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

[math]\displaystyle{ \sum \vec {F}_{ext} = \vec{F}(v) = m \frac{dv}{dt} }[/math]

[math]\displaystyle{ \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]\displaystyle{ F(v) \approx v^n }[/math]

If [math]\displaystyle{ n }[/math] is unity then the velocity is exponentially approaching zero.

[math]\displaystyle{ F(v) = -bv }[/math]: negative sign indicates a retarding force and [math]\displaystyle{ b }[/math] is a proportionality constant

[math]\displaystyle{ \sum \vec {F}_{ext} = -bv = m \frac{dv}{dt} }[/math]

[math]\displaystyle{ \Rightarrow \int_{v_i}^{v_f} \frac{dv}{v} = \int_{t_i}^{t_f} \frac{-b}{m}dt }[/math]

[math]\displaystyle{ \ln\frac{v_f}{v_i} = \frac{-b}{m}t }[/math]; [math]\displaystyle{ t_i \equiv 0 }[/math]

[math]\displaystyle{ v_f = v_i e^{-\frac{b}{m}t} }[/math]

The displacement is given by

[math]\displaystyle{ x = \int_0^t v_i e^{-\frac{b}{m}t} dt }[/math]

[math]\displaystyle{ = \left . v_i \left ( \frac {e^{-\frac{b}{m}t}}{-\frac{b}{m}} \right ) \right |_0^t }[/math]

[math]\displaystyle{ = \left . v_i \left ( -\frac{m}{b} e^{-\frac{b}{m}t} \right ) \right |_0^t }[/math]

[math]\displaystyle{ = \left . v_i \left ( \frac{m}{b} e^{-\frac{b}{m}t} \right ) \right |_t^0 }[/math]

[math]\displaystyle{ = v_i \left ( \frac{m}{b} e^{-\frac{b}{m}0} -\frac{m}{b} e^{-\frac{b}{m}t} \right ) }[/math]

[math]\displaystyle{ = \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]\displaystyle{ \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]\displaystyle{ \frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v\frac{dv}{dy} }[/math]

The integral becomes

[math]\displaystyle{ mg -bv^2 = m v\frac{dv}{dy} }[/math]

[math]\displaystyle{ \int_{y_i}^{y_f} dy = \int_{v_i}^{v_f} m \frac{dv}{\left ( mg -bv^2 \right ) } }[/math]

[math]\displaystyle{ y = \int_{v_i}^{v_f} \frac{dv}{\left ( g -\frac{b}{m}v^2 \right ) } }[/math]

let [math]\displaystyle{ u = g -\frac{b}{m}v^2 }[/math]

then [math]\displaystyle{ du = -2\frac{b}{m}v dv }[/math]

[math]\displaystyle{ 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]\displaystyle{ y =\frac{m}{2b} \ln \left ( \frac {g -\frac{b}{m}v_i^2}{g -\frac{b}{m}v_f^2} \right ) }[/math]

Forest_UCM_NLM_BlockOnIncline

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]\displaystyle{ \vec{v} = v_x \hat i + v_y \hat j }[/math]

[math]\displaystyle{ \vec{B} = B \hat k }[/math]

Lorentz Force

[math]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ \vec{F} = q \left (v_y B \hat i - v_x B \hat j \right ) }[/math]

Apply Newton's 2nd Law

[math]\displaystyle{ ma_x = qv_yB }[/math]

[math]\displaystyle{ ma_y = -qv_x B }[/math]

[math]\displaystyle{ 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]\displaystyle{ \omega=\frac{qB}{m} }[/math] = fundamental cyclotron frequency

Then we have two coupled equations

[math]\displaystyle{ \dot{v}_x = \omega v_y }[/math]

[math]\displaystyle{ \dot{v}_y = - \omega v_x }[/math]

determine the velocity as a function of time

let

[math]\displaystyle{ v^* = v_x + i v_y }[/math] = complex variable used to change variables

[math]\displaystyle{ \dot{v}^* = \dot{v}_x + i \dot{v}_y }[/math]

[math]\displaystyle{ = \omega v_y + i (-\omega v_x) }[/math]

[math]\displaystyle{ = -i \omega \left ( \omega v_x +i\omega v_y \right ) }[/math]

[math]\displaystyle{ = -i \omega v^* }[/math]

[math]\displaystyle{ \Rightarrow }[/math]

[math]\displaystyle{ v^* = Ae^{-i\omega t} }[/math]

the complex variable solution may be written in terms of [math]\displaystyle{ \sin }[/math] and [math]\displaystyle{ \cos }[/math]

[math]\displaystyle{ v_x +i v_y = A \left ( \cos(\omega t) - i \sin ( \omega t) \right ) }[/math]

The above expression indicates that [math]\displaystyle{ v_x }[/math] and [math]\displaystyle{ 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]\displaystyle{ v_{\perp} }[/math] such that

[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ x^* = x + i y }[/math]

Using the definitions of velocity

[math]\displaystyle{ x^* = \int v^* dt = \int v_{\perp}e^{-i\omega t} dt }[/math]

[math]\displaystyle{ = \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]\displaystyle{ 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]\displaystyle{ r = \left | x^* \right | = \frac{v_{perp}}{\omega} = \frac{mv_{perp}}{qB} }[/math]

[math]\displaystyle{ r = \frac{p}{qB} }[/math]

[math]\displaystyle{ 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