Estimate the terminal speed of a wooden sphere (density 0.770 g/cm3) falling through air, if its radius is 9.00 cm and its drag coefficient is 0.500. (The density of air is 1.20 kg/m3.)
(b) From what height would a freely falling object reach this speed in the absence of air resistance?
To solve this problem, we can use the following expression for terminal velocity:
v = √[(2 * m * g) / (ρ * A * C)]
where v is the terminal velocity, m is the mass of the object, g is the acceleration due to gravity (approximately 9.81 m/s²), ρ is the density of the fluid (air) in kg/m³, A is the cross-sectional area of the object, and C is the drag coefficient.
First, we need to find the mass of the wooden sphere. The volume of a sphere is given by the formula:
V = (4 / 3) * π * r³
where r is the radius of the sphere.
For a sphere with a radius of 9.00 cm, the volume is:
V = (4 / 3) * π * (0.09 m)³ ≈ 3.05 x 10⁻³ m³
Next, we need to convert the density of wood from g/cm³ to kg/m³:
density_wood = 0.770 g/cm³ * (1 kg / 1000 g) * (100 cm / 1 m)³ = 770 kg/m³
Now we can find the mass of the sphere:
mass = density_wood * volume ≈ 770 kg/m³ * 3.05 x 10⁻³ m³ ≈ 2.35 kg
The cross-sectional area of a sphere is given by the formula:
A = π * r²
For a sphere with a radius of 9.00 cm, the area is:
A = π * (0.09 m)² ≈ 0.0254 m²
Now we can plug the values into the terminal velocity equation:
v = √[(2 * 2.35 kg * 9.81 m/s²) / (1.20 kg/m³ * 0.0254 m² * 0.500)] ≈ 47.9 m/s
The terminal speed of the wooden sphere falling through air is approximately 47.9 m/s.
(b) To find the height from which an object would reach this speed in the absence of air resistance, we can use the following equation:
v² = 2 * g * h
where v is the terminal velocity, g is the acceleration due to gravity, and h is the height.
Rearrange the equation to solve for h:
h = v² / (2 * g) ≈ (47.9 m/s)² / (2 * 9.81 m/s²) ≈ 116.2 m
In the absence of air resistance, a freely falling object would reach the terminal speed of 47.9 m/s from a height of approximately 116.2 m.
To estimate the terminal speed of the wooden sphere, we can use the following equation:
Terminal speed = (2 * m * g) / (ρ * A * Cd)
where:
- m is the mass of the sphere
- g is the acceleration due to gravity
- ρ is the density of air
- A is the cross-sectional area of the sphere
- Cd is the drag coefficient
First, we need to find the mass of the sphere. The mass can be calculated using the formula:
m = (4/3) * π * r^3 * ρ
where:
- r is the radius of the sphere
- ρ is the density of the sphere
Given that the radius of the sphere is 9.00 cm and the density is 0.770 g/cm^3, we can convert the units to get:
r = 9.00 cm = 0.09 m
ρ = 0.770 g/cm^3 = 770 kg/m^3
Using these values, we can calculate the mass of the sphere:
m = (4/3) * π * (0.09)^3 * 770
Next, we can calculate the cross-sectional area of the sphere:
A = π * r^2
Substituting the values, we get:
A = π * (0.09)^2
Lastly, we can substitute the known values into the terminal speed equation:
Terminal speed = (2 * m * g) / (ρ * A * Cd)
Given that the density of air is 1.20 kg/m^3 and the drag coefficient is 0.500, we can substitute these values and solve for the terminal speed.
To calculate the height from which a freely falling object would reach this speed in the absence of air resistance, we can use the equation for gravitational potential energy:
Potential energy = m * g * h
where:
- m is the mass of the object
- g is the acceleration due to gravity
- h is the height from which the object falls
Equating the potential energy to the kinetic energy at terminal speed, we can solve for the height:
(m * g * h) = (1/2) * (m * v^2)
where:
- m is the mass of the object
- g is the acceleration due to gravity
- h is the height from which the object falls
- v is the terminal speed
We can substitute the known values, including the mass of the sphere, acceleration due to gravity, and terminal speed, and solve for h to find the height at which the object would reach terminal speed in the absence of air resistance.
To estimate the terminal speed of the wooden sphere, we can use the concept of terminal velocity. Terminal velocity is the maximum velocity a falling object can reach when the drag force equals the gravitational force pulling it downwards.
To calculate the terminal speed, we need to find the drag force acting on the sphere. The drag force can be calculated using the formula:
F_drag = (1/2) * ρ * C * A * v^2
Where:
- F_drag is the drag force
- ρ is the density of the air
- C is the drag coefficient
- A is the cross-sectional area of the sphere
- v is the velocity of the sphere
First, let's convert the density of the air from kg/m3 to g/cm3:
Density of air = 1.20 kg/m3 = 1.20 * 1000 g/m3 = 1200 g/m3 = 0.0012 g/cm3
Next, we need to calculate the cross-sectional area of the sphere. The cross-sectional area of a sphere can be calculated using the formula:
A = π * r^2
Where:
- A is the cross-sectional area
- π is a constant (approximately 3.14)
- r is the radius of the sphere
A = 3.14 * (9.00 cm)^2 = 254.34 cm^2
Now we can calculate the drag force:
F_drag = (1/2) * 0.0012 g/cm3 * 0.500 * 254.34 cm^2 * v^2
The gravitational force acting on the wooden sphere can be calculated using the formula:
F_gravity = m * g
Where:
- F_gravity is the gravitational force
- m is the mass of the sphere
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
The mass of the wooden sphere can be calculated using its density and volume:
m = ρ * V
Where:
- m is the mass of the sphere
- ρ is the density of the sphere
- V is the volume of the sphere
The volume of a sphere can be calculated using the formula:
V = (4/3) * π * r^3
Now we can calculate the mass and gravitational force:
V = (4/3) * 3.14 * (9.00 cm)^3 = 3053.628 cm^3
m = 0.770 g/cm3 * 3053.628 cm^3 = 2351.839 g
F_gravity = 2351.839 g * 9.8 m/s^2 = 23044.64 g cm/s^2
Finally, we can equate the drag force to the gravitational force to find the terminal velocity:
(1/2) * 0.0012 g/cm3 * 0.500 * 254.34 cm^2 * v^2 = 23044.64 g cm/s^2
Simplifying the equation, we have:
v^2 = (2 * 23044.64 g cm/s^2) / (0.0012 g/cm3 * 0.500 * 254.34 cm^2)
v^2 = 772891666.7 cm^2/s^2
v = sqrt(772891666.7) cm/s ≈ 17514.18 cm/s
Therefore, the estimated terminal speed of the wooden sphere falling through air is approximately 17514.18 cm/s.
Moving on to the second part of the question, we need to find the height from which a freely falling object would reach this speed in the absence of air resistance.
In the absence of air resistance, the freely falling object would only be subject to the gravitational force, and it would continue to accelerate as it falls. The velocity of a freely falling object can be calculated using the equation:
v = sqrt(2 * g * h)
Where:
- v is the velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- h is the height from which it falls
To find the height, we can rearrange the equation and solve for h:
h = (v^2) / (2 * g)
h = (17514.18 cm/s)^2 / (2 * 9.8 m/s^2)
h ≈ 151637.08 cm ≈ 1516.37 m
Therefore, a freely falling object would reach a speed of approximately 17514.18 cm/s (or 175.14 m/s) when falling from a height of approximately 1516.37 meters in the absence of air resistance.