Two dimensional Kinematics

Ka-Lok Ng

Department of Bioinformatics

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Noble Prize in Physics 2007 – "for the discovery of Giant Magnetoresistance“

http://www.sciscape.org/news_detail.php?news_id=2277

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Noble Prize in Physics 2006 – Cosmic Microwave Background Radiation

http://www.sciscape.org/news_detail.php?news_id=2115

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Noble Prize in Physics 2005 – Laserhttp://nobelprizes.com/nobel/nobel.htmlhttp://memo.cgu.edu.tw/yun-ju/CGUWeb/SciKnow/PhyNews/NobelPhy2005.htm

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http://www.sciscape.org/

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Quark Asymptotic Freedom 夸克漸近自由

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http://www.sciscape.org/

The Noble Prize in Physiology or Medicine 2005

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Translation of coordinate system

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yy

axx

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'a x’

x

y y’

x x’

z

y

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zdm

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ydm

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2

2

2

2

2

2

For simplicity, assume x // x’Equation of motion in X

CoordinateSystem transformation

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Translation of coordinate system

'2

2

'2

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'2

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y

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Assume m is fixed, and X’ is fixed a is fixed.

Proof

''' ,, zzyyxx FFFFFF Vector decompostion

For RHS, we find that

Equation of motion in X’ ?

LHS

Newton’s Law of Motion is invariant under coordinate system translation.Homogeneity of space.

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Rotation of coordinate system

Express the rotated system X’, with respect to X

z’ = z

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Rotation of coordinate system

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cossin

sincos

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cossin'

sincos'

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By vector decomposition

Equation of motion in X’ ?

Calculate the LHS and RHS of X’

LHS

RHS of X’

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RHS = LHS !

Newton’s Law of Motion is invariant under coordinate system rotation. Isotopic of space.

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Fig . Ubiquitin-mediated protein degradation

1. The E1 enzyme activates the ubiquitin molecule. This reaction requires energy in the form of ATP.

2. The ubiquitin molecule is transferred to a different

enzyme, E2.

3. The E3 enzyme can recognise the protein target which is to be destroyed. The E2-ubiquitin complex binds so near to the protein target that the actual ubiquitin label can be transferred from E2 to the target.

4. The E3 enzyme now releases the ubiquitin-labelled

protein.

5. This last step is repeated until the protein has a short chain of ubiquitin molecules attached to itself. This ubiquitin chain is recognised in the opening of the proteasome. The ubiquitin label is disconnected and the protein is admitted and chopped into small pieces.

14http://nobelprize.org/medicine/laureates/2004/press.html

嗅覺系統

15The Olfactory System

上皮細胞

小球

bulb - 球狀的部分

有味的分子

僧帽狀細胞

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The smallest atomic clock

17http://www.sciam.com/article.cfm?chanID=sa003&articleID=0003EA53-CB42-1130-B53F83414B7F4945

The smallest atomic clock

18http://physicsweb.org/articles/news/8/9/2

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Kinematics in two dimensions

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Two identicalballs

Notice that the vertical displacement are the same.

Horizontal and verticle motions are independent.

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Vectors

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Vector addition – head-to-tail method

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Find A + B + C

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A + B + C = R

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Vector addtion in different order, C + A + B = A + B + C

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Negative of a vector B- has the same length but opposite direction

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Vector subtraction

Vector addition

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Graphical method

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Analytical methods of- vector addition and subtraction

Ax = A cos Ay = A sin

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Projectile motionMotion of an object move through air subject to gravity

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ay = -9.8 m/s2

ax = 0

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ax = 0x = x0 + vxtvx = v0x = constant

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ay = -9.8 m/s2

y = y0 + vytvy = v0y – ayty = y0 + v0yt – ayt2/2vy2 = voy2 – 2ay (y –y0)

1D 2D

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Motion in 2D

Suppose a vehicle in your game has a current velocity of 10m/s at 530 when it gets accelerated at a rate of 5m/s2 at 300. how fast will it be going after 3 seconds?

Solution1. Decompose the velocity and acceleration into the x and y axis2. v = 10*[cos53, sin53], a = 5[cos30, sin30] = [6.0, 8.0], a = [4.3, 2.5] vf = vi + at = [6.0, 8.0] + 3*[4.3, 2.5] vf = [18.9, 15.5] unit m/s

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V0 = 70 m/sIgnite the shell when it reachesthe highest point(a) Find the height at which the shell explode(b) How much time elapse between the launch of the shell and the explosion(c) How far the downrange did the shell explode ?

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Calculate the boat’s velocity relative to an observer on the shore.

A boat trying to head straightacross a river will actually move diagonally relative tothe shore as shown.

Calculate the boat’s relative velocityto an observer on the shore, if the velocity of the boat is 0.750 m/s relative to the river and the Speed of the river is 1.20 m/s to the right.

Vector sum(Vboat)earth= (Vboat)river + (Vriver)earth

(Vboat)earth = (1.22+ 0.752)0.5

= 1.42 m/s

=tan-1(0.75/1.2) = 32.00

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An airplane heading straight north is instead carried to the west and slowed down by wind. The plan does not move relative to the ground in the direction it points, rather, it moves in the direction of its total velocity [solid arrow].

Calculate the wind velocity for the situation shown in the figure at the right. The plane is moving at 45.0 m/s due north relative to the air, while its velocity relative to the

ground is 38.0 m/s in a direction 20.0° west of north. (Vtot)X=(Vplane)earth = (Vplane)wind + (Vwind)earth

Find Vw= (Vwind)earth

(Vtot)X = (Vw)X 38 cos(110°) = - 13.0 m/s

(Vtot)Y=Vp+(Vw)Y38 sin(110°)=45+ (Vw)Y

(Vw)Y = -9.29 m/sVw = [(Vw)X

2 + (Vw)Y2 ]0.5 = 16.0 m/s

= tan -1 ((Vw)Y / (Vw)X ) = 35.6° (south of west)

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(Vknife)earth= (Vknife)boat + (Vboat)earth

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