Work done in submerging a weightless sphere, I am getting exactly half the correct, can someone point out my mistake!

Question

This is the question: A sphere of radius 0.4 m and having negligible weight is floating in a large freshwater lake. How much work is required to completely submerge the sphere? The density of the water is 1000 kg/m3

My answer:

Imagine the position of the sphere when it is floating on the water and no force has been applied yet, we will cut this sphere into slices of height $dh$ that are parallel to the water surface.

Finding the work done on the bottom hemisphere: Consider a slice that is $h$ units below the center of the sphere, its radius (radius of the slice, not the sphere), is given by the Pythagoras theorem $$r=\sqrt{0.4^2-h^2}=\sqrt{0.16-h^2}$$

The Volume of this slice is $\pi r^2\hspace{1mm}dh=\pi \left(0.16-h^2\right)\hspace{1mm}dh$

Every slice will move down by 0.8m, but the slices will experience no force when they are moving in the air, a constant force proportional to their volume will act once they are in the water.

A slice that is $h$ units below the center, will spend $0.4-h$ meters in the air and $0.8-(0.4-h)=0.4+h$ meters in the water

So, the work done on this slice is $$W^*=F\cdot d=9.8\cdot 1000\pi\left(0.16-h^2\right)\hspace{1mm}dh\cdot (0.4+h)$$

Note that: $F=\rho g V$ ($\rho$ is density of water and $V$ is the volume displaced by the slice)

Integrate from 0 to 0.4, to get complete work done on the lower hemisphere

$$W=9800\pi \int_0^{0.4}(0.4+h)\left(0.16-h^2\right)\hspace{1mm}dh\longrightarrow\left(\text{Equation 1}\right)$$

Finding the work done on the upper hemisphere: Consider a slice that is $h$ units above the center of the sphere, its radius is $$r=\sqrt{0.4^2-h^2}=\sqrt{0.16-h^2}$$

Volume of this slice is $\pi r^2\hspace{1mm}dh=\pi\left(0.16-h^2\right)\hspace{1mm}dh$

A slice that is $h$ units above the center, will spend $0.4+h$ meters in the air and $0.8-(0.4-h)=0.4-h$ meters in the water

So, the work done on this slice is $$W^*=F\cdot d=9.8\cdot1000\pi \left(0.16-h^2\right)\hspace{1mm}dh\cdot (0.4-h)$$

Integrate from 0 to 0.4, to get complete work done on the lower hemisphere

$$W_2=9800\pi \int_0^{0.4}(0.4-h)\left(0.16-h^2\right)\hspace{1mm}dh\longrightarrow\left(\text{Equation 2}\right)$$

After adding $\left(\text{Equation 1}\right)$ and $\left(\text{Equation 2}\right)$, we get

$$W=W_1+W_2=9800\pi \int_0^{0.4}(0.4-h+0.4+h)\left(0.16-h^2\right)\hspace{1mm}dh$$

$$=0.8\cdot9800\pi\int_0^{0.4}0.16-h^2\hspace{1mm}dh$$

$$=0.8\cdot9800\pi\bigg[0.16h-\dfrac{h^3}{3}\bigg]_0^{0.4}$$

$$=0.8\cdot9800\pi\bigg[0.16\cdot0.4-\dfrac{0.4^3}{3}\bigg]-0\approx1051\text{ J}$$

The correct answer is obviously $\bigg[\dfrac{4}{3}\pi\cdot 0.4^3\cdot1000\bigg]\cdot9.8\cdot0.8\approx2102\text{ Joules}$

But I need to do this by using calculus because this is in a calculus textbook with very little physics background.

Answer 1

The correct answer is obviously $\bigg[\dfrac{4}{3}\pi\cdot 0.4^3\cdot1000\bigg]\cdot9.8\cdot0.8\approx2102\text{ Joules}\mathrm{~[sic]}$

This is the correct answer to submerge the sphere and give it continued downward velocity from applying the full equilibrium submersion force when it was only entering the water, i.e., when less force was needed for continued quasistatic downward movement. In other words, a constant force accelerates the sphere substantially, giving it kinetic energy as well as the potential energy from being submerged.

To maintain slow movement—so that only the potential energy is relevant—we intuitively know to push gently initially when submerging something, ramping up the load during submersion. This makes the difference between the integral of the work applied corresponding to a right triangle (whose base equals the displacement and height equals the end force) vs. corresponding to a rectangle that contains that right triangle, with double the area.

This factor-of-two question comes up frequently when someone is dealing with work and energy in the case of a restoring force. There must be a dozen questions on this site alone in the context of hanging weights on springs and charging capacitors with a constant voltage.

EDIT: Perhaps two dozen. See Calculating elastic energy constant, Wrong calculation of work done on a spring, how is it wrong?, Work done by battery and potential energy of a capacitor, What happens to half of the energy in a circuit with a capacitor?, Saving energy while charging capacitor, Mass dropped on a spring, Getting 2 different answers when finding spring constant $k$ when gravity is involved, If string is stretched just by weight, where does the gravitational potential energy goes if only half is converted to elastic potential energy?, Why the work done on the spring is different?, Where does the half of Potential energy go?, Strange factor of 2 in Hooke's Law, Why is energy absorbed by the battery when the plates of a parallel plate capacitor are pulled apart?, Energy stored in a capacitor vs Work that the battery does, What is maximum compression of an ideal spring?, Does work done on a spring = elastic potential energy?, Where does half of the work done in charging a capacitor is dissipated?, Approaches to Sand on a Conveyor Belt, Different methods give different answers to the compression of spring, The lowering of a block using a spring and the net work done by each force involved

Answer 2

The correct answer is obviously $\bigg[\dfrac{4}{3}\pi\cdot 0.4^3\cdot1000\bigg]\cdot9.8\cdot0.8\approx2102\text{ Joules}$

No, this is not obviously the correct answer.

The answer that is (obviously) correct is:

$$\bigg[\dfrac{4}{3}\pi\cdot 0.4^3\cdot1000\bigg]\cdot9.8\cdot0.4\approx1051\text{ Joules}$$

Which, conveniently, corresponds to the answer you got by the more roundabout route.

Answer 3

To make the statement of the problem less ambiguous, you can think of an alternative problem. Consider a rigid sphere of uniform density in a fluid which is only affected by the buoyancy force of the fluid. This means neglecting the weight of the sphere and any buoyancy force from the air, which in principle could matter substantially if the density of the sphere is low enough. The sphere is initially static and fully submerged, in such a way that the top of the sphere is tangential to the fluid's surface. In this situation, it is easy to see that the sphere will be pushed up by the fluid until it is fully expelled with non-$0$ vertical velocity. We can then ask how much work is done by the buoyancy force from this initial configuration to the point in time in which the sphere has fully exited the fluid. Naturally, the amount of kinetic energy acquired by the sphere matches the amount of work done by the net force acting on the body, which, in this case, is just the buoyancy force.

This is equivalent to asking the amount of work done by an external vertical force to push a (weightless) sphere inside the fluid, leaving it just fully submerged and preserving whichever kinetic energy it initially had. Under this constraint of maintaining the amount of kinetic energy, the work done by the net force, external plus buoyancy, is necessarily $0$, so each contribution matches in magnitude.

Let's come back to the problem of a static submerged body that exits the fluid, and let's try to explore the problem of a more general rigid body. Let's say the density of the fluid is $\rho_f$, and let's say the body has uniform density $\rho_0$. Let's also say that the body enjoys enough symmetry around the vertical axis for the center of buoyancy to lie in the same vertical as the center of mass throughout the whole expulsion of the body. This implies the buoyancy won't produce a torque around the center of mass. We shall take this vertical axis as the $y$ axis to match the notation of the exercise, making it also the origin of the $x$-$z$ plane. The horizontal cross section area then fully describes the shape of the body and determines the position of the center of mass $Y_\textrm{CM}$.

Since the body accelerates, a frame attached to the center of mass is not inertial. Let's pick instead the surface of the fluid where the symmetry axes intercepts as the origin of our lab frame. Since the body is submerged initially, we have that the position of the center of mass is $\mathbf{R_\mathrm{CM}}(t) = \left(0, Y_\textrm{CM}(t), 0\right)$ with $Y_\textrm{CM}(0) < 0$. The equation of motion is then \begin{equation} \rho_0 V \ddot{Y}_\textrm{CM} = \rho_f V_s(Y_\textrm{CM}) g \end{equation} where $V$ is the total volume of the object and $V_s(Y_\textrm{CM})$ is the submerged volume of the object. To compute these volumes we use the horizontal cross section of the body $A$. For convenience, we write $A$ as a function of the vertical distance $y_\mathrm{CM}$ measured from the center of mass. Let's say the shape of the body extends vertically a length $l_p$ above the center of mass, defining the top of the body, and the bottom of the body is located at a distance $l_m$ below the center of mass. We have then \begin{equation} V = \int_{-l_m}^{l_p} A(y) dy \, . \end{equation} More importantly, we can write also the intial conditions \begin{eqnarray} Y_\textrm{CM}(0) &=& -l_p \\ \dot{Y}_\textrm{CM}(0) &=& 0 \, , \end{eqnarray} and a expression for the submerged volume \begin{equation} V_s(Y_\textrm{CM}) = \begin{cases} V &, Y_\textrm{CM} \leq -l_p \\ \int_{-l_m}^{- Y_\textrm{CM}} A(y) dy &, -l_p \leq Y_\textrm{CM} \leq l_m \\ 0 &, l_m \leq Y_\textrm{CM} \end{cases} \, . \end{equation}

The work done by the buoyancy force from the initial state to the point in which the body exits the fluid is then \begin{eqnarray} W_B &=& g \rho_f \int_{-l_p}^{l_m} V_s(Y_\textrm{CM}) dY_\textrm{CM} \\ &=& g \rho_f \int_{-l_p}^{l_m} dY_\textrm{CM} \int_{-l_m}^{- Y_\textrm{CM}} A(y) dy \\ &=& g \rho_f \int_{-l_m}^{l_p} dy' \int_{-l_m}^{y'} A(y) dy \, , \end{eqnarray} where we just made the substitution $y' = - Y_\textrm{CM}$. Interestingly, this expression can be simplified to find a very nice result. Using integration by parts, we pick \begin{equation} u = \int_{-l_m}^{y'} A(y) dy \ \ \ , \ \ \ dv = dy' \, , \end{equation} to get \begin{equation} du = A(y') dy' \ \ \ , \ \ \ v = y' \, . \end{equation} This gives us \begin{eqnarray} W_B &=& g \rho_f \left[ \left. y' \int_{-l_m}^{y'} A(y) dy \right|_{y'= -l_m}^{y' = l_p} - \int_{-l_m}^{l_p} y' A(y') dy' \right] \\ &=& g \rho_f \left[ l_p V - \int_{-l_m}^{l_p} y' A(y') dy' \right] \, . \end{eqnarray} But the second term in the square brackets is, up to a constant factor, the position of the center of mass measured in the center of mass frame. This vanishes and we conclude \begin{eqnarray} W_B = g \rho_f l_p V \, . \end{eqnarray} In particular, for the sphere we have $l_p = R$, matching OP's result.

Answer 4

You immerse the upper part with the same force as the lover part, so your result is for two half spheres immersed side by side, do of course your result is half. For the upper half you have the force to keep the lower part under water + the force for the new slices.