Diving Cone

Relevant wiki: Conservation of Mechanical Energy

We know that the center of mass of a right circular cone is at a height of $\frac{h}{4}$ from the base. When the cone goes into the water, the water displaces and comes to the surface. Thus, the cone is losing the gravitational potential energy which is gained by the water. The kinetic energy is zero at the beginning and the end of the process, thus using the conservation of energy, we get.

$m_{\text{cone}} g h = m_{\text{water displaced}} g \dfrac{h}{4}$

$\dfrac{m_{\text{cone}}}{m_{\text{water displaced}}} = \dfrac{\rho_{\text{cone}}}{\rho_{\text{water}}} = \dfrac{1}{4} =0.25$

Let us call:

• $m$ as the cone mass
• $R$ as the cone base radius
• $H$ as the cone height
• $y$ as the portion of the cone height which is submerged
• $r$ as the radius of the base of the cone sector which is submerged
• $v$ as the cone vertical velocity downwards
• $\rho_{H2O}$ as the water density
• $\rho_{cone}$ as the cone density

First, by triangle similarity:

$\color{#20A900} \boxed{ \large \displaystyle r = \frac{R}{H} y }$

Writing the differential equation of the cone:

$\large \displaystyle m \frac{dv}{dt} = mg - \rho_{H2O} g \frac{\pi r^2 y}{3}$

$\large \displaystyle \frac{dv}{dt} = g - \rho_{H2O} g \frac{\pi R^2}{3mH^2} y^3$

Let us define:

$\color{#D61F06} \boxed{\large \displaystyle k \triangleq \rho_{H2O} g \frac{\pi R^2}{3mH^2} }$

$\large \displaystyle \frac{dv}{dt} = g - k y^3$

$\large \displaystyle \frac{dv}{dy} \frac{dy}{dt} = g - k y^3$

$\large \displaystyle \frac{dv}{dy} v = g - k y^3$

$\large \displaystyle v dv = \left( g - k y^3 \right) dy$

Since at $y=0$ , $v$ is also $0$ (cone begins at rest):

$\large \displaystyle \int_0^v v dv = \int_0^y \left( g - k y^3 \right) dy$

$\color{#20A900} \boxed{ \large \displaystyle \frac{v^2}{2} = gy - \frac{ky^4}{4} }$

At $y = H$ (instant in which cone fully submerged), $v$ is also equal to $0$ :

$\large \displaystyle 0 = gH - \frac{kH^4}{4}$

$\large \displaystyle H^3 = \frac{4g}{k}$

$\large \displaystyle H^3 = 4g \frac{3mH^2}{\pi R^2 g \rho_{H2O} }$

$\large \displaystyle \frac{m}{\frac{\pi R^2 H}{3}} = \frac{\rho_{H2O}}{4}$

$\large \displaystyle \rho_{cone} = \frac{\rho_{H2O}}{4}$

$\color{#3D99F6} \boxed{ \large \displaystyle \frac{\rho_{cone}}{\rho_{H2O}} = \frac14 = 0.25}$

applying thekinetic energy theorem

At the place where the cone ends up, there is initially water. That water is pushed toward the surface. Thus the water gains potential energy in the amount $\Delta U_w = mgy = V \rho_w \cdot g \cdot \tfrac14 h = (\tfrac14 \rho_w) Vgh.$ The factor $\tfrac14$ is due to the fact that the center of mass of a solid cone lies at $\tfrac34$ from its vertex.*

Meanwhile, the cone loses potential energy because its center of mass drops the cone's height. Thus $\Delta U_c = -mgh = \rho_c Vgh.$ Since the mechanical energy is conserved and no kinetic energy was gained or lost, we have $\Delta U_w + \Delta U_c = 0 \ \ \ \therefore \ \ \ \tfrac14\rho_w = \rho_c,$ showing that the cone's density is $\boxed{0.25}$ times the density of the water.

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• Proof: let the radius of the cone by $R$ and the height $h$ . Divide the cone into infinitesimal disks; the disk at height $y$ will have radius $r(y) = Ry/h$ . Therefore $\langle y\rangle = \frac{\int_0^h y r^2 dy}{\int_0^h r^2 dy} \\ = \frac{\int_0^h y (Ry/h)^2 dy}{\int_0^h (Ry/h)^2 dy} \\ = \frac{\int_0^h y^3 dy}{\int_0^h y^2 dy} = \frac{\tfrac14 h^4}{\tfrac13 h^3} = \frac 34 h.$