Fluids mechanics

A sphere of radius RR, is placed in a hole of radius rr (R>rR>r), at the bottom of a beaker. What is the minimum mass of sphere required for it to not leave the hole for any amount of liquid of density ρ\rho poured into the beaker?

Note by Azimuddin Sheikh
7 months, 3 weeks ago

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@Azimuddin Sheikh I think I will post a problem on this

The sphere of radius RR sits in a hole of radius rr at the bottom of a container. Suppose that the center of the sphere is at the origin (x,y,z)=(0,0,0)(x,y,z) = (0,0,0). Let ϕ\phi be the angle with respect to the positive zz axis, and let θ\theta be the angle with respect to the positive xx axis.

Water fills the container from the bottom to some height. Let ϕ2\phi_2 correspond to the bottom contact point between the water and the sphere, and let ϕ1\phi_1 correspond to the top contact point between the water and the sphere.

The expression for ϕ2\phi_2 in terms of rr and RR is:

ϕ2=π2+acos(rR)\phi_2 = \frac{\pi}{2} + acos \Big( \frac{r}{R}\Big )

The expression for the zz coordinate of an arbitrary point on the sphere is:

z=Rcos(ϕ)z = R \, cos(\phi)

Here, ϕ=0\phi = 0 corresponds to the top of the sphere. The pressure is:

P=ρgR[cosϕ1cosϕ]P = \rho \, g \, R \, [cos \phi_1 - cos \phi]

The infinitesimal sphere surface area is:

dA=R2sinϕdθdϕdA = R^2 \, sin \phi \, d \theta \, d \phi

The infinitesimal pressure force magnitude is:

dF=PdA=ρgR3sinϕ[cosϕ1cosϕ]dθdϕdF = P \, dA = \rho \, g \, R^3 \, sin \phi \, [cos \phi_1 - cos \phi] \, d \theta \, d \phi

The infinitesimal force in the zz direction is:

dFz=dFzR=dF(cosϕ)=ρgR3sinϕcosϕ[cosϕ1cosϕ]dθdϕdF_z = dF \, \frac{-z}{R} = dF \, (-cos \phi) = -\rho \, g \, R^3 \, sin \phi \, cos \phi \, [cos \phi_1 - cos \phi] \, d \theta \, d \phi

The total force in the zz direction is:

Fz=ρgR3ϕ1ϕ202πsinϕcosϕ[cosϕ1cosϕ]dθdϕ=2πρgR3ϕ1ϕ2sinϕcosϕ[cosϕ1cosϕ]dϕ=πρgR33[cosϕ1cosϕ2]2[cosϕ1+2cosϕ2]F_z = -\rho \, g \, R^3 \int_{\phi_1}^{\phi_2} \int_0^{2 \pi} sin \phi \, cos \phi \, [cos \phi_1 - cos \phi] \, d \theta \, d \phi \\ = - 2 \pi \, \rho \, g \, R^3 \int_{\phi_1}^{\phi_2} sin \phi \, cos \phi \, [cos \phi_1 - cos \phi] \, d \phi \\ = -\frac{\pi \, \rho \, g \, R^3}{3} [ cos \phi_1 - cos \phi_2]^2 \, [cos \phi_1 + 2 cos \phi_2]

As a sanity check, suppose the hole had zero width, and that the water was filled to the top of the sphere. In this case, ϕ1=0\phi_1 = 0 and ϕ2=π\phi_2 = \pi. The FzF_z expression reduces to:

Fz=πρgR33[1+1]2[12]=4πρgR33=Vsphereρg=weight of displaced waterF_z = -\frac{\pi \, \rho \, g \, R^3}{3} [ 1 + 1]^2 \, [1 - 2] \\ = \frac{4 \pi \, \rho \, g \, R^3}{3} = V_{sphere} \, \rho \, g = \text{weight of displaced water}

In general, we take ϕ2\phi_2 as a given, and we must find the value of ϕ1\phi_1 which maximizes the upward pressure force. I will leave this as an exercise for the reader, but it turns out that the upward pressure force is maximized when the following condition is satisfied:

ϕ1+ϕ2=π\phi_1 + \phi_2 = \pi

So the minimum weight of the sphere, to ensure that it covers the hole unconditionally, is:

Wspheremin=πρgR33[cosϕ1cosϕ2]2[cosϕ1+2cosϕ2]ϕ2=π2+acos(rR)ϕ1=πϕ2=π2acos(rR)W_{sphere-min} = -\frac{\pi \, \rho \, g \, R^3}{3} [ cos \phi_1 - cos \phi_2]^2 \, [cos \phi_1 + 2 cos \phi_2] \\ \phi_2 = \frac{\pi}{2} + acos \Big( \frac{r}{R}\Big ) \\ \phi_1 = \pi - \phi_2 = \frac{\pi}{2} - acos \Big( \frac{r}{R}\Big )

Steven Chase - 7 months, 3 weeks ago

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@Steven Chase Sir thx for your method great way .!! . yeah sure post a problem sir.

Azimuddin Sheikh - 7 months, 3 weeks ago

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@Steven Chase Sir BTW can't it be much simplified with the help.of conservation of energy ? Without using force and integration method ?? Also sir if u have some time can u pls try this problem also out ?? https://brilliant.org/discussions/thread/classical-mechanics-problem-3/?ref_id=1563236

Azimuddin Sheikh - 7 months, 3 weeks ago

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@Azimuddin Sheikh This is a stasis problem, so how would energy conservation be relevant?

Steven Chase - 7 months, 3 weeks ago

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@Steven Chase Oh yes sir nvm was thinking about some other problem . BTW sir may u try this one if u have some time ?https://brilliant.org/discussions/thread/classical-mechanics-problem-3/?ref_id=1563236

Azimuddin Sheikh - 7 months, 3 weeks ago

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@steven chase sir pls help ,ideas pls. @ram mohith pls help ?

Azimuddin Sheikh - 7 months, 3 weeks ago

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@Steven Chase and @Ram Mohith you may want to check this note.......

Aaghaz Mahajan - 7 months, 3 weeks ago

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This is an interesting one. I'll start thinking about it.

Steven Chase - 7 months, 3 weeks ago

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Good to know about that its interesting 😊

Azimuddin Sheikh - 7 months, 3 weeks ago

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How can a sphere of radius RR be placed in a hole of radius rr if R>rR >r. I didn't understand the first part of the question. Is there any diagram.

Ram Mohith - 7 months, 3 weeks ago

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@Ram Mohith bro , in a beaker there is hole in the basement where a sphere of radius R is placed , off course some part of sphere is outside in air and other inside the liquid

Azimuddin Sheikh - 7 months, 3 weeks ago

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