rho[/tex]Vg

The Attempt at a Solution

Tell me if I'm going about this the wrong way. To describe the motion of the bubble, which is essentially finding the function of position with respect to time, I equated F = ma = Fb - Ff. Then integrating knowing that the initial conditions for velocity and position are 0, perhaps this could be it?

As I researched, however, I had found that the initial acceleration for the bubble starting at rest to be 2g, whereas, by my method, i would only get 1g.

Other considerations I had heard from discussion are that the mass of the air inside the bubble is negligible, so am i supposed to use the mass of water in contact to the surface when calculating "m" ? In addition I had also heard something about work, but i have no idea how that plays in this problem.

As for the second part, I thought about how...since the bubble is not a rigid body, the molecules themselves are in a circular flow that ultimately act as a .. buffer of some sort? i know I am not making sense but i know there should be a difference between fluid in contact with fluid and fluid in contact with a solid.

 

[ehild]

Do not forget about Archimedes' principle when counting forces.

ehild

 

[kerimek]

Dear student,

Your approach is on the right track, but there are a few things to consider in order to fully describe the motion of the bubble through water.

First, let's look at the forces acting on the bubble. As you correctly stated, we have the buoyant force, Fb, which is equal to the weight of the displaced liquid. We also have the drag force, Ff, which is given by the equation -4πηRv. This drag force is due to the friction between the bubble and the surrounding liquid, and it is always opposing the motion of the bubble.

Now, we can set up our equation F = ma = Fb - Ff. Since we are assuming that the bubble's diameter remains unchanged, we can also assume that its volume remains constant. This means that the buoyant force, Fb, will also remain constant. However, the drag force, Ff, will vary with the velocity of the bubble. This means that as the bubble moves faster, the drag force will increase, and as it slows down, the drag force will decrease.

Using this information, we can set up a differential equation to describe the motion of the bubble:

ma = Fb - Ff

Since m is constant (assuming the mass of the air inside the bubble is negligible), we can rewrite this as:

a = (Fb - Ff)/m

Now, we can substitute our expressions for Fb and Ff:

a = (ρVg - (-4πηRv))/m

Since V and m are both constant, we can combine them into one term, ρV. This gives us:

a = (ρVg + 4πηRv)/m

Now, we can use the formula F = ma to rewrite this as a differential equation for the velocity of the bubble with respect to time:

v' = (ρVg + 4πηRv)/m

This is a first-order, linear, non-homogeneous differential equation. We can solve it using standard methods, such as separation of variables or integrating factors. Once we have solved for v(t), we can then find the position of the bubble as a function of time by integrating v(t).

As for the second part of the problem, the reason for the difference in the numerical coefficients is due to the shape of the objects moving through the liquid. A solid sphere has a much smoother surface than a bubble