Latest revision as of 17:45, 8 September 2014

Air Resistance (A Damping force that depends on velocity (F(v)))

Newton's second law

Consider the impact on solving Newton's second law when there is an external Force that is velocity dependent

Frictional forces tend to be proportional to a fixed power of velocity

Linear air resistance (n=1) arises from the viscous drag of the medium through which the object is falling.

Quadratic air resistance (n=2) arises from the objects continual collision with the medium that causes the elements in the medium to accelerate.

Air resistance for rain drops or ball bearings in oil tends to be more linear while canon balls and people falling through the air tends to be more quadratic.

Example: A Sphere moving through air at STP

Linear
Quadratic

Thus in order for the above ratio to be near unity, D is very small like a raindrop and has a small velocity < 1 m/s.

Linear Air Resistance

Forest_UCM_PnCP_LinAirRes

quadratic friction

Forest_UCM_PnCP_QuadAirRes

Another block on incline example

Forest_UCM_NLM_BlockOnIncline

Projecile Motion

Forest_UCM_PnCP_ProjMotion

Charged Particle in uniform B-Field

Forest_UCM_PnCP_QubUniBfield

Forest_Ugrad_ClassicalMechanics

@@ Line 17: / Line 17: @@
 Linear air resistance (n=1) arises from the viscous drag of the medium through which the object is falling.
 Quadratic air resistance (n=2) arises from the objects continual collision with the medium that causes the elements in the medium to accelerate.
-==Linear Air Resistance==
-If <math>n</math> is unity then the velocity is exponentially approaching zero.
+Air resistance for rain drops or ball bearings in oil tends to be more linear while canon balls and people falling through the air tends to be more quadratic.
-:<math>F(v) = -bv</math>: negative sign indicates a retarding force and <math>b</math> is a proportionality constant
-:<math>\sum \vec {F}_{ext} = -bv = m \frac{dv}{dt}</math>
+Example:  A Sphere moving through air at STP
-: <math>\Rightarrow  \int_{v_i}^{v_f} \frac{dv}{v} = \int_{t_i}^{t_f} \frac{-b}{m}dt</math>
-:<math>\ln\frac{v_f}{v_i} = \frac{-b}{m}t</math>; <math>t_i \equiv 0</math>
-: <math>v_f = v_i e^{-\frac{b}{m}t}</math>
-The displacement is given by
+;Linear:
+:<math>F_f = bv = \beta D v = \left ( 1.6 \times 10^{-4} \frac{N \cdot s}{m^2}\right ) D v</math>
+;Quadratic:
+:<math>F_f = cv^2 = \gamma D^2 v^2 = \left ( 2.5 \times 10^{-1} \frac{N \cdot s^2}{m^4}\right ) D^2 v^2</math>
-:<math>x = \int_0^t v_i e^{-\frac{b}{m}t} dt</math>
+:<math>\frac{F_f(\mbox{Quadratic})}{F_f(\mbox{linear})} = \left (1.6 \times 10 ^{2} \frac{s}{m^2} \right ) D v</math>
-:: <math>=  \left . v_i \left ( \frac {e^{-\frac{b}{m}t}}{-\frac{b}{m}} \right ) \right |_0^t</math>
-:: <math>=  \left . v_i \left ( -\frac{m}{b} e^{-\frac{b}{m}t} \right ) \right |_0^t</math>
-:: <math>=  \left . v_i \left ( \frac{m}{b} e^{-\frac{b}{m}t} \right ) \right |_t^0</math>
-:: <math>=  v_i \left ( \frac{m}{b} e^{-\frac{b}{m}0} -\frac{m}{b} e^{-\frac{b}{m}t} \right ) </math>
-:: <math>=  \frac{m}{b} v_i \left ( 1-e^{-\frac{b}{m}t} \right )</math>
-==Example: falling object with air friction==
+Thus in order for the above ratio to be near unity,<math> Dv < 10^{-3} \Rightarrow</math> D is very small like a raindrop and has a small velocity < 1 m/s.
-Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity squared
+==Linear Air Resistance==
-:<math>\sum \vec{F}_{ext} = mg -bv^2 = m \frac{dv}{dt}</math>
-Find the fall distance
-Here is a trick to convert the integral over time to one over distance so you don't need to integrate twice as inthe previous example
-:<math>\frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v\frac{dv}{dy}</math>
-The integral becomes
-:<math>mg -bv^2 = m v\frac{dv}{dy}</math>
+[[Forest_UCM_PnCP_LinAirRes]]
-:<math>\int_{y_i}^{y_f} dy  = \int_{v_i}^{v_f} m \frac{dv}{\left ( mg -bv^2 \right ) }</math>
-:<math>y  = \int_{v_i}^{v_f} \frac{dv}{\left ( g -\frac{b}{m}v^2 \right ) }</math>
+==quadratic friction==
-let <math>u = g -\frac{b}{m}v^2</math>
-then <math>du = -2\frac{b}{m}v dv</math>
+[[Forest_UCM_PnCP_QuadAirRes]]
-:<math>y  =\int_{v_i}^{v_f} \frac{-m}{2b} \frac{du}{u } = \frac{b}{m} \int_{v_f}^{v_i} \ln {g -\frac{b}{m}v^2} =</math>
-:<math>y  =\frac{m}{2b}  \ln \left ( \frac {g -\frac{b}{m}v_i^2}{g -\frac{b}{m}v_f^2} \right ) </math>
 ==Another block on incline example==
@@ Line 69: / Line 47: @@
 [[Forest_UCM_NLM_BlockOnIncline]]
-=Charged Particle in uniform B-Field=
+=Projecile Motion=
-Consider a charged particle moving the x-y plane in the presence of a uniform magnetic field with field lines in the z-dierection.
-:<math>\vec{v} = v_x \hat i + v_y \hat j</math>
+[[Forest_UCM_PnCP_ProjMotion]]
-:<math>\vec{B} = B \hat k</math>
+=Charged Particle in uniform B-Field=
-;Lorentz Force
+[[Forest_UCM_PnCP_QubUniBfield]]
-:<math>\vec{F} = q \vec{E} + q\vec{v} \times \vec{B}</math>
-;Note: the work done by a magnetic field is zero if the particle's kinetic energy (mass and velocity) don't change.
-:<math>W = \Delta K.E.</math>
-No work is done on a charged particle forced to move in a fixed circular orbit by a magnetic field (cyclotron)
-:<math>\vec{F} = m \vec{a} = q \vec{v} \times \vec{B} =  q\left ( \begin{matrix} \hat i  & \hat j & \hat k \\ v_x   & v_y &0 \\ 0 &0 & B  \end{matrix} \right )</math>
-:<math>\vec{F} = q \left (v_y B \hat i - v_x B \hat j \right )</math>
-==Apply Newton's 2nd Law==
-:<math>ma_x = qv_yB</math>
-:<math>ma_y = -qv_x B</math>
-:<math>ma_z = 0</math>
-;Motion in the z-direction has no acceleration and therefor constant (zero) velocity.
-;Motion in the x-y plane is circular
-Let
-:<math>\omega=\frac{qB}{m}</math> = fundamental cyclotron frequency
-Then we have two coupled equations
-:<math>\dot{v}_x = \omega v_y</math>
-:<math>\dot{v}_y = - \omega v_x</math>
-==determine the velocity as a function of time==
-let
-:<math>v^* = v_x + i v_y</math> = complex variable used to change variables
-:<math>\dot{v}^* = \dot{v}_x + i \dot{v}_y</math>
-:: <math>= \omega v_y + i (-\omega v_x)</math>
-:: <math>= -i \omega \left ( \omega v_x +i\omega v_y \right )</math>
-:: <math>= -i \omega v^*</math>
-:<math>\Rightarrow</math>
-::<math>v^* = Ae^{-i\omega t}</math>
-the complex variable solution may be written in terms of <math>\sin</math> and <math>\cos</math>
-:<math>v_x +i v_y = A \left ( \cos(\omega t) - i \sin ( \omega t) \right )</math>
-The above expression indicates that <math>v_x</math> and <math>v_y</math> oscillate at the same frequency but are 90 degrees out of phase.  This is characteristic of circular motion with a magnitude of <math>v_{\perp}</math> such that
-:<math>v^* = v_{\perp}e^{-i\omega t}</math>
-==Determine the position as a function of time==
-To determine the position as a function of time we need to integrate the solution above for the velocity as a function of time
-:<math>v^* = v_{\perp}e^{-i\omega t}</math>
-Using the same trick used to determine the velocity, define a position function using complex variable such that
-:<math>x^* = x + i y</math>
-Using the definitions of velocity
-: <math>x^* = \int v^* dt = \int v_{\perp}e^{-i\omega t} dt</math>
-:: <math>= \frac{v_{\perp}}{i \omega} e^{-i\omega t} </math>
-The position is also composed of two oscillating components that are out of phase by 90 degrees
-:<math>x^* = x + i y= \frac{v_{\perp}}{i \omega} e^{-i\omega t} = -i\frac{v_{perp}}{\omega} \left ( \cos(\omega t) - \sin(\omega t) \right )</math>
-The radius of the circular orbit is given by
-:<math>r = \left | x^* \right | = \frac{v_{perp}}{\omega} = \frac{mv_{perp}}{qB}</math>
-:<math>r = \frac{p}{qB}</math>
-::<math>p=qBr</math>
-The momentum is proportional to the charge, magnetic field, and radius
-http://hep.physics.wayne.edu/~harr/courses/5200/f07/lecture10.htm
-http://www.physics.sfsu.edu/~lea/courses/grad/motion.PDF
-http://physics.ucsd.edu/students/courses/summer2009/session1/physics2b/CH29.pdf
-http://cnx.org/contents/77faa148-866e-4e96-8d6e-1858487a520f@9
 [[Forest_Ugrad_ClassicalMechanics]]

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