A thin spherical shell of mass 4.00 kg and diameter 0.200 m is filled with helium (density ρ = 0.180 kg/m3). It is then released from rest on the bottom of a pool of water that is 4.00 m deep. (a) Neglecting frictional effects, show that the shell rises with constant acceleration and determine the value of that acceleration. 5pts (b) How long will it take for the top of the shell to reach the water surface?
To determine the value of the constant acceleration of the shell, we can use Newton's second law of motion, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration.
(a) First, let's find the buoyant force acting on the shell when it is submerged in water.
The buoyant force (F_b) is given by the formula:
F_b = ρ_fluid * V_displaced * g
where ρ_fluid is the density of the fluid (water in this case), V_displaced is the volume of the fluid displaced by the object, and g is the acceleration due to gravity.
The volume of the fluid displaced by the spherical shell is equal to the volume of the shell itself. The volume of a sphere is given by the formula:
V_shell = (4/3) * π * r^3
where r is the radius of the sphere. Since the diameter is given, we can find the radius:
r = diameter / 2 = 0.200 m / 2 = 0.100 m
Now, let's calculate the volume of the shell:
V_shell = (4/3) * π * (0.100 m)^3 ≈ 0.00419 m^3
Next, let's find the weight of the shell when submerged in water. The weight (F_g) is given by the formula:
F_g = m * g
where m is the mass of the shell and g is the acceleration due to gravity.
F_g = 4.00 kg * 9.8 m/s^2 ≈ 39.2 N
Since the shell is in equilibrium when submerged, the buoyant force is equal to the weight of the shell:
F_b = F_g
ρ_fluid * V_shell * g = m * g
ρ_fluid * V_shell = m
Now we can find the mass of the fluid displaced by the shell:
m_fluid_displaced = ρ_fluid * V_shell
m_fluid_displaced = 0.180 kg/m^3 * 0.00419 m^3 ≈ 0.000755 kg
The net force acting on the shell when submerged is the difference between the buoyant force and the weight of the fluid displaced:
F_net = F_b - F_g
F_net = m_fluid_displaced * g
Now, we can find the acceleration of the shell using Newton's second law:
F_net = m_shell * a
m_fluid_displaced * g = m_shell * a
Substituting the known values:
0.000755 kg * 9.8 m/s^2 = 4.00 kg * a
Simplifying, we find:
0.007399 N = 4.00 kg * a
Dividing both sides by 4.00 kg, we get:
a ≈ 0.0018497 m/s^2
So, the value of the constant acceleration of the shell is approximately 0.0018497 m/s^2.
(b) To determine the time it will take for the top of the shell to reach the water surface, we can use the equation of motion:
s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity (which is 0 since the shell is released from rest), a is the acceleration, and t is the time.
In this case, the displacement is the depth of the water (4.00 m), and we need to find the time.
s = 4.00 m
u = 0 m/s
a = 0.0018497 m/s^2
Plugging in the values into the equation, we have:
4.00 = 0 + (1/2)(0.0018497)t^2
Simplifying further, we find:
8.00 = 0.00092485t^2
Dividing both sides by 0.00092485, we get:
t^2 ≈ 8663.40
Taking the square root of both sides, we find:
t ≈ 93.11 seconds
So, it will take approximately 93.11 seconds for the top of the shell to reach the water surface.