Äîêóìåíò âçÿò èç êýøà ïîèñêîâîé ìàøèíû. Àäðåñ îðèãèíàëüíîãî äîêóìåíòà : http://www.mso.anu.edu.au/pfrancis/phys1101/Lectures/L07/Lecture07.pdf Äàòà èçìåíåíèÿ: Mon Mar 7 03:51:26 2011 Äàòà èíäåêñèðîâàíèÿ: Tue Oct 2 14:27:35 2012 Êîäèðîâêà: Ïîèñêîâûå ñëîâà: þæíàÿ àòëàíòè÷åñêàÿ àíîìàëèÿ

Motion in an Circle and Oscillation
Two Special Cases

Monday, 7 March 2011


Course news
· · ·
Monday, 7 March 2011

Labs start tomorrow Clickers will now be used for assessment. You need to have a "U" in front of your student number in the clicker. If you can't join the class, come see me now! Class reps - introduce yourselves.


Course Reps Nominated
· · · · ·
Monday, 7 March 2011

Samantha Cheah Raj Srilakshmi Ellen Rykers Lachlan McGinness Sarah Biddle


Momentum and Force
· ·
We will talk about two special cases circular motion and oscillation. Then we will start dealing with the general case.

Monday, 7 March 2011


Circular motion
·
Remember - if a force is applied that is always sideways, an object will move in a circle.

Monday, 7 March 2011


Example - Orbits
·
If one object (say the Space Station) is in a circular orbit around another, much larger object (say the Earth), the larger object's gravity must be supplying the necessary (centripetal) force to keep the space station moving in a circle.

Monday, 7 March 2011


Velocity v

GM m Gravitational Force F = r2
Dis tance

r

Monday, 7 March 2011


To stay in a circular orbit, this gravitational force mv must supply the necessary centripetal force F =

2

r

so...

GM m mv = r2 r

2

Cancel masses and one of the r's GM 2 =v r Rearrange to find v

v=
Monday, 7 March 2011



GM r


Centrifugal Force

·

A particularly confusing topic

Monday, 7 March 2011


Clicker Question
You are a passenger in a car and not wearing your seatbelt. Without increasing or decreasing its speed, the car makes a sharp right turn, and you find yourself colliding with the left-hand door. Which is the correct analysis of the situation? 1. Before and after the collision, there is a leftward force pushing you into the door 2. Starting at the time of the collision, the door exerts a leftward force on you 3. Both of the above 4. Neither of the above

Monday, 7 March 2011


The Answer
· ·
The door exerts a force on you. You are trying to continue moving in a straight line, and the door pushes into you sideways, forcing you to turn.

Monday, 7 March 2011


Centrifugal Force
· · ·
Centrifugal force is even more imaginary than centripetal force. There is no outward force when you go around a circle. You are just trying to continue in a straight line and being prevented from doing so by some force (which might be due to gravity or friction or the door, acts towards the centre and has magnitude mv 2 r

Monday, 7 March 2011


SImilarly for "g"-forces
· ·
When you speed up or slow down there is no "g"-force.You are being pushed by your chair or the dashboard. This push is what is changing your speed.

Monday, 7 March 2011


Crucial Facts
· · ·
Monday, 7 March 2011

Special case - a force that is constant in magnitude but perpendicular to the motion. Result - motion in a circle. The force points at the centre of the circle.

mv F= r

2


Spring force
·
This is another special case - a situation you almost never meet in the real world, but which can be solved without the need for a computer.

Monday, 7 March 2011


Spring Forces
SImple Harmonic Motion

Monday, 7 March 2011


"Ideal Spring"
· ·
It has a "Natural" or "unextended" length. Whenever you pull it away from this length by a distance D, it exerts an opposing force F = -kD

where k is the "Spring Constant"
Monday, 7 March 2011


Vertical weight calculation
· ·
Example - a 50 g weight is hung from a spring of constant k=3.0 N/m. By how much does it stretch?

Monday, 7 March 2011


Draw a free-body diagram for the weight
·
This is a diagram just showing the weight, as a dot, and the forces ACTING ON IT Spring force

Weight

Monday, 7 March 2011


As it's hanging still...
·
Forces must balance. So the weight and the spring force must be the same Spring force

mg = kD
Rearrange to get D

mg D= k

Weight

So this gives how much the spring stretched.
Monday, 7 March 2011


Why are we worried about this?
· · · ·
Monday, 7 March 2011

Because while ideal springs are rare, forces which always pull towards a point are common. Such as chemical bonds Any elastic behaviour So it's worth getting used to this sort of force.


Motion attached to a spring
· ·
We've seen how to calculate a static situation with a spring. But what if something is moving while attached to a spring?

Monday, 7 March 2011


Vertical spring-mass system
·
VPython simulation, spring_vertical.py

Monday, 7 March 2011


Oscillation
· · ·
The net force is towards the equilibrium position. It accelerates towards it. But thanks to momentum, it overshoots. The force is now backwards and slows it to a halt.

Monday, 7 March 2011


Energy
·
A constant interplay between kinetic and spring energy (with a little gravitational potential energy thrown in for good measure)

Monday, 7 March 2011


Very general behaviour
· ·
Whenever you get any sort of force which tends to push things back into place. Usually need a computer to solve exact motion, but if you assume the spring is ideal (seldom the case in reality) you can solve it.

Monday, 7 March 2011


Analytic Solution
· · ·
Monday, 7 March 2011

I'll show you the mathematical solution in this idealised case. But first - what would you expect to determine how rapidly it oscillates? What makes it oscillate faster?


Clicker Question
· · ·
What makes it oscillate faster? The spring constant? The mass?

Monday, 7 March 2011


Answer
· ·

A stiffer spring - pushes back harder A lighter mass - accelerates faster.

Monday, 7 March 2011


Horizontal

· ·
Monday, 7 March 2011

To make the maths simpler, let's take a horizontal spring-mass system, with the mass sliding along a frictionless surface. (the result is the same as for a vertical system but the argument is a bit simpler)


Coordinates
x

· ·
Monday, 7 March 2011

Let's call the position of the weight x, and measure it from the spring's rest position. (Once again you can use any axes you like and will get the same result, but it makes the calculation messier).


Force
·
x The only horizontal force acting is the spring force F = -kx

As we know the force, we can work out the acceleration using F=ma

k a=- x m

Monday, 7 March 2011


Calculus
· · ·
Monday, 7 March 2011

So we know the acceleration. But what about the velocity v or position x? Luckily, we know that acceleration is defined as the rate of change of velocity. So

k dv =- x a= dt m


Position
·
And velocity is defined as the rate of dx change of position, so

v=

dt
2

This means that acceleration a is

dv a= = dt

d

So acceleration is what you get when you differentiate position twice with respect to time.
Monday, 7 March 2011



dx dt

dt



dx =2 dt


So we now know that...
dx k =- x 2 dt m
2

· ·
Monday, 7 March 2011

k and m are just constants. So this is telling us that if you differentiate x twice, you get x back again, albeit multiplied by a constant. Can you think of any functions that when you differentiate themselves twice are unchanged (apart from a constant?)


What appears in its own second differential?
· · ·
How about Cosine? Let's try x = A cos(t), where A and (omega) are constants, currently unknown. Let's try differentiating this twice

x = A cos ( t) dx = -A sin ( t) dt 2 dx 2 2 = -A cos ( t) = - x dt2
Monday, 7 March 2011


It works!
·
Compare

dx 2 = - x dt2 dx k =- x 2 dt m =
2

2

with the spring acceleration equation we got earlier

Identical, as long as we make
Monday, 7 March 2011

k m


x = A cos ( t) where

So the answer is. ..
=

k m

· · ·
Monday, 7 March 2011

This whole derivation should remind you of the projectile motion one. Write down F=ma, and integrate twice to get position versus time. This is called "simple harmonic motion"


What are and A?
x!

Period T!

Amplitude A!

t!

A is the amplitude of the oscillation - how far it goes ON ONE SIDE of the equilibrium position
Monday, 7 March 2011


x!

Period T!

Amplitude A!

?
t!

is the angular frequency, and is measured in radians per second. As 2 radians is a complete circle, this corresponds to the period T above.
Monday, 7 March 2011


· ·

Period and Frequency
So the angular frequency

=

k m

The period T (time to repeat) is

2 T=

The frequency (in cycles per second, also known as Hertz, Hz) is 1

f=

T

=

2

Monday, 7 March 2011


Very Useful
·
You will come back to these quantities time and time again, as they are fundamental in waves, interference and all sorts of vibrations.

Monday, 7 March 2011


Resonance
· · ·
Monday, 7 March 2011

One final feature. An oscillating system like this is peculiarly responsive to outside wiggles at its natural frequency. This is called resonance.


Vpython simulation

Monday, 7 March 2011


Tacoma Narrows

Monday, 7 March 2011


SkyMapper
· ·
Currently being commissioned Has a resonance problem. The cryocoolers are resonating with the secondary mirror (we think)

Monday, 7 March 2011


Carbon Dioxide

·

VPython simulation

Monday, 7 March 2011


Key Points
· ·
Whenever you get a force that pushes back towards some equilibrium position, you probably get vibrations. You can work out the frequency of oscillations if you know how strong the restoring force and how big the inertia of whatever is being vibrated.

Monday, 7 March 2011