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.

Question

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.

Answer 1

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.

Answer 2

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