Difference between revisions of "Forest UCM PnCP"

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:<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  =\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>
:<math>y  =\frac{m}{2b}  \ln \left ( \frac {g -\frac{b}{m}v_i^2}{g -\frac{b}{m}v_f^2} \right ) </math>
=Charged Particle in uniform B-Field=
=Charged Particle in uniform B-Field=

Revision as of 18:12, 29 August 2014

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 = \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]

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

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

[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]


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]
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


[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


[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]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]

The momentum is proportional to the charge, magnetic field, and radius