Forest UCM PnCP

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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

[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]

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.

Linear Air Resistance

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]

Another block on incline example


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