What is the drag coefficient of the cue ball

AI Thread Summary
The discussion focuses on calculating the drag coefficient (Cd) of a cue ball that reaches a terminal velocity of 1.55 m/s in water. The user correctly identifies the formula for Cd and calculates the sphere's area using the diameter, but struggles with determining the drag force. They note that at terminal velocity, the gravitational force equals the drag force, leading to confusion about the correct conversion of area to meters. The user emphasizes the importance of using the cross-sectional area (πr²) for accurate calculations. Ultimately, they express uncertainty about their calculations and seek clarification on the area conversion.
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Homework Statement



hile playing pool on a cruise ship, you accidentally hit the cue ball over the railing and it falls into the ocean below. If the cue ball (a sphere of diameter = 57.15 mm, m = 160 g) reaches a terminal velocity of 1.55 m/s in the water (), what is the drag coefficient of the cue ball in water?

Homework Equations



Cd= Fd/ 1/2 pv^2A
Cd= coef of drag
Fd= drag force
p = mass density of the fluid
V= speed
A = area of the sphere


The Attempt at a Solution



Ok from the above equation i am able to determine all of the items except Drag force

I got the area from pi * d^2 for the sphere (pi*55.15^2= 10260.82)
adn the rest you can fill in

I am stuck trying to figure out the drag force
 
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At terminal velocity the gravity force downwards is equal to the drag force upward.
 


even when in water?
 


so M * G ---) 1.568 = drag force?
 


air is less dense than water is why i ask
 


also converting the Area to m 10.26m is that correct?
 


if my numbers are correct i arrange the problem

1.56 / (1/2 * 1000*1.55^2 * 10.26)

doing the math Coefficient of drag = 1.26e-4?
 


i feel like I am making a mistake in converting the reference area since the object is falling
 


It's the cross sectional area, so pi r^2 is correct, it would be safest to work in metres
 
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