P1X Dynamics & Relativity : Newton & Einstein Chris Parkes October 2007 Dynamics Motion Forces – Newton’s Laws Simple Harmonic Motion Circular Motion Part I - “I frame no hypotheses; for whatever is not deduced from the phenomena is to be called a hypothesis; and hypotheses, whether metaphysical or physical, whether of occult qualities or mechanical, have no place in experimental philosophy.” READ the textbook! section numbers in syllabus

Motion Position [m] Velocity [ms -1 ] –Rate of change of position Acceleration [ms -2 ] –Rate of change of velocity t x v t dx dt e.g 0 a 0 0

Equations of motion in 1D –Initially (t=0) at x 0 –Initial velocity u, –acceleration a, s=ut+1/2 at 2, where s is displacement from initial position v=u+at Differentiate w.r.t. time: v 2 =u 2 +2 as

2D motion: vector quantities Position is a vector –r, (x,y) or (r,  ) –Cartesian or cylindrical polar co- ordinates –For 3D would specify z also Right angle triangle x=r cos , y=r sin  r 2 =x 2 +y 2, tan  = y/x Scalar: 1 number Vector: magnitude & direction, >1 number 0 X Y x y r 

vector addition c=a+b c x = a x +b x c y = a y +b y scalar product x y a b c can use unit vectors i,j i vector length 1 in x direction j vector length 1 in y direction finding the angle between two vectors a,b, lengths of a,b Result is a scalar a b 

Vector product e.g. Find a vector perpendicular to two vectors a b c  Right-handed Co-ordinate system

Velocity and acceleration vectors Position changes with time Rate of change of r is velocity –How much is the change in a very small amount of time  t 0 X Y x r(t) r(t+  t) Limit at  t  0

Projectiles Motion of a thrown / fired object mass m under gravity x y x,y,t v  Velocity components: v x =v cos  v y =v sin  x direction y direction a: v=u+at: s=ut+0.5at 2 : a x =0 a y =-g v x =vcos  + a x t = vcos  v y =vsin  - gt This describes the motion, now we can use it to solve problems x=(vcos  )ty= vtsin  -0.5gt 2 Force: -mg in y direction acceleration: -g in y direction

Relative Velocity 2D V boat 2m/s V Alice 1m/s V relative to shore Relative Velocity 1D e.g. Alice walks forwards along a boat at 1m/s and the boat moves at 2m/s. What is Alice’s velocity as seen by Bob ? If Bob is on the boat it is just 1 m/s If Bob is on the shore it is 1+2=3m/s If Bob is on a boat passing in the opposite direction….. and the earth is spinning… Velocity relative to an observer e.g. Alice walks across the boat at 1m/s. As seen on the shore: θ

Changing co-ordinate system vt Frame S (shore) Frame S’ (boat) v boat w.r.t shore (x’,y’) Define the frame of reference – the co-ordinate system – in which you are measuring the relative motion. x x’ Equations for (stationary) Alice’s position on boat w.r.t shore i.e. the co-ordinate transformation from frame S to S’ Assuming S and S’ coincide at t=0 : Known as Gallilean transformations As we will see, these simple relations do not hold in special relativity y

First Law –A body continues in a state of rest or uniform motion unless there are forces acting on it. No external force means no change in velocity Second Law –A net force F acting on a body of mass m [kg] produces an acceleration a = F /m [ms -2 ] Relates motion to its cause F = ma units of F: kg.m.s -2, called Newtons [N] Newton’s laws We described the motion, position, velocity, acceleration, now look at the underlying causes

Third Law –The force exerted by A on B is equal and opposite to the force exerted by B on A Block on table Weight (a Force) FbFb FaFa Force exerted by block on table is F a Force exerted by table on block is F b F a =-F b (Both equal to weight) Examples of Forces weight of body from gravity (mg), - remember m is the mass, mg is the force (weight) tension, compression Friction,

Force Components  Force is a Vector Resultant from vector sum Resolve into perpendicular components

Free Body Diagram Apply Newton’s laws to particular body Only forces acting on the body matter –Net Force Separate problem into each body Body 1 Tension In rope Block weightFriction Body 2 Tension in rope Block Weight e.g. Supporting Force from plane (normal force)

Tension & Compression Tension –Pulling force - flexible or rigid String, rope, chain and bars Compression –Pushing force Bars Tension & compression act in BOTH directions. –Imagine string cut –Two equal & opposite forces – the tension mg

A contact force resisting sliding –Origin is chemical forces between atoms in the two surfaces. Static Friction (f s ) –Must be overcome before an objects starts to move Kinetic Friction (f k ) –The resisting force once sliding has started does not depend on speed Friction mg N F f s or f k

Simple Harmonic Motion Occurs for any system with Linear restoring Force »Same form as Hooke’s law –Hence Newton’s 2 nd –Satisfied by sinusoidal expression –Substitute in to find  Oscillating system that can be described by sinusoidal function Pendulum, mass on a spring, electromagnetic waves (E&B fields)… or A is the oscillation amplitude  is the angular frequency  in radians/sec Period Sec for 1 cycle Frequency Hz, cycles/sec

SHM General Form A is the oscillation amplitude - Maximum displacement Displacement Oscillation frequency Phase (offset of sine wave in time) 

SHM Examples 1) Mass on a spring Let weight hang on spring Pull down by distance x –Let go! In equilibrium F=-kL’=mg L’ x Restoring Force F=-kx Energy: (assuming spring has negligible mass) potential energy of spring But total energy conserved At maximum of oscillation, when x=A and v=0 TotalSimilarly, for all SHM (Q. : pendulum energy?)

SHM Examples 2) Simple Pendulum but if  is small Working along swing: Hence, Newton 2: c.f. this with F=-kx on previous slide and Angular frequency for simple pendulum, small deflection x mg sin   mg L Not actually SHM, proportional to sin , not  Mass on a string

Circular Motion x y =t=t R t=0 s 360 o = 2  radians 180 o =  radians 90 o =  /2 radians Acceleration Rotate in circle with constant angular speed  R – radius of circle s – distance moved along circumference  =  t, angle  (radians) = s/R Co-ordinates x= R cos  = R cos  t y= R sin  = R sin  t Velocity N.B. similarity with S.H.M eqn 1D projection of a circle is SHM

Magnitude and direction of motion And direction of velocity vector v Is tangential to the circle  v  And direction of acceleration vector a a Velocity v=  R Acceleration a=  2 R=(  R) 2 /R=v 2 /R a= -  2 r  Acceleration is towards centre of circle

Force towards centre of circle Particle is accelerating –So must be a Force Accelerating towards centre of circle –So force is towards centre of circle F=ma= mv 2 /R in direction –r or using unit vector Examples of central Force 1.Tension in a rope 2.Banked Corner 3.Gravity acting on a satellite

Gravitational Force Myth of Newton & apple. He realised gravity is universal same for planets and apples Newton’s law of Gravity Inverse square law 1/r 2, r distance between masses The gravitational constant G = 6.67 x Nm 2 /kg 2 F F m1m1 m2m2 r Gravity on earth’s surface OrHence, m E =5.97x10 24 kg, R E =6378km Mass, radius of earth Explains motion of planets, moons and tides Any two masses m 1,m 2 attract each other with a gravitational force:

Satellites N.B. general solution is an ellipse not a circle - planets travel in ellipses around sun M m R Distance in one revolution s = 2  R, in time period T, v=s/T T 2  R 3, Kepler’s 3 rd Law Special case of satellites – Geostationary orbit Stay above same point on earth T=24 hours Centripetal Force provided by Gravity

Dynamics I – Key Points 1.1D motion, 2D motion as vectors –s=ut+1/2 at 2 v=u+atv 2 =u 2 +2 as –Projectiles, 2D motion analysed in components 2.Newton’s laws –F = ma –Action & reaction 3.SHM 4.Circular motion (R,  ) Oscillating system that can be described by sinusoidal function Force towards centre of circle