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P1X*Dynamics & Relativity: Newton & Einstein Part I - “I frame no hypotheses; Dynamics Motion READ the textbook! section numbers in syllabus Forces Energy & Momentum Conservation 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.” Simple Harmonic Motion Circular Motion http://ppewww.ph.gla.ac.uk/~parkes/teaching/DynRel/DynRel.html October 2004 Chris Parkes Motion • Position [m] • Velocity [ms-1] x e.g dx dx v dt – Rate of change of position 0 2 • Acceleration [ms-2] a dv d dt – Rate of change of velocity dt x v 2 0 a 0 dt t t Equations of motion in 1D – Initially (t=0) at x0 – Initial velocity u, – acceleration a, x x0 ut at 1 2 Differentiate w.r.t. time: dx v u at dt 2 d x aa 2 dt s=ut+1/2 at2, 2 where s is displacement from initial position v=u+at v 2 (u at ) 2 u 2 2uat a 2t 2 v 2 u 2 2a(ut 12 at 2 ) v2=u2+2 as 2D motion: vector quantities • Position is a vector – r, (x,y) or (r, ) – Cartesian or cylindrical polar coordinates – For 3D would specify z also Scalar: 1 number Vector: magnitude & direction, >1 number Y • Right angle triangle x=r cos , y=r sin r2=x2+y2, tan = y/x r 0 x y X vector addition • c=a+b y cx= ax +bx cy= ay +by b can use unit vectors i,j a c i vector length 1 in x direction x j vector length 1 in y direction scalar product finding the angle between two vectors a b ab cos a x bx a y by a,b, lengths of a,b Result is a scalar a xbx a y by a b cos 2 2 2 2 ab a x a y bx by a b Vector product e.g. Find a vector perpendicular to two vectors c ab c a b sin iˆ c a b ax bx ˆj ay by kˆ a y bz a z by a z a z bx a x bz bz a x by a y bx c Right-handed Co-ordinate system b a Velocity and acceleration vectors • Position changes with time Y • Rate of change of r is velocity – How much is the change in a very small amount of time t d r r (t t ) r (t ) Limit at t0 v dt t 0 yd xd yv , xv td td d v v(t t ) v(t ) d 2 r a 2 dt t dt vd y vd ya , x xa td td r(t) x r(t+t) X Projectiles Motion of a thrown / fired object mass m under gravity y Velocity components: v vx=v cos x,y,t x a: v=u+at: x direction ax=0 vx=vcos + axt = vcos s=ut+0.5at2: x=(vcos )t vy=v sin Force: -mg in y direction acceleration: -g in y direction y direction ay=-g vy=vsin - gt y= vtsin -0.5gt2 This describes the motion, now we can use it to solve problems 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. V boat 1m/s As seen on the shore: V 1 2 5m / s 2 2 tan 2 / 1, 63.4 V relative to shore V Alice 2m/s Changing co-ordinate system Define the frame of reference – the co-ordinate system – in which you are measuring the relative motion. (x’,y’) y Frame S (shore) vt Frame S’ (boat) v boat w.r.t shore 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 : x x'vt Known as Gallilean transformations As we will see, these simple relations do not hold in y y' special relativity We described the motion, position, velocity, acceleration, now look at the underlying causes Newton’s laws • 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] • Third Law – The force exerted by A on B is equal and opposite to the forceFexerted by B on A b •Force exerted by block on table is Fa Block on table Fa=-Fb Weight (a Force) Fa •Force exerted by table on block is Fb (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, Tension & Compression • Tension – Pulling force - flexible or rigid • String, rope, chain and bars mg • Compression – Pushing force • Bars mg mg • Tension & compression act in BOTH directions. – Imagine string cut – Two equal & opposite forces – the tension Friction • A contact force resisting sliding – Origin is chemical forces between atoms in the two surfaces. • Static Friction (fs) – Must be overcome before an objects starts to move • Kinetic Friction (fk) – The resisting force once sliding has started • does not depend on speed N fs or fk F mg fs s N fk k N Linear Momentum Conservation • Define momentum p=mv d p d (mv) nd • Newton’s 2 law actually F dt dt m ddtv ma • So, with no external forces, momentum is dp conserved. F 0, dt 0, p const Also true for net forces on groups of particles • e.g. two body collision on frictionless If F F 0, surface in 1D then p p const i i i i before m1 m2 0 ms-1 Initial momentum: m1 v0 = m1v1+ m2v2 : final momentum after v0 m1 m2 v2 v1 For 2D remember momentum is a VECTOR, must apply conservation, separately for x and y velocity components Energy Conservation •Energy can neither be created nor destroyed •Energy can be converted from one form to another • Need to consider all possible forms of energy in a system e.g: – – – – – Kinetic energy (1/2 mv2) Potential energy (gravitational mgh, electrostatic) Electromagnetic energy Work done on the system Heat (1st law of thermodynamics of Lord Kelvin) • Friction Heat Energy measured in Joules [J] Collision revisited m1 v1 m2 • We identify two types of collisions – Elastic: momentum and kinetic energy conserved Initial K.E.: ½m1 v02 = ½ m1v12+ ½ m2v22 : final K.E. – Inelastic: momentum is conserved, kinetic energy is not • Kinetic energy is transformed into other forms of energy See lecture example for cases of elastic solution Newton’s cradle 1. m1>m2 2. m1<m2 3. m1=m2 v2 Impulse • Change in momentum from a force acting for a short amount of time (dt) Impulse J p 2 p1 Fdt Where, p1 initial momentum p2 final momentum • NB: Just Newton 2nd law rewritten F p 2 p1 dt dp dt m dv ma dt Approximating derivative Impulse is measured in Ns. change in momentum is measured in kg m/s. since a Newton is a kg m/s2 these are equivalent Q) Estimate the impulse For Greg Rusedski’s serve [150 mph]? Work & Energy Work is the change in energy that results from applying a force • 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 P dW dt So, for constant Force Power P F x F [Js-1Watts] dF x dx F F v dt dt • Energy can be converted into work – Electrical, chemical,Or letting the mgh of – weight fall (gravitational) water • Hydro-electric power station s This stored energy has the potential to do work Potential We are dealing with changes in energy • choose an arbitrary 0, and look at p.e. This was gravitational p.e., another example : Energy h 0 W F x (mg)h 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 Work done by a variable force Consider small distance dx over which force is constant F(x) Work W=Fx dx X dx So, total work is sum W F dx F ( x)dx 0 X 0 Graph of F vs x, F integral is area under graph work done = area For spring,F(x)=-kx: x F X dx X X 0 0 X W F ( x)dx kxdx [ 12 kx2 ]0X 12 kX 2 Stretched spring stores P.E. ½kX2 Work - Energy dU W U U Fdx or F dx 2 1 d [ 2 2 kx ] 1 kx e.g. spring U 2 kx F dx Conservative & Dissipative Forces • For a system conserving K.E. + P.E., then – Conservative forces dU F dx • 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 Simple Harmonic Motion Oscillating system that can be described by sinusoidal function Pendulum, mass on a spring, electromagnetic waves (E&B fields)… • Occurs for any system with F k x Linear restoring Force » Same form as Hooke’s law d2x k F ma 2 x dt m – Hence Newton’s 2nd – Satisfied by sinusoidal expression x A sin t or x A cos t A is the oscillation amplitude is the angular frequency – Substitute in to find dx d 2x x A sin t A cos t 2 A 2 sin t dt k k m m 2 dt in radians/sec Frequencyf Period T 1 Hz, cycles/sec 2 Sec for 1 cycle f SHM Examples 1) Simple Pendulum • Mass on a string If is small x mg sin mg Working Horizontally: sin x l F mg sin mg Lx c.f. this with F=-kx on previous slide Hence, Newton 2: and g l d 2x g x 2 dt l Angular frequency for simple pendulum, small deflection SHM Examples 2) Mass on a spring • Let weight hang on spring • Pull down by distance x – Let go! L’ In equilibrium F=-kL’=mg Energy: x K .E. 12 mv2 U 12 kx2 Restoring Force F=-kx k m (assuming spring has negligible mass) potential energy of spring But total energy conserved At maximum 2of oscillation, when x=A and v=0 Total E 12 kA Similarly, for all SHM (Q. : pendulum energy?) Circular Motion 360o = 2 radians 180o = radians 90o = /2 radians • 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 •Acceleration d v x ( R cos t ) R sin t dt d v y ( R sin t ) R cos t dt d d a x (v x ) ( R sin t ) R 2 cos t dt dt d d a y (v y ) ( R cos wt ) R 2 sin t dt dt y R s =t x t=0 N.B. similarity with S.H.M eqn 1D projection of a circle is SHM Magnitude and direction of motion •Velocity v 2 vx v y R 2 w2 sin 2 t R 2 2 cos 2 t 2 R 2 2 2 v=R tan And direction of velocity vector v Is tangential to the circle vy vx cos t 1 sin t tan 90o •Acceleration 2 a a ax a y 2 v 2 R 2 w4 cos 2 t R 2 4 sin 2 t 4 R 2 a= 2R=(R)2/R=v2/R And direction of acceleration vector a a= -2r a x 2 x a y 2 y Acceleration is towards centre of circle Angular Momentum • For a body moving in a circle of radius r at speed v, the angular momentum is L=r (mv) = mr2= I (using v=R) I is called moment of inertia The rate of change of angular momentum is dL d dv dt (r mv) mr mr a dt dt r ma r F r s – The product rF is called the torque of the Force • Work done by force is Fs =(Fr)(s/r) = Torque angle in radians Power d Torque Torque dt = rate of doing work = Torque Angular velocity Force towards centre of circle • Particle is accelerating – • • 1. 2. 3. So must be a Force Accelerating towards centre of circle – So force is towards centre of circle F=ma= mv2/R in direction –r 2 v or using unit vector F m rˆ r Examples of central Force Tension in a rope Banked Corner Gravity acting on a satellite Gravitational Force Myth of Newton & apple. He realised gravity is universal same for planets and apples •Any two masses m1,m2 attract each other with a gravitational force: F F m1m2 F G 2 r r m2 m1 Newton’s law of Gravity Inverse square law 1/r2, r distance between masses The gravitational constant G = 6.67 x 10-11 Nm2/kg2 •Explains motion of planets, moons and tides 24kg, m =5.97x10 mE m GmE E Gravity on m F G RE=6378km 2 2 earth’s surface RE RE Mass, radius of earth GmE 1 9.81ms Or F mg Hence, g 2 RE N.B. general solution is an ellipse not a circle - planets travel in ellipses around sun Satellites •Centripetal Force provided by Gravity Mm mv2 F G 2 R R M 2 M v G v G R R m R M Distance in one revolution s = 2R, in time period T, v=s/T R T 2R / v 2R GM T2R3 , Kepler’s 3rd Law •Special case of satellites – Geostationary orbit •Stay above same point on earth T=24 hours 3 24 60 60 2 R 42,000km R2 GM E Moment of Inertia • Have seen corresponding angular quantities masses m for linear quantities I mi ri 2 r 2dm distance from i rotation axis r – x; v; pL – Mass also has an equivalent: moment of Inertia, I – Linear K.E.: K .E. mv – Rotating body v, mI: K .E. I – Or p=mv becomes: L I Conservation of ang. mom.: I11 I 2 2 e.g. frisbee solid sphere hula-hoop 2 2 I 1 M (R R ) 1 2 2 1 2 pc hard disk I MR 1 2 2 neutron star I 52 MR 2 R 2 space station 1 2 R1 R2 R 2 Dynamics Top Five 1. 1D motion, 2D motion as vectors – – s=ut+1/2 at2 v=u+at v2=u2+2 as Projectiles, 2D motion analysed in components 2. Newton’s laws – F = ma 3. Conservation Laws • • 4. Energy (P.E., K.E….) and momentum Elastic/Inelastic collisions SHM, Circular motion v2 F m rˆ x A sin t r 5. Angular momentum • L=r (mv) = mr2= I • Moment of inertia