1. P1X*Dynamics & Relativity : Newton & Einstein Chris Parkes October 2004 Dynamics Motion Forces Energy & Momentum Conservation Simple Harmonic Motion Circular Motion http://ppewww.ph.gla.ac.uk/~parkes/teaching/DynRel/DynRel.html 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

2. 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

3. 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

4. 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 

5. 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 

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

7. 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

8. Projectiles Force: -mg in y direction acceleration: -g in y direction 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

9. Relative Velocity 1D e.g. Alice walks forwards along a boat at 1m/s and the boat moves at 2 m/s. what is Alices’ 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 Relative Velocity 2D e.g. Alice walks across the boat at 1m/s. As seen on the shore: V boat 1m/s V Alice 2m/s V relative to shore

10. 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

11. 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

12. 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,

13. 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

14. 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

15. Linear Momentum Conservation Define momentum p=mv Newton’s 2 nd law actually So, with no external forces, momentum is conserved. e.g. two body collision on frictionless surface in 1D before after m1m1 m2m2 m1m1 m2m2 v0v0 0 ms -1 v1v1 v2v2 For 2D remember momentum is a VECTOR, must apply conservation, separately for x and y velocity components Initial momentum: m 1 v 0 = m 1 v 1 + m 2 v 2 : final momentum Also true for net forces on groups of particles If then

16. Energy Conservation Need to consider all possible forms of energy in a system e.g: –Kinetic energy (1/2 mv 2 ) –Potential energy (gravitational mgh, electrostatic) –Electromagnetic energy –Work done on the system –Heat (1 st law of thermodynamics of Lord Kelvin) Friction  Heat Energy can neither be created nor destroyed Energy can be converted from one form to another Energy measured in Joules [J]

17. Collision revisited We identify two types of collisions –Elastic: momentum and kinetic energy conserved –Inelastic: momentum is conserved, kinetic energy is not Kinetic energy is transformed into other forms of energy Initial K.E.: ½m 1 v 0 2 = ½ m 1 v 1 2 + ½ m 2 v 2 2 : final K.E. m1m1 v1v1 m2m2 v2v2 See lecture example for cases of elastic solution 1.m 1 >m 2 2.m 1 <m 2 3.m 1 =m 2 Newton’s cradle

18. Impulse Change in momentum from a force acting for a short amount of time (dt) NB: Just Newton 2 nd law rewritten Where, p 1 initial momentum p 2 final momentum Approximating derivative Impulse is measured in Ns. change in momentum is measured in kg m/s. since a Newton is a kg m/s 2 these are equivalent Q) Estimate the impulse For Greg Rusedski’s serve [150 mph]?

19. Work & Energy Work = Force F times Distance s, units of Joules[J] –More precisely W=F.x –F,x Vectors so W=F x cos  e.g. raise a 10kg weight 2m F=mg=10*9.8 N, W=Fx=98*2=196 Nm=196J The rate of doing work is the Power [Js-1  Watts] Energy can be converted into work –Electrical, chemical,Or letting the –weight fall (gravitational) Hydro-electric power station Work is the change in energy that results from applying a force F s x F  mgh of water So, for constant Force

20. This stored energy has the potential to do work Potential Energy We are dealing with changes in energy 0 h choose an arbitrary 0, and look at  p.e. This was gravitational p.e., another example : Stored energy in a Spring Do work on a spring to compress it or expand it Hooke’s law BUT, Force depends on extension x Work done by a variable force

21. Consider small distance dx over which force is constant F(x) dx Work W=F x dx So, total work is sum 0 X Graph of F vs x, integral is area under graph work done = area F X dx For spring,F(x)=-kx: F x X Stretched spring stores P.E. ½kX 2

22. Work - Energy For a system conserving K.E. + P.E., then –Conservative forces But if a system changes energy in some other way (“dissipative forces”) –Friction changes energy to heat Then the relation no longer holds – the amount of work done will depend on the path taken against the frictional force or e.g. spring Conservative & Dissipative Forces

23. 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

24. SHM Examples Mass on a string 1) Simple Pendulum If  is small Working Horizontally: Hence, Newton 2: c.f. this with F=-kx on previous slide x mg sin   mgand Angular frequency for simple pendulum, small deflection

25. SHM Examples 2) 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?)

26. 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

27. 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

28. For a body moving in a circle of radius r at speed v, the angular momentum is L=r  (mv) = mr 2  = I  The rate of change of angular momentum is –The product r  F is called the torque of the Force Work done by force is F  s =(Fr)  (s/r) = Torque  angle in radians Power = rate of doing work = Torque  Angular velocity Angular Momentum (using v=  R) I is called moment of inertia  s r

29. 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

30. 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 10 -11 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:

31. 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

32. Moment of Inertia Have seen corresponding angular quantities for linear quantities –x  ; v  ; p  L –Mass also has an equivalent: moment of Inertia, I –Linear K.E.: –Rotating body v , m  I: –Or p=mv becomes: Conservation of ang. mom.: e.g. frisbeesolid sphere hula-hoop pc hard diskneutron star space station R  R   R1R1 R2R2 masses m distance from rotation axis r

33. Dynamics Top Five 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 3.Conservation Laws Energy (P.E., K.E….) and momentum Elastic/Inelastic collisions 4.SHM, Circular motion 5.Angular momentum L=r  (mv) = mr 2  = I  Moment of inertia