Äîêóìåíò âçÿò èç êýøà ïîèñêîâîé ìàøèíû. Àäðåñ îðèãèíàëüíîãî äîêóìåíòà : http://www.mso.anu.edu.au/pfrancis/phys1101/Lectures/L09/Lecture09.pdf Äàòà èçìåíåíèÿ: Thu Mar 10 08:44:46 2011 Äàòà èíäåêñèðîâàíèÿ: Tue Oct 2 14:21:49 2012 Êîäèðîâêà: Ïîèñêîâûå ñëîâà: http www.astronomy.ru forum index.php topic 4644.0.html

Iteration and Problem Solving Strategies
How to solve anything!

Thursday, 10 March 2011


How to work out really complicated motion
· · ·
Break it up into little tiny steps. Use an approximate method for each step. Add them all up.

Thursday, 10 March 2011


Vertical spring-mass
· Time t with t+t · Position x with x+vt · Velocity v with v + g - k x t
m

Thursday, 10 March 2011


Let's go
· Start off with t=0, x=0, v=0 · Apply our equations:
­ New value of t is t + t = 0+0.1 = 0.1 ­ New value of x is x + v t = 0+0â0.1 = 0 ­ New value of v is

k 5 v + g - x t = 0 + 9.8 - 0 â 0.1 = 0.98 m 0.1

Thursday, 10 March 2011


So after 0.1 seconds...
· According to our method, the position hasn't changed (still zero) but the velocity has increased to 0.98 m/s. · Now do this again, using these new numbers as the starting parameters..

Thursday, 10 March 2011


Second iteration
· Start off with t=0.1, x=0, v=0.98 · Apply our equations:
­ New value of t is t + t = 0.1+0.1 = 0.2 ­ New value of x is x + v t = 0+0.98â0.1 = 0.098 ­ New value of v is

k v + g - m

5 x t = 0.98 + 9.8 - 0.1

0 â 0.1 = 1.96

Thursday, 10 March 2011


Third iteration
· Start off with t=0.2, x=0.098, v=1.96 · Apply our equations:
­ New value of t is t + t = 0.2+0.1 = 0.3 ­ New value of x is x + v t = 0.098+1.96â0.1 = 0.294 ­ New value of v is

k v + g - m
Thursday, 10 March 2011

5 x t = 1.96 + 9.8 - 0.098 â 0.1 = 2.45 0.1


Fourth iteration
· Start off with t=0.3, x=0.294, v=2.45 · Apply our equations:
­ New value of t is t + t = 0.3+0.1 = 0.4 ­ New value of x is x + v t = 0.294+2.45â0.1 = 0.539 ­ New value of v is

k 5 v + g - x t = 2.45 + 9.8 - 0.294 â 0.1 = 1.96 m 0.1
Thursday, 10 March 2011


And so on...
· Do the calculations for each step, and then use the results as the input for the next step. · That's what iteration means! · What results do we get?

Thursday, 10 March 2011


Results for first few iterations (steps)
t 0 0.1 0.2 0.3 0.4 0.5 x 0 0 0.098 0.294 0.539 0.735 v 0 0.98 1.96 2.45 1.96 0.245

Thursday, 10 March 2011


A graph of the first twenty interations...

Thursday, 10 March 2011


Good and bad
· If you remember - the correct solution is an oscillation. · Our iteration has correctly produced an oscillation. · But it has the amplitude steadily increasing - which is wrong. · Springs don't do that!
Thursday, 10 March 2011


Our time step was too big.
· · ·
The approximation (that the speed and velocity are approximately constant within each time-step) wasn't good enough. If we make our time-step smaller... (say 0.01 sec)... We have to do a lot more steps...

Thursday, 10 March 2011


But it gets better...

Thursday, 10 March 2011


And if we make our time step smaller still - say 0.001 sec...

Thursday, 10 March 2011


Really rather good...
· But I needed to do 2000 steps (iterations) to get the last plot. · Which would have been very tedious and error-prone had I not used a computer... · Luckily we have computers and doing those 2000 steps took less than 0.1 sec...

Thursday, 10 March 2011


But it's painful

· ·

So do it by computer! Example python program

Thursday, 10 March 2011


So even this crude approximation...
· It pretty good with small timesteps. · And with the speed of modern computers, small timesteps are not much of a problem. · Using a better (more complicated) approximation to the motion in each timestep will mean that you can get away with bigger timesteps. · But each timestep needs more calculations to evaluate - so overall you may not be better off.

Thursday, 10 March 2011


Let's try an example
· A spaceship near a black hole...

x
Thursday, 10 March 2011


What forces apply?
· In this case there is only one - the gravitational force. · Being in space there is no friction or drag, so...

GMm F= 2 x
x
Thursday, 10 March 2011

Where M is the mass of the black hole and m the mass of the spaceship.


What variables will we track?
· · ·
Time, position (x) and velocity (v) as before. For one-dimensional problems it will always be these. In 3D, you will need to track vector position and vector velocity.

Thursday, 10 March 2011


Iteration equations
· For time: t + t (as before) · For position: x+ v t (as before) · But what about for velocity?

GMm F= 2 x

Write it down... What will the velocity be at the end of a time-step? x increases away from the black hole. Velocity is (as always) rate of change of x.

Thursday, 10 March 2011


Clicker Question

·

What is the new velocity?

Thursday, 10 March 2011


The answer...
· Gravity works to decrease the (outwards) velocity

GM v - 2 t x

Thursday, 10 March 2011


Let's chose some values
· Mass of the black hole = 1031 kg · Starting distance = 1,000,000 km · Starting speed = 2000 km/s away (You've been blasting away from it as hard as you could - but now your fuel has run out... Is your speed great enough to escape?)

Thursday, 10 March 2011


Python simulation

And then VPython simulation

Thursday, 10 March 2011


Summary
· Divide up your problem into little tiny steps. · Write down an approximate set of equations for each step · Plug numbers into these formulae over and over again - taking the output from one step as the input to the next.

Thursday, 10 March 2011


Chaos
· · ·
You can get extremely complicated results from this... TIny changes in the starting positions can cause huge changes in the outcomes. This is the hallmark of "Chaos"

Thursday, 10 March 2011


Computer Lab
· · ·
You will get to practice iteration in the computer lab. This is one of four rotations - check in which week you are doing it. Venue is different - BOZO112

Thursday, 10 March 2011


Contact Forces
Whenever one object touches another...

Thursday, 10 March 2011


Peculiarly tricky
· ·

Because they can point in different directions Because there is no simple formula to work them out

Thursday, 10 March 2011


Is there really a force when you sit on something?
· Newton's laws say there must be...
Must be an equal and opposite upward force or I'd be accelerating

Gravitational force downwards, due to Earth
Thursday, 10 March 2011


But how can a chair push?
· A force is "push" or · But you d walls, the normally thought of as a "pull" on't normally think of chairs, floor pushing?

Thursday, 10 March 2011


Imagine replacing the chair...
· With a spring...

Thursday, 10 March 2011


What would happen as I sat down?
· My weight would compress the spring. · As I put more and more weight on it, the spring would compress more. · And the more you compress a spring, the harder it pushes back. · Eventually I would have compressed it so much that it would push back on me as much as my weight pushes down on it.
Thursday, 10 March 2011


This is where contact forces come from
· At an atomic level, supposedly solid things (like chairs) are made of atoms stuck together by stretchy chemical bonds. · These chemical bonds behave much like springs. · When you apply a force to something, they bend and push back.

Thursday, 10 March 2011


A brick on a table

Thursday, 10 March 2011


Normal Force
· This is the explanation of normal force. · Whenever you apply a force to a solid surface, it will push back with just enough force to stop you from sinking into that surface. · Unless you push hard enough to break the solid surface.

Thursday, 10 March 2011


How do you work it out?
· · ·
If you knew how much you were sinking into the surface, and the spring constant of the surface, you could use the spring equation. But you usually don't. Instead, work backwards from the motion. If an object is not sinking into a surface or leaping off it, the component of the forces perpendicular (normal) to the surface must add up to zero. The normal force is whatever you need to make this happen!

Thursday, 10 March 2011


Perpendicular
· It's called the "normal force" because it is perpendicular ("normal") to the surface. · How do you work it out? · Usually by elimination. Work out all the other forces on some object. · Add up (vector sum) these forces. · Work out the component perpendicular to the surface. · The normal force will be equal and opposite.

Thursday, 10 March 2011


For example
You are dragging a box of mass M along the floor at a constant speed. You do it by pulling on a rope with force T. What is the normal force?



Thursday, 10 March 2011


Draw a free-body diagram
· ·
Show the box as a dot Show only the forces that act ON THE BOX

Thursday, 10 March 2011


Free-body diagram...
T Friction F Normal force N Weight Mg

Now what's the equation for the normal force?
Thursday, 10 March 2011


Free-body diagram...
Forces perpendicular to the surface (vertical in this case) must balance. So -

T sin( ) + N = Mg N = Mg - T sin( )
Friction F Normal force N

T

(as it's not accelerating, horizontal forces must balance too - so

Weight Mg

F = T cos( )

Thursday, 10 March 2011


Next time

·

Friction - another contact force.

Thursday, 10 March 2011


Key points
· ·
Whenever objects touch, there is a contact force. The normal force is usually whatever is needed to stop the objects moving into each other or springing apart.

Thursday, 10 March 2011