Kinematics - study of motion HIGHER PHYSICS 1A UNSW SESSION s o t See S&J , ,

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1 1 Kinematics - study of motion HIGHER PHYSICS 1A UNSW SESSION 1 01 s Joe Wolfe s o t See S&J.1-.6, , Is this straightforward, or are there subtleties? See Physclips Chs &3 and support pages Measure lengths to get (relatie) positions Measure durations to get (relatie) times How? At first, we use rulers and repeated cycles for space and time. First we count ratios, then indirect methods But what are time and space? Here are some subtleties: What happens when we moe the clocks and rulers around? Does this change them? If we use 'identical' ones? How do we calibrate them? Can you (always) gie an object a position or an eent a time? Can you do so more than once? Can you make continuous measurements? What is a point? A particle? Kinematics We shall go ery quickly oer some bits as they are assumed knowledge* Motion with constant acceleration a y = constant a y = d y dt the definition of acceleration, so, by integrating, y = a y dt subscript y for y direction = a y t + const what is the constant? Use the initial state: at t = 0, y = y0 y = y0 + a y t (i) common conention: use subscript 0 for t = 0 y = y dt again, using the definition y = dy dt. Substitute (i) gies: = y0 + a y t and integration gies = y0 t + 1 a yt + constant again, use the initial state at t = 0 y = y 0 + y0 t + 1 a yt. (ii) Definition of constant acceleration and rearrange: t = ( 0 )/a y ii) gies y - y 0 = y0 t + 1 a yt substitute from t = ( y - y0 )/ a y a y (y - y 0 ) = y0 ( y - y0 ) + 1 ( y - y0 ) a y (y - y 0 ) = y y0 (iii) You may also remember these as = u +at (i) Δs = ut + 1 at (ii) u = as (iii) Howeer, we shall be looking at motion in both x and y directions so you will need subscripts * What if this is not clear to you? See the kinematics chapter in the text in Serway & Jewett Constant Acceleration, a multimedia chapter in Physclips See also Calculus, a supporting page in Physclips

2 Example. Joe runs at constant speed 6 ms -1 towards a stationary bus. When he is 30 m from it, the bus accelerates away at m.s -. Can he oertake it? " Joe runs at constant speed 6 ms -1 " > J = 6 ms -1, aj = 0 " stationary bus " & " accelerates away at m.s - " > b0 = 0, ab = m.s - " When he is 30 m from it " > xb0 - xj0 = 30 m " Can he oertake it?" > Does xj = xb at any t? x b = x b0 + b0 t + 1 a bt (i) (the pale terms are zero, but I thought about them!) x J = x J0 + J0 t + 1 a Jt (ii) it's easiest to remember the general form and lose terms Eliminate x Jo by choice of origin, x bo = 30 m Draw a diagram or two. Which of the following graphs is correct? x b = x J substitute from (i) and (ii) gies: In kinematics, displacement time graphs are useful! What does oertake mean? How to write it mathematically? 1 a bt Jo t + x bo = 0 Recall quadratic: solutions are b ± b 4ac a so t = Jo ± Jo - a b x bo a b are the units correct? Put in numbers = 6 * * 30 ms -1 à -e no (real) solutions, no oertaking. what would the imaginary solutions mean? Example. Ball 1 thrown ertically up at 5 ms -1 from 0 m aboe ground. Simultaneously, ball thrown ertically down at 5 ms -1 from 0 m aboe ground. What are their speeds when they hit the ground, and what is the time interal between collisions? We are gien: o1 = 5 ms -1 o = - 5 ms -1 y o = 0 m Translate the question: When y = 0, t =? Use (ii) to get t 1 and t. Use (i) to get y1 and y.

3 3 How much do you lose if you miss the starter's gun? Two runners at different times (Δt apart). During the (constant) acceleration phase, when are they a distance Δx apart? x 1 = ½at ½ x = ½a(t') t' = t Δt x 1 x = ½at ½a(t Δt) Sole this for t: x 1 x = (aδt)t ½a(Δt) t = (x 1 x + ½a(Δt) )/aδt A trickier but real example. Oer a marked kilometre, on a flat track with no wind, solar car sunswift slows from 0 = 70 k.p.h. to = 50 k.p.h. Assume that a = k (turbulent drag only). What is k? Gien x (= 1 km), 0,, need k. Must relate these. Must eliminate a. Must introduce x. a = k and by definition a = d dt so d dt = k But this has t. Can eliminate dt using dx = dt. Multiply both sides by dt: d = k.(dt) = k.dx so d Remember that = k.dx Finally and x! Nearly there. d(ln y) dy = 1 y so Separate ariables (i.e. get on one side, x on the other): d = d(ln ) Physclips has an introduction to calculus d(ln ) = k dx ln = kdx = k x + const ln 0 = k x + const = k*0 + const ln ln 0 = ln 0 = kx finally it looks easy: just integrate both sides! What constant?once again, use initial conditions at x = 0, = 0 subtract these and remember ln (a/b) = ln a ln b so k = ln (/ 0 )/x (It's also interesting that = 0 e kx ) Physclips does this and more in detail

4 4 Vectors hae direction and magnitude e.g. displacement, elocity, acceleration, force, spin, electric field are all ectors displacement = m towards door; wind elocity is 3.7 ms -1 at 31 E. of N., acceleration is 9.8 ms - up, Force is 4 N in +e x direction etc (cf Scalars: mass, length, heat, temperature..., which also hae magnitude, but no direction) Notation: a in most texts ã when hand writing a in my notes, to combine the aboe a in Halliday, Resnick and Walker Addition x' z c = a + b b a y' y put them head to tail to add. Subtraction z a Does a_ + b_ = b_ + a_? d = a + ( b) = a b to subtract, rewrite the equation: a_ b_ = a_ + ( b_ ) or a_ b_ = d_ > a_ = d_ + b_ See Vectors, a supporting page in Physclips Consider the displacement m North What sort of quantity is m North? What does m mean? What does North mean? Should/could I write m North? How big is North? magnitude direction ( m) distance (in the direction of North) b We hae a word for "in the direction of North" (It's just "North"). y Think: (a b) is what I hae to add b to in order to get a. Wouldn't it be useful to hae a shorthand for "in the positie x direction"? head to head to subtract ectors Consider adding m North and m East So let's introduce

5 5 Vector components and unit ectors y a x a = a a y ĵ x î + a y ĵ ĵ a y x ˆ a i x î a x = a cos θ, a y = a sin θ a x is the component of a in the x direction it is a scalar î is unit ector: magnitude of 1 in x direction. î means "in the positie x direction" a = a x î + a y ĵ i.e. means a x in the positie x direction plus a y in the positie y direction a = a x + a y θ = tan -1 a y a x so we can conert back to magnitude and direction Addition by components c_ = a_ + b_ = (a x î+ a y ĵ ) + (b x î+ b y ĵ ) c x î + c y j_ = (a x + b x ) î + (a y + b y ) ĵ c x = a x + b x and c y = a y + b y so we hae two scalar equations from one ector one ector eqn in n dimensions > n independent algebraic eqns. important case: 0 = 0 î + 0 ĵ e.g. mechanical if a_ + b_ = 0, equilibrium a x + b x = 0 and a y + b y = 0 more on this later in Newton's laws. Resoling ectors is often useful. Choose conenient axes: more on this later in Newton's laws Component of W_ in direction of plane = - Wsinθ Total force in y' direction = - Wsinθ = m d y' dt Newton's second law requires that components in normal direction add to zero: N Wcosθ = 0

6 6 In three dimensions: z (up) a x y (North) ˆk ĵ ˆi x (East) r_ = r x î + r y ĵ + r z ˆk (sometimes just i, j, k) right hand conention: î ĵ ˆk in dir n of thumb index middle fingers of right hand î kˆ ĵ Pythagoras' theorem in three dimensions z a a z a y x x h a y What is magnitude of a? Hypotenuse h: h = a x + a y. Now look at triangle h,a z,a: a = h + a z = a x + a y + a z. a = a x + a y + a z Not in our syllabus, but, for your interest, in four dimensions: we write j = 1 and use c as the natural unit of elocity. Two eents at (e x1,e y1,e z1,jct 1 ) and (e x,e y,e z,jct ) are separated by (e x e x1 ) +(e y e y1 ) +(e z e z1 ) (ct ct 1 )

7 7 Uniform circular motion Write θ = ωt. where ω = constant ω is the angular elocity y (t+δt) Δ (t) (t) r (t+δt) (t+δt) Δθ Δθ r (t) θ x As Δt and Δθ > 0, triangles are ery narrow so Δ_ > right angles to _ lim $ Δ' a_ = Δt > & ) From the triangle, we see it is parallel to r, i.e. towards centre. We call it 0% Δt ( centripetal acceleration (and centripetal acceleration = 1*radial acceleration). We need magnitudes, so Δs = rδθ = ds dt = r dθ dt definition of angle defininition of speed = rω Use arc straight line of triangle, here for Δ_ : Δ_ Δθ (Interesting: notice that Δ_ =/ Δ _ ) lim Δt >0 a_ = d_ dt = dθ dt a = r d_ = dθ = ω so a_ = ω r_ = ω r but a_ // r_

8 8 Example. Car traelling at goes oer hill with ertical radius r = 30 m (>> height of car) at summit. It doesn't slow down. How high must be for the car to lose contact with the ground at the summit? The only force pulling the car down is graity, so the downwards acceleration cannot be greater than g. So the centripetal acceleration must be less than or equal to g (or equal to g for the critical alue). This example on Physclips

9 9 Example What is the acceleration of this theatre? Due to rotation of the Earth: a rot = ω r = π T r (T Earth ~ 4hrs, r Sydney to Earth's axis ~ 5300 km) = π 3.9*3600 s = 8 mm.s - ( m) Due to Earth's orbit around the sun: a orb = π T r (Torbit, r orbit ) = π 365.4*4*3600 s ( m) = 6 mm.s - So, een though the earth moes rapidly, we cannot feel this motion: it's a tiny acceleration Due to Sun's orbit around the centre of the galaxy: r ~ 10 0 m, T ~ s a galactic ~ 0.1 nm.s - Example r = (A.sin ωt) î + (A.cos ωt) ĵ + (Bt) ˆk where A and B are constants. What shape is r_? What is a_? a = d dt r = -(Aω.sin ωt) î - (Aω.cos ωt) ĵ so a is in xy plane and a = a x + a y =... = Aω = constant (but direction is always changing)

10 Example A cockroach on a turntable crawls in the radial direction (initially the x direction) at speed. The turntable rotates at ω anticlockwise. Describe his path in i, j coordinates. ω r θ t ωt We can specify his position with the coordinates (r,θ), where r = t, and θ = ωt. x = r cos θ, y = r sin θ path is What is this shape? r(t) = t cos ωt i + t sin ωt j Relatie elocities Gallilean/Newtonian relatiity watch for hidden assumptions origin of frame (x',y') is at _ r f and moes with _ f with respect to the (x,y) frame. No rotation. y y' r r' r f x' x Subraction of ectors (see aboe): Directly from the geometry, we write _ r' = r_ _ r f and, by definition, ' _ = d dt _ r' So we differentiate both sides to gie d dt _ r' = d dt r_ d dt _ r f ' _ = _ _ f It's important to understand this. Think of this example: elocity of wind oer the ground = elocity of wind with respect to me + my elocity oer the ground or elocity of boat oer the ground = elocity of rier oer ground plus elocity of boat oer rier

11 3 Example A boat heads East at 8 km.hr -1. The current flows South at 6 km.hr -1. What is the boat's elocity relatie to the earth? c b = boat + current To add the ectors, draw them head-to-tail. b θ c = + b c magnitude: = b + c = (8 km.hr -1 ) + (6 km.hr -1 ) = (8 + 6 ) (km.hr -1 ) = ( ) (km.hr -1 ) = 10 km.hr -1 direction: θ = tan-1 6 km.hr-1 8 km.hr -1 = tan = 37 Answer: 10 km.hr -1 at 40 South of East

12 4 Reminder about ector subtraction Draw the ectors head to head to subtract them. z a b d = a + ( b) = a b y to subtract, rewrite the equation: a_ b_ = a_ + ( b_ ) i.e. we can also rearrange subtraction so that it becomes addition Example A sailor wants to trael East at 8 km.hr -1. The current flows South at 6 km.hr -1. What direction must she head, and what speed should she make relatie to the water? = boat + current c boat = current To subtract the ectors, draw them head-to-head. magnitude: b = + c b θ c direction: θ = (8 km.hr -1 ) + (6 km.hr -1 ) = 10 km.hr -1 = tan -1 c = 37 = tan-1 6 km.hr-1 8 km.hr -1 She must head 40 North of East and trael at 10 km.hr -1 with respect to the water. Puzzle Rier flows East at 10 km/hr. Sailing boat traels East down the rier. Can the boat trael faster a) with no wind? b) with 10 km/hr wind from West? Consider motion relatie to the water. b) ' w = 0 while a) 10 km/hr headwind. Not in syllabus. See Physclips Physics of Sailing if interested.

13 5 A SE wind blows at 30 km/hr. If you are traelling North, how fast must you trael before the wind is coming (i) exactly from your right? (ii) From 30 E of N? From the ground: w N or y E or x From your frame: ' w Relatie motion: (add el wrt me to my el to get el wrt ground) w = you + ' w Either) y' x' or so ' w = w you Do it algebraically. Gien: w = w cos 45 i + w sin 45 j you = 0 i + you j ' w = ' wx i + 0 j y direction: w sin 45 = you + 0 you = w sin 45 = 1 km/hr x direction: w cos 45 = 0 + ' wx ' wx = w cos 45 = 1 km/hr from E Or) Do it with ector diagrams: ' w you 45 w Same answers, but directly. Second case: ' w is 30 E of N ' w 30 y' x' x direction: w cos 45 = ' w cos 60. y direction: you = w sin 45 + ' w sin 60 equations, unknowns, first gies ' w = w cos 45 cos 60 = 4 km/hr Substitute this in second equation: you = 60 km/hr

14 6 Projectiles Question. A man fires a gun horizontally. At the same time he drops a bullet. Which hits the ground first? Explain your reasoning. (To simplify, let's say it happens on the moon.) F = ma. So a force in y direction does not cause an acceleration in x direction. Independence of x and y motion. (in air, there are other forces) Question. A man shoots at a coconut. At the instant that he fires, the coconut falls. What happens? h L θ obious method (c for coconut, b for bullet). From the triangle: h = L tan θ Now write down the heights of both y c = h 1 gt y b = yo t 1 gt Will they miss? What is the difference between the heights? y c y b = h yo t = h o sinθ.t L = t o cos θ which gies T = The bullet has no horizontal acceleration, so L o cos θ substitute gies y c y b = h o sinθ.l o cos θ = h L tan θ and from the geometry aboe = 0 Alternatiely: consider a frame of reference falling with g. g The non-egetarian ersion of this problem is ery old one, called The Monkey and the Hunter. Physclips has a page on this with a film clip, using a toy monkey. No real monkeys were etc.

15 7 Projectiles Without air, a y = g = constant. a x = 0 (Galileo: independence of horiz. & ert motion) y 0 a θ Strategy: kinematics eqn (ii) gies y(t), x(t): R x (ii) > y = y o + yo t 1 gt. x = x o + x t. Eliminate t gies y(x) For range R, y(r) = 0 Rearrange to get R = R(θ) Find maximum in R(θ) (ii) > y = y o + yo t 1 gt. motion with constant acceleration (ii) > x = x o + x t. motion with no acceleration Choose axes so that x o = y o = 0 and use (ii) to eliminate t (i.e. substitute t = x/ x ): y = yo x 1 x g x x y = 0 when x = 0 or R (*) > yo R = 1 x g R Set R θ x (**) yo = o sin θ, x = o cos θ > R = R( o, θ) = 0 to obtain θ for maximum range. We had yo R = 1 x g R x (*) Now we hae parabolic cure in x, not t, this time. (**) sole (**) for R: so R = x yo g = 0 sin θ. 0 cos θ g = 0 sin θ g R θ = 0 cos θ g R θ (check units) = 0 when θ = 90 so θ = 45 substitute gies: sin θ cos θ =

16 8 Example The human cannon of Circus Oz has a muzzle elocity o. For a new trick, they will fire the human canonball into a horizontal teflon tube at height h aboe the canon mouth. To aoid damage to the canonball, he must arrie with purely horizontal elocity. Calculate the position of the canon and its angle to the ertical. i) draw a diagram ii) iii) put in symbols for quantities translate question θ o h L gien h, o and final y = 0 Find L and θ Think about this: chose θ to get highest point correct, then choose L so as to hit the target. Relate h, y and yo. yo depends on θ. During flight, the acceleration is g upwards. The desired y is zero, so (think = u + aδs) 0 = y = yo + a y (Δy) = o cos θ gh o cos θ = gh cos θ = gh o θ = cos 1 Find time of flight. x t = L. 0 = y = yo + a y t t = o cos θ g L = o t sin θ. = o sin θ o cos θ g gh o <optional> = o g sin θ where θ = cos 1 gh o simplify optional

17 9 Projectile in 3D Example. Throw object from origin at 0 ms -1, 30 East and at 45 to horizontal. Take x = East, y = North, describe its motion as a function of t in î, ĵ, ˆk notation. z (up) a x ˆk ĵ 30 y (North) horiz ˆi zo = o sin 45 = 14 ms -1 x (East) horiz = o cos 45 = 14 ms -1 (= const) x = horiz cos 60 = 7 ms -1 (= const) y = horiz cos 30 = 1 ms -1 (= const) r_ = r x î + r y ĵ + r z ˆk = (7 ms -1 ).t î + (1 ms -1 ).t ĵ + (14 ms -1 )t - 1 (9.8 ms- )t ˆk But normally we should just make it a D problem. Example. Electron enters a uniform electric field at (x,y,t)=(0,0,0), with elocity x i. The field extends from x = 0 to x = L, but is zero for x < 0 and x > L. In the field, the electron is accelerated at a y j. Write an equation for the position of the electron for x > L. Hint: diide the problem into parts y a y j x i + yf j x i L,y f y Outer parts with linear motion, middle one with projectile motion First region x 0 (t 0) = x i = const easy Second region 0 x L (0 t L/ x ) constant acceleration in the y direction. We need final position & elocity x = x 0 + x0 t + 1 a xt x = x0 + a x t y = y 0 + y0 t + 1 a yt y = y0 + a y t y f = 1 a yt f = 1 a y L x yf = a y L x Third region x L x = x t (still no acceleration in x direction) y = y f + yf t L x position is x t i + 1 a y L + ay L x t L x j x now all we need to do is to substitute for yf: