A solid right circular cone of uniform mass density is initially at rest above a body of water, so that its vertex is just touching the water's surface with its axis of symmetry along the vertical.

Now, the cone falls into the water, and has zero speed at the instant it becomes fully submerged. What is the ratio of the density of the cone to the density of the water? Submit your answer to 2 decimal places.

Details and Assumptions:

There is an ambient downward gravitational field.
Assume that the buoyant force is the only force exerted by the water on the cone.

R = radius of cone

height of cone = h
draft of cone = x =distance below surface

area of base = pi R^2
area of water plane = pi r^2

x/h = r/R geometry

weight of cone = rho g(1/3)pi R^2 h
weight of water displaced = 1000 g(1/3)pi r^2 x

force up = weight of water displaced - weight of cone = -m d^2x/dt^2

= -[rho (1/3) pi R^2 h]d^2x/dt^2
so
1000 g(1/3)pi r^2 x-rho g(1/3)pi R^2 h=-[rho (1/3) pi R^2 h]d^2x/dt^2

1000 g r^2 x-rho g R^2 h=-[rho R^2 h]d^2x/dt^2

[rho R^2 h]d^2x/dt^2 + 1000 g r^2 x=rho g R^2 h

r and x change with time, everything else is constant
however by geometry r = c x = (R/h)x
so
[rho R^2 h]d^2x/dt^2 + 1000 g (R/h)^2 x^3=rho g R^2 h
call rho/1000 = density ratio we want = p
[p R^2 h]d^2x/dt^2 + g (R/h)^2 x^3= p g R^2 h

C1 d^2x/dt^2 + C2 x^3 = C3
at t = 0
x = 0 and dx/dt = 0, d^2x/dt^2 = g

at x = h, dx/dt = 0

I do not know of a closed form solution and would have to do it numerically.

I need a number----a ratio. I can't understand this.

To solve this problem, we need to consider the balance of forces acting on the cone when it is fully submerged in water.

First, let's analyze the forces acting on the cone. The gravitational force acting on the cone can be represented as the weight of the cone, which depends on its mass and the acceleration due to gravity. Since the problem states that the cone has uniform mass density, we can assume that the mass is proportional to the volume of the cone.

The buoyant force is the force exerted by the water on the cone. According to Archimedes' principle, the buoyant force is equal to the weight of the fluid displaced by the immersed object. In this case, the fluid displaced is water, and the buoyant force acts in the upward direction.

When the cone is fully submerged, the buoyant force is greater than or equal to the weight of the cone. However, since the cone has zero speed at the instant it becomes fully submerged, we can conclude that the buoyant force is equal to the weight of the cone.

Let's derive an expression for the weight of the cone. The weight of an object is given by its mass multiplied by the acceleration due to gravity (W = mg). In this case, the mass of the cone is proportional to its volume, which can be written as V = (1/3)πr²h, where r is the base radius of the cone and h is its height.

If the density of the cone is denoted as ρ_c and the density of the water is denoted as ρ_w, then the mass of the cone can be written as m = ρ_cV. Thus, the weight of the cone is W = ρ_cVg, where g is the acceleration due to gravity.

Equating the weight of the cone to the buoyant force, we have:

ρ_cVg = ρ_wVg

Canceling out the V and g terms, we find:

ρ_c = ρ_w

Therefore, the ratio of the density of the cone to the density of the water is 1.00.

So, the answer to the question is 1.00.