Fluids mechanics

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?

#Mechanics

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Comments

@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:

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

Steven Chase - 2 years, 3 months ago

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