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I am not quit sure what is your meaning, but your notation somehow didn't make sense to me, in the $\dot{\vec{r}}^2 \cdot \hat{e_r} $. In the expansion, there is a vector square which is a vector? and without the cross term? Then had it inner product with the 45 degree vectors. This algebraic process makes no sense at all. Therefore, I try to image the ...


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The problem arises because $v$ is not being treated as a vector. With $v$ a vector $\vec{v}$ and $g$ a vector $\vec{g}$, the equation could be written as $$ m\frac{\mathrm d\vec{v}}{\mathrm dt}=\frac{b\vec{v}(\vec{v}\cdot \vec{v})}{(\vec{v}\cdot \vec{v})^{1/2}}-m \vec{g}.$$ It's less messy to just split the problem into two parts: the upward part of the ...


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HINT: Your expression for $y(t)$ implies that $y(0) = (g+\alpha v_0)/\alpha^2$. Is this what you intended?


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To Niels' answer I will add a couple points. At low speeds engine friction is the dominant mechanism for energy loss. At higher speeds it is atmospheric drag. Air has mass. For a car to pass, air must be pushed out of the way. It must be accelerated in front of the car, flow around the car, fill in the void behind, and come to rest after the car has passed. ...


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Mass has nothing to do with aerodynamic drag, except in the trivial sense that larger vehicles are heavier and have more drag- but not because they are heavier, only because they are larger. When cars are made lighter, they become more economical because it requires less work to accelerate them to speed and less work to drive them up hills. When cars are ...


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An infinitesimal amount of work is given by: $$\text{d}W_{loss}=F_D\text{d}x\tag{1}$$ The drag force is given by: $$F_D=\frac12 \rho A C_Dv^2\tag{2}$$ Now find a relation between $\text{d}x$ and $\text{d}v$: $$v=\frac{\text{d}x}{\text{d}t}=\frac{\text{d}x}{\text{d}v}\frac{\text{d}v}{\text{d}t}=a\frac{\text{d}x}{\text{d}v}$$ Because $\frac{\text{d}v}{\text{d}...


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It's similar question if you would ask - Why kinetic energy is defined as $E_k = 1/2~m v^2$. The answer is below. Elementary work is : $$ \begin{align} dW &= F \cdot dr \\&= F\frac {dr}{dt} dt \\&=Fv~dt \\&=m\frac {dv}{dt}v~dt \\&=mv~dv \end{align} $$ Now integrate both sides : $$ \int dW = m \int v~dv $$ Which gives : $$W=E_k=1/2~mv^2$...


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I believe it was initially shown that the drag force is proportional (under certain conditions) to the dynamic pressure $q$ and the frontal (projected) area $A$. The proportionality constant of this relationship is what we call a drag coefficient $C_D$. $$F_D\propto qA\quad\Leftrightarrow\quad F_D=C_DqA$$ Nice and neat. It turns out that this paramter $C_D$ ...


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Drag coefficient formula $$ C_d = \frac{2F_d}{p u^2A} $$ does not restrict length of the body. As you increase length, skin friction will drive your $C_d$ to infinity. There is no shape with highest $C_d$, but you can get any unreasonable value by increasing length of the body. Of course it will be drag coefficient for infinitely high Reynolds number :-P


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Does a drag coefficient of exactly one correspond to any significant shape? Cd = 1 and Cd = F / (1/2 * A * v^2 * ro) <=> F = (1/2) A * v^2 * ro That can occur with different combinations of: fluid, area, speed and drag. So it does not imply a particular shape. Is Cd a result of the arbitrary choice of the definitions of the SI units? Note that Cd has ...


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I am not sure what exactly you want to do but you can study motion in terms of translation of center of mass and rotation about center of mass in cases of combined translation and rotation.


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You have not factored in mass: you state in your question that the force of drag is 0.24*v^2. In your code, however, you use this as an acceleration, applying it alongside the 9.8 from gravity. This only works if the mass is one kilogram. Modifying your function to take mass as an argument: def jump(mass): t = 0 dt = 0.01 v = 0 x = 728 #...


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All of the reasons you say are correct. The bullet bumps into air molecules, speeding them up. This slows the bullet. The air molecules are bumped forward, crowding into air already present. More molecules are now in the region than would normally be. The density goes up. Since that region has extra molecules bumping into a neighboring region, the force ...


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The problem is you are confusing definitions and the way you are connecting dependencies. Aerodynamic drag is dependent on speed, and speed can be dependent on many things. It seems to me that the falling object would exert a force on each particle in the air that is causing the drag force, and that this force would depend on the mass of the object? I am ...


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In fluid dynamics, drag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid.[1] This can exist between two fluid layers (or surfaces) or a fluid and a solid surface. Unlike other ...


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yes molecules are a little frenzy / vibrate, but their average velocity is 0 (for example the air in a room doesn't move from a macroscopic view). Likely the author is assuming that the falling object is dragging with him(they don't bounce) the air molecules , that start with an average velocity v = 0 and then accelerate at the same speed of the falling ...


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Tho solve the quadratically damped harmonic oscillator, you need two equations: $$ \ddot y + \frac b m \dot y^2 + \frac D m = 0\ \ \ \ \dot y < 0 $$ $$ \ddot y - \frac b m \dot y^2 + \frac D m = 0\ \ \ \ \dot y > 0 $$ so that the damping force is always opposing the velocity. A standard brute force method is to use a power series: $$ y(t) = \sum_{n=0}^{...


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There are (as far as I know) no closed form solutions, and certainly none tht are simple, to this differential equation. As @G.Smith points out, there is no reason to assume every differential equation has a closed form solution, and certainly not a non-linear equation like this one. You may obtain a solution using something like WolframAlpha, or integrate ...


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Yes, we can. Projectile motion can be understood, as the name suggests, the motion experienced by a body projected into air, whose path is usually curved. When we talk about kinematics, we do not consider the nature of object, so whenever a statement reads, "a body is projected" we assume it to be a point mass, which is theoretically dimensionless....


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Yes I think we can. Source: University Physics - Sears & Zemansky For a clearer explanation we can also go for the origin of the word 'projectile'. "body projected or impelled forward by force," 1660s, from Modern Latin projectilis. Usually used for a missile fired from a gun. Hope this Helps.


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