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
- [math]\sum \vec {F}_{ext} = \vec{F}(v) = m \frac{dv}{dt}[/math]
- [math]\Rightarrow \int_{v_i}^{v_f} \frac{dv}{F(v)} = \int_{t_i}^{t_f} \frac{dt}{m}[/math]
Frictional forces tend to be proportional to a fixed power of velocity
- [math]F(v) \approx v^n[/math]
If [math]n[/math] is unity then the velocity is exponentially approaching zero.
- [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]
- [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
- [math]x = \int_0^t v_i e^{-\frac{b}{m}t} dt[/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
Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity squared
- [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]
- [math]\int_{y_i}^{y_f} dy = m \frac{dv}{\left ( mg -bv^2 \right ) }[/math]
- [math]\sum \vec{F}_{ext} = mg -bv^2 = m \frac{dv}{dt}[/math]
Charged Particle in uniform B-Field
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]
- [math]\vec{B} = B \hat k[/math]
- Lorentz Force
- [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