From c95f5efd9122c5308c88a6596186f0b41c6dcc8f Mon Sep 17 00:00:00 2001 From: Krishna Date: Thu, 26 Sep 2024 20:38:18 -0500 Subject: [PATCH] a bit more before the test --- content/physics/Forces.md | 40 ++++++++++++++++++++++++++++++++++++++- 1 file changed, 39 insertions(+), 1 deletion(-) diff --git a/content/physics/Forces.md b/content/physics/Forces.md index fb9ba45..1a2e7ec 100644 --- a/content/physics/Forces.md +++ b/content/physics/Forces.md @@ -201,6 +201,26 @@ $$ \frac{V_{\mathrm{fl}}}{V_{\mathrm{obj}}}=\frac{\rho_\mathrm{obj}}{\rho_\mathrm{fl}} $$ +## Stokes' law + +**Stokes' law** is an empirical law for the frictional forces, or drag force, exerted on spherical objects with very small Reynolds numbers in a *viscous* fluid. + +$$ +\mathrm{Re}=\frac{uL}{\nu}=\frac{\rho uL}{\eta} +$$ + +- $\rho$ is the density of the fluid +- $u$ is the flow speed +- $L$ is the *characteristic length* or reference scale for the rest of the number. For a flat plate, it is the length of the plate (the distance the fluid may travel unrestricted). For a hollow cylinder or pipe, it is the diameter of the pipe. For a sphere, it is its diameter +- $\eta$ (pronounced *eta*) is the *dynamic viscosity* of the fluid +- **Stokes' law only works when $\mathrm{Re}<1$** +- $\nu$ is the *kinematic viscosity* of the fluid. + +Stokes' law also only functions for spheres. It's usually written as, with all of the aforementioned variables translating perfectly: +$$ +||\vec{F}_d||=6\pi\eta Rv +$$ + # Circular motion An object that moves in a circle at a constant speed $||\vec{v}||$ experiences **uniform circular motion**. The most important part of that sentence is the constant speed part; the measurement $||\vec{v}||$ is reference to the center of the circle. @@ -211,7 +231,8 @@ As seen above, a velocity for every point on the path of the circle's motion exi $$ a_R=\lim_{\Delta t\to0}{\frac{\Delta \vec{v}}{\Delta t}}=\frac{v^2}{r} $$ -$a_R$ is the radial acceleration, otherwise known as the centripetal acceleration. It's easier not to consider centripetal acceleration as a result of the imaginary centripetal force if instead thought of as acceleration experienced by the object radially, or on the radius of motion. +$a_R$ is the radial acceleration, otherwise known as the centripetal acceleration. It's easier not to consider centripetal acceleration as a result of the imaginary centripetal force if instead thought of as acceleration experienced by the object radially, or on the radius of motion. A Reynolds number is calculated like so (**you don't need to know this**): + $$ T=\frac{1}{f} $$ @@ -234,8 +255,25 @@ It's important to note that the acceleration and velocity vectors at an instant > $v=\frac{2\pi r}{T}$. \ > Since the ball makes 2 revolutions in a second, the period $T$ must be $\tfrac{1}{2}$. $v=\frac{2\pi(0.6\mathrm{m})}{0.5}=7.54$, the unit of that output being meters/second. Substituting that value for the speed $v$ in the acceleration formula, acceleration is $94.7 \mathrm{m}/\mathrm{s}^2$ +# Momentum +> [!NOTE]+ +> Calculus ahead. If you're not there, move on or start back at the top. It hardly matters +The momentum of a single particle is written as: +$$ +\vec{p}=m\vec{v} +$$ +This is a vector quantity, since velocity is a vector, so a constant multiplied by a vector is still a vector of the same dimensions. Momentum is also incredibly related to force. In differential form, as it was first written, *Newton's $2^\mathrm{nd}$ law* states: +$$ +F=\frac{\mathrm{d}p}{\mathrm{d}t} +$$ + +Additionally - if the net force experienced by an object at any time $t$ is expressed as $F(t)$: +$$ +\Delta p = J = \int_{t_1}^{t_2}{F(t)\mathrm{d}t} +$$ +$j$ is also something called the *impulse.*