Forest UCM PnCP

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

$\sum \vec {F}_{ext} = \vec{F}(v) = m \frac{dv}{dt}$

$\Rightarrow \int_{v_i}^{v_f} \frac{dv}{F(v)} = \int_{t_i}^{t_f} \frac{dt}{m}$

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

$F(v) \approx v^n$

If $n$ is unity then the velocity is exponentially approaching zero.

$F(v) = -bv$: negative sign indicates a retarding force and $b$ is a proportionality constant

$\sum \vec {F}_{ext} = -bv = m \frac{dv}{dt}$

$\Rightarrow \int_{v_i}^{v_f} \frac{dv}{v} = \int_{t_i}^{t_f} \frac{-b}{m}dt$

$\ln\frac{v_f}{v_i} = \frac{-b}{m}t$; $t_i \equiv 0$

$v_f = v_i e^{-\frac{b}{m}t}$

The displacement is given by

$x = \int_0^t v_i e^{-\frac{b}{m}t} dt$

$= \left . v_i \left ( \frac {e^{-\frac{b}{m}t}}{-\frac{b}{m}} \right ) \right |_0^t$

$= \left . v_i \left ( -\frac{m}{b} e^{-\frac{b}{m}t} \right ) \right |_0^t$

$= \left . v_i \left ( \frac{m}{b} e^{-\frac{b}{m}t} \right ) \right |_t^0$

$= v_i \left ( \frac{m}{b} e^{-\frac{b}{m}0} -\frac{m}{b} e^{-\frac{b}{m}t} \right )$

$= \frac{m}{b} v_i \left ( 1-e^{-\frac{b}{m}t} \right )$

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

$\sum \vec{F}_{ext} = mg -bv^2 = m \frac{dv}{dt}$

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

$\frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v\frac{dv}{dy}$

The integral becomes

$mg -bv^2 = m v\frac{dv}{dy}$

$\int_{y_i}^{y_f} dy = \int_{v_i}^{v_f} m \frac{dv}{\left ( mg -bv^2 \right ) }$

$y = \int_{v_i}^{v_f} \frac{dv}{\left ( g -\frac{b}{m}v^2 \right ) }$

let $u = g -\frac{b}{m}v^2$

then $du = -2\frac{b}{m}v dv$

$y =\int_{v_i}^{v_f} \frac{-m}{2b} \frac{du}{u } = \frac{b}{m} \int_{v_f}^{v_i} \ln {g -\frac{b}{m}v^2} =$

$y =\frac{m}{2b} ln \left ( \frac {g -\frac{b}{m}v_i^2}{g -\frac{b}{m}v_f^2} \right ) =$

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.

$\vec{v} = v_x \hat i + v_y \hat j$

$\vec{B} = B \hat k$

Lorentz Force

$\vec{F} = q \vec{E} + q\vec{v} \times \vec{B}$

Note

the work done by a magnetic field is zero if the particle's kinetic energy (mass and velocity) don't change.

$W = \Delta K.E.$

No work is done on a charged particle forced to move in a fixed circular orbit by a magnetic field (cyclotron)

$\vec{F} = m \vec{a} = q \vec{v} \times \vec{B} = q\left ( \begin{matrix} \hat i & \hat j & \hat k \\ v_x & v_y &0 \\ 0 &0 & B \end{matrix} \right )$

$\vec{F} = q \left (v_y B \hat i - v_x B \hat j \right )$

Apply Newton's 2nd Law

$ma_x = qv_yB$

$ma_y = -qv_x B$

$ma_z = 0$

Motion in the z-direction has no acceleration and therefor constant (zero) velocity.

Motion in the x-y plane is circular

Let

$\omega=\frac{qB}{m}$ = fundamental cyclotron frequency

Then we have two coupled equations

$\dot{v}_x = \omega v_y$

$\dot{v}_y = - \omega v_x$

determine the velocity as a function of time

let

$v^* = v_x + i v_y$ = complex variable used to change variables

$\dot{v}^* = \dot{v}_x + i \dot{v}_y$

$= \omega v_y + i (-\omega v_x)$

$= -i \omega \left ( \omega v_x +i\omega v_y \right )$

$= -i \omega v^*$

$\Rightarrow$

$v^* = Ae^{-i\omega t}$

the complex variable solution may be written in terms of $\sin$ and $\cos$

$v_x +i v_y = A \left ( \cos(\omega t) - i \sin ( \omega t) \right )$

The above expression indicates that $v_x$ and $v_y$ oscillate at the same frequency but are 90 degrees out of phase. This is characteristic of circular motion with a magnitude of $v_{\perp}$ such that

$v^* = v_{\perp}e^{-i\omega t}$

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

$v^* = v_{\perp}e^{-i\omega t}$

Using the same trick used to determine the velocity, define a position function using complex variable such that

$x^* = x + i y$

Using the definitions of velocity

$x^* = \int v^* dt = \int v_{\perp}e^{-i\omega t} dt$

$= \frac{v_{\perp}}{i \omega} e^{-i\omega t}$

The position is also composed of two oscillating components that are out of phase by 90 degrees

$x^* = x + i y= \frac{v_{\perp}}{i \omega} e^{-i\omega t} = -i\frac{v_{perp}}{\omega} \left ( \cos(\omega t) - \sin(\omega t) \right )$

The radius of the circular orbit is given by

$r = \left | x^* \right | = \frac{v_{perp}}{\omega} = \frac{mv_{perp}}{qB}$

$r = \frac{p}{qB}$

$p=qBr$

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