A sphere of radius \(R\), is placed in a hole of radius \(r\) (\(R>r\)), 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 \(\rho\) poured into the beaker?

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

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

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

The expression for \(\phi_2\) in terms of \(r\) and \(R\) is:

The expression for the \(z\) coordinate of an arbitrary point on the sphere is:

\[z = R \, cos(\phi)\]

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

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

The infinitesimal sphere surface area is:

\[dA = R^2 \, sin \phi \, d \theta \, d \phi\]

The infinitesimal pressure force magnitude is:

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

The infinitesimal force in the \(z\) direction is:

\[dF_z = dF \, \frac{-z}{R} = dF \, (-cos \phi) = -\rho \, g \, R^3 \, sin \phi \, cos \phi \, [cos \phi_1 - cos \phi] \, d \theta \, d \phi\]

The total force in the \(z\) direction is:

\[F_z = -\rho \, g \, R^3 \int_{\phi_1}^{\phi_2} \int_0^{2 \pi} sin \phi \, cos \phi \, [cos \phi_1 - cos \phi] \, d \theta \, d \phi \\
= - 2 \pi \, \rho \, g \, R^3 \int_{\phi_1}^{\phi_2} sin \phi \, cos \phi \, [cos \phi_1 - cos \phi] \, d \phi \\
= -\frac{\pi \, \rho \, g \, R^3}{3} [ cos \phi_1 - cos \phi_2]^2 \, [cos \phi_1 + 2 cos \phi_2]\]

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, \(\phi_1 = 0\) and \(\phi_2 = \pi\). The \(F_z \) expression reduces to:

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

In general, we take \(\phi_2 \) as a given, and we must find the value of \(\phi_1 \) 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:

\[\phi_1 + \phi_2 = \pi\]

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

@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

@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

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

@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

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

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

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

The expression for \(\phi_2\) in terms of \(r\) and \(R\) is:

\[\phi_2 = \frac{\pi}{2} + acos \Big( \frac{r}{R}\Big )\]

The expression for the \(z\) coordinate of an arbitrary point on the sphere is:

\[z = R \, cos(\phi)\]

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

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

The infinitesimal sphere surface area is:

\[dA = R^2 \, sin \phi \, d \theta \, d \phi\]

The infinitesimal pressure force magnitude is:

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

The infinitesimal force in the \(z\) direction is:

\[dF_z = dF \, \frac{-z}{R} = dF \, (-cos \phi) = -\rho \, g \, R^3 \, sin \phi \, cos \phi \, [cos \phi_1 - cos \phi] \, d \theta \, d \phi\]

The total force in the \(z\) direction is:

\[F_z = -\rho \, g \, R^3 \int_{\phi_1}^{\phi_2} \int_0^{2 \pi} sin \phi \, cos \phi \, [cos \phi_1 - cos \phi] \, d \theta \, d \phi \\ = - 2 \pi \, \rho \, g \, R^3 \int_{\phi_1}^{\phi_2} sin \phi \, cos \phi \, [cos \phi_1 - cos \phi] \, d \phi \\ = -\frac{\pi \, \rho \, g \, R^3}{3} [ cos \phi_1 - cos \phi_2]^2 \, [cos \phi_1 + 2 cos \phi_2]\]

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, \(\phi_1 = 0\) and \(\phi_2 = \pi\). The \(F_z \) expression reduces to:

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

In general, we take \(\phi_2 \) as a given, and we must find the value of \(\phi_1 \) 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:

\[\phi_1 + \phi_2 = \pi\]

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

\[W_{sphere-min} = -\frac{\pi \, \rho \, g \, R^3}{3} [ cos \phi_1 - cos \phi_2]^2 \, [cos \phi_1 + 2 cos \phi_2] \\ \phi_2 = \frac{\pi}{2} + acos \Big( \frac{r}{R}\Big ) \\ \phi_1 = \pi - \phi_2 = \frac{\pi}{2} - acos \Big( \frac{r}{R}\Big ) \]

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

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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

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

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

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

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Good to know about that its interesting 😊

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

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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

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