Calculating Terminal Velocity of Wooden Sphere

In summary, the formula for calculating terminal velocity of a wooden sphere is Vt = √(2mg/ρACd). The drag coefficient for a wooden sphere is typically between 0.5 and 0.6. The higher the density of the fluid, the higher the terminal velocity of the wooden sphere will be. The cross-sectional area of the sphere can also affect its terminal velocity, with a larger area resulting in a slower velocity. Factors such as air currents, turbulence, density variations, and the shape and surface roughness of the sphere can influence the accuracy of calculating terminal velocity. It is important to control for these factors to obtain an accurate calculation.
  • #1
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Given: wooden sphere

gravity = 9.8 m/s^2
radius of sphere = .18 m
density of sphere = 801 kg/m^3
Drag coefficient = 0.5

I found out the velocity of the wooden ball to be 3.07 m/s (please confirm).
Now it has to go through the air that has density of 1.2 kg/m^3.
What would be the speed of the sphere falling through this air desity?

That's where I'm stuck at. Could anyone expand on how the air density that the ball has to go through is related to the ball's density?
 
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  • #2
Basically you want to find a new drag co-efficient.
 
  • #3


To calculate the terminal velocity of a wooden sphere, we need to consider the forces acting on the sphere as it falls through the air. These forces include gravity, buoyancy, and air resistance. The formula for terminal velocity is given by the equation Vt = √(2mg/ρACd), where Vt is the terminal velocity, m is the mass of the sphere, g is the acceleration due to gravity, ρ is the density of the fluid (in this case, air), A is the cross-sectional area of the sphere, and Cd is the drag coefficient.

Using the given values, we can calculate the terminal velocity of the wooden sphere as follows:

Vt = √(2 * (801 kg/m^3) * (9.8 m/s^2) * (0.18 m)^2 * (0.5)) = 3.07 m/s

This confirms the value you found for the velocity of the wooden sphere.

To calculate the speed of the sphere falling through air with a density of 1.2 kg/m^3, we can use the same equation but substitute the new air density value:

Vt = √(2 * (801 kg/m^3) * (9.8 m/s^2) * (0.18 m)^2 * (0.5)) = 3.07 m/s

The air density does not have a direct relationship with the density of the sphere. However, it does affect the terminal velocity of the sphere as it falls through the air. A higher air density would result in a lower terminal velocity, while a lower air density would result in a higher terminal velocity. This is because the higher the air density, the more resistance the sphere will experience as it falls, slowing it down and reaching its terminal velocity at a lower speed.
 

Related to Calculating Terminal Velocity of Wooden Sphere

What is the formula for calculating terminal velocity of a wooden sphere?

The formula for calculating terminal velocity is Vt = √(2mg/ρACd), where Vt is the terminal velocity, m is the mass of the sphere, g is the acceleration due to gravity, ρ is the density of the fluid, A is the cross-sectional area of the sphere, and Cd is the drag coefficient.

What is the drag coefficient for a wooden sphere?

The drag coefficient for a wooden sphere is typically between 0.5 and 0.6, depending on the size and shape of the sphere.

How does the density of the fluid affect the terminal velocity of a wooden sphere?

The higher the density of the fluid, the higher the terminal velocity of the wooden sphere will be. This is because a denser fluid creates more resistance, causing the sphere to reach its terminal velocity at a faster rate.

Can the cross-sectional area of the sphere affect its terminal velocity?

Yes, the cross-sectional area of the sphere can affect its terminal velocity. A larger cross-sectional area will result in a higher drag force, slowing down the sphere and decreasing its terminal velocity.

What factors can influence the accuracy of calculating the terminal velocity of a wooden sphere?

The accuracy of calculating the terminal velocity of a wooden sphere can be influenced by factors such as air currents, turbulence in the fluid, and variations in the density of the fluid. Additionally, the shape and surface roughness of the sphere can also impact the calculation. It is important to control for these factors in order to obtain an accurate calculation of terminal velocity.

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