Difference between revisions of "Forest UCM PnCP"

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:<math>v_t = \frac{mg}{b}</math>
 
:<math>v_t = \frac{mg}{b}</math>
 
:<math> v_t -v = \frac{1}{b} \frac{dv}{dt}</math>
 
:<math> v_t -v = \frac{1}{b} \frac{dv}{dt}</math>
:<math>  -b dt= \frac{dv}{v_t -v} </math>
+
:<math>  b dt= \frac{dv}{v_t -v} </math>
 +
:<math>  -b dt= \frac{dv}{v -v_t} </math>
  
 
==Example: falling object with quadratic  air friction==
 
==Example: falling object with quadratic  air friction==

Revision as of 13:35, 31 August 2014

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

Fext=F(v)=mdvdt
vfvidvF(v)=tftidtm


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

F(v)vn


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

Horizontal motion

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
Fext=bv=mdvdt
vfvidvv=tftibmdt
lnvfvi=bmt; ti0
vf=viebmt

The displacement is given by

x=t0viebmtdt
=vi(ebmtbm)|t0
=vi(mbebmt)|t0
=vi(mbebmt)|0t
=vi(mbebm0mbebmt)
=mbvi(1ebmt)

Example: falling object with linear air friction

Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity

Fext=mgbv=mdvdt

let

vt=mgb
vtv=1bdvdt
bdt=dvvtv
bdt=dvvvt

Example: falling object with quadratic air friction

Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity squared

Fext=mgbv2=mdvdt

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

dvdt=dvdydydt=vdvdy

The integral becomes

mgbv2=mvdvdy
yfyidy=vfvimdv(mgbv2)
y=vfvidv(gbmv2)


let u=gbmv2

then du=2bmvdv

y=vfvim2bduu=bmvivflngbmv2=
y=m2bln(gbmv2igbmv2f)

Another block on incline example

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.

v=vxˆi+vyˆj
B=Bˆk


Lorentz Force
F=qE+qv×B
Note
the work done by a magnetic field is zero if the particle's kinetic energy (mass and velocity) don't change.
W=ΔK.E.

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


F=ma=qv×B=q(ˆiˆjˆkvxvy000B)
F=q(vyBˆivxBˆj)

Apply Newton's 2nd Law

max=qvyB
may=qvxB
maz=0


Motion in the z-direction has no acceleration and therefor constant (zero) velocity.
Motion in the x-y plane is circular

Let

ω=qBm = fundamental cyclotron frequency

Then we have two coupled equations

˙vx=ωvy
˙vy=ωvx

determine the velocity as a function of time

let

v=vx+ivy = complex variable used to change variables
˙v=˙vx+i˙vy
=ωvy+i(ωvx)
=iω(ωvx+iωvy)
=iωv
v=Aeiωt

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

vx+ivy=A(cos(ωt)isin(ωt))

The above expression indicates that vx and vy oscillate at the same frequency but are 90 degrees out of phase. This is characteristic of circular motion with a magnitude of v such that

v=veiωt

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=veiωt

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

x=x+iy

Using the definitions of velocity

x=vdt=veiωtdt
=viωeiωt

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

x=x+iy=viωeiωt=ivperpω(cos(ωt)sin(ωt))

The radius of the circular orbit is given by

r=|x|=vperpω=mvperpqB
r=pqB
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