Pressure at the center of an accelerating sphere

[Nathan B]

Homework Statement

Consider a glass sphere completely filled with water except for a small air bubble at 1 atmosphere. The sphere is accelerating to the right. What is the pressure at the center of the sphere?
Let Pe = Atmospheric pressure
r = radius of sphere
pw = density of water
a = acceleration of sphere

Homework Equations

I've previously solved this problem successfully with Pascal's law P = Pe + pw*g*h

The Attempt at a Solution

A successful solution was found with the equation P (center of sphere) = Pe + pw * sqrt (g^2 + a^2) r

While this solution rendered the correct answer, my professor told me that summing the acceleration with gravity is not a conceptually correct way of solving this problem, and that it should be solved with forces and pressure differences. Could someone please show me how that might be done?

 

[Charles Link]

The equation that you need is the force per unit volume equals minus the pressure gradient: $\vec{f}_v=-\nabla P$ . You also know (basically F=ma) that $\vec{f}_v=\delta \,\vec{a}= \delta \, a \, \hat{x}$ where $\delta$ is the density (kg/m^3) etc. With the small air bubble at 1 atm., I think that puts boundary conditions on the internal pressure at the rightmost glass surface. One other item to consider would be gravity. The gravity equations require that $-\nabla P -\delta g \hat{z}=0$. (static equilibrium from the gravity. The pressure gradient force plus the gravity force per unit volume must add to zero). Perhaps the gravity could be ignored for at least the first calculation of the pressure from the acceleration. Otherwise, you have $-\nabla P=\delta \, a \, \hat{x}+\delta \, g \, \hat{z}$. For the gradient operator in the first equation without the gravity, you only need the $\frac{\partial{P}}{\partial{x}}$ term. $\\$ Editing: @Nathan B Your approach in the OP to introduce gravity as well I think simplifies the solution: There is a symmetry in this differential equation if you let $\vec{u}=\delta \, a \, \hat{x} +\delta \, g \, \hat{z}$ and consider the gradient along the $\vec{u}$ direction, and integrate $dP$ back to the center along this line. In fact, doing this gives the exact same answer that you gave in the OP.

 

[Nathan B]

After working at it for a bit I came up with this solution with forces. Still not with pressure gradients, but it works.

ΣF_y = 0
ΣF_x = ma
As the buoyant force supplies all force in the x direction, F_Bx = ma
F_B = ma/cos θ = ma/(ma/√((ma)²+F_G²)) = √((ma)²+(mg)²) = m√(a²+g²)
Let P = gauge pressure and A = Area
P = m√(a²+g²)/A
m/A = ρh
P = ρh√(a²+g²)
Let P_a = absolute pressure and P_o = atmospheric pressure

P_a = P_o + ρh√(a²+g²)
Which is the same answer I got originally, just achieved in a somewhat more rigorous manner.

 

[Charles Link]

It looks good what you even did in the first posting. To do it with gradients, you get $dP/dr=-\delta \, \sqrt{a^2+g^2}$ where $r$ is in the $\vec{u}$ direction, and I'm no longer writing it as a partial derivative. (The $r$ is now basically the height in the $\vec{u}$ direction with $\vec{u}=\delta \, a \, \hat{x}+\delta \, g \, \hat{z}$). Integrating: $P(0)=P(R)+\int\limits_{0}^{R} (\frac{dP}{dr}) \, dr=P(R)+\delta \, \sqrt{a^2+g^2} \, R$ which is the same result that you got. In doing these buoyancy problems, it can be helpful to use the pressure gradient form. $\\$ Editing: Meanwhile your approach of treating the presence of a gravitational field as equivalent to an upward acceleration of $g$ is completely accurate and should be seen as a valid solution. Alternatively, an acceleration of $a$ to the right could be seen as equivalent to a gravitational field with that magnitude pulling to the left. Editing: The only other thing I can think of that he may be looking for is a derivation of the form: $[P(x)-P(x+\Delta x)]A=\delta \, \Delta x \, A \, a$ which gives $-\frac{dP}{dx}=\delta \, a$, which is a simplified form of $-\nabla P=\delta \, \vec{a}$.