1 Particle dynaics Physics 1A, UNSW Newton's laws: S & J: Ch 5.1 5.9, 6.1 force, ass, acceleration also weight Physclips Chapter 5 Friction - coefficients of friction Physclips Chapter 6 Hooke's Law Dynaics of circular otion Question. Top view of ball. What is its trajectory after it leaves the race? e d c b a? then what? Aristotle: v_ = 0 is "natural" state (not in syllabus) Galileo & Newton: a_ = 0 is "natural" state Galileo: what if we reove the side of the bowl? Newton's Laws First law "zero (total) force zero acceleration" (It's actually a bit ore subtle. More forally, we should say: If Σ F_ = 0, there exist reference fraes in which a_ = 0 called Inertial fraes What is an inertial frae? One in which Newton's laws are true. observation: w.r.t. these fraes, distant stars don't accelerate Experient in the foyer
2 In inertial fraes: Second law Σ F_ = a_ Σ is iportant: it is the total force that deterines acceleration Σ F x = a x Σ F y = a y Σ F z = a z 3D > 3 equations 1st law is special case of 2nd What does the 2nd law ean? Σ F_ = i a_ and W_ = g g_ are i and g necessarily the sae? called inertial and gravitational asses F_ = a_ a_ is already defined, but this leaves us with a puzzle: i) Does this equation define? ii) Does this law define F_? iii) iv) v) How? Is it a physical law? All of the above? --------------------------------------------------------------- i) Given one ass, we could calibrate any forces by easuring the a they produced. ii) iii) Siilarly, for any one F, we could calibrate any 's by the accelerations produced The 2nd Law is the observation that the 's and F's thus defined are consistent. eg Having used standard to calibrate F, now produce 2F (eg two identical F systes). Is a now doubled? Every such experient is a test of Newton's second law. Or, for those who want it logically: NeNewton 1: "Every body persists in its state of rest or of unifor otion in a straight line unless it is copelled to change that state by forces ipressed on it." postulate An inertial frae of reference is one in which Newton's 1st law is true. Such fraes exist (and with respect to these fraes, distant stars don't accelerate) if Σ F = 0, _a = 0 w.r.t. distant stars. Force causes acceleration. F // a, F a definition observation: definition
3 Another way of writing Newton 2: To any body ay be ascribed a (scalar) constant, ass, such that the acceleration produced in two bodies by a given force is inversely proportional to their asses, i.e. for sae F, 2 1 = a 1 a 2 We already have etre, second, choose a standard body for kg, then choose units of F (Newtons) such that Σ F = a Newton's first and second laws (this eqn. is laws 1&2, definition of ass and units of force) So, how big are Newtons? Newton 3: "To every action there is always opposed an equal reaction; or the utual actions of two bodies upon each other are always equal and directed to contrary parts" Or Forces always occur in pairs, F and F, one acting on each of a pair of interacting bodies. Third law F AB = F BA Why so? What would it be like if internal forces didn t add to zero? Iportant conclusion: internal forces in a syste add to zero. So we can now write the 1st and 2nd laws: Σ F external = a Total external force = a
4 Exaple Where is centre of earth-oon orbit? F e = F = F g equal & opposite each akes a circle about coon centre of ass F g = a = ω 2 r F g = e a e = e ω 2 r e NB sign conventions r r e = e = 5.98 1024 kg 7.36 10 22 kg = 81.3 (i) earth-oon distance r e + r = 3.85 10 8 (ii) (two equations, two unknowns) r e (1 + 81.3) = 3.85 10 8 gives r = 3.80 10 8, r e = 4.7 10 6 = 4700 k centre of both orbits is inside earth (later we'll see that it is the centre of ass of the two)
5 Using Newton's laws How do we use the to solve probles? Newton's 2nd ΣF = a the Σ is iportant: in principle, we have to consider the all. Newton's 3rd (forces coe in pairs, F and F), so: internal forces add to zero. They don't affect otion Therefore, applying Newton's 2nd, we use ΣF external = a draw diagras ('free body diagras') to show only the external forces on the body of interest Newton's second law is a vector equation write coponents of Newton's second law in 2 (or 3) directions Exaple. As the bus takes a steady turn with radius 8 at constant speed, you notice that a ass on a string hangs at 30 to the vertical. How fast is the bus going? Draw a diagra with physics string 30 T a T W W The ass (and the bus) are in circular otion, so we can apply Newton 2: F = a = v2 horiz r Only the tension has a horizontal coponent, so (the acceleration is centripetal) T sin 30 = v2 r (i) Need one ore eqn: ass is not falling down, ie vertical acceleration = 0, so T cos 30 = g (ii) Again, we have two equations in two unknowns Dived (i) by (ii) to eliinate T: tan 30 = v2 r 1 g > v = gr tan 30 = 6.7 /s > 20 kph. Don t stop yet! First, check the diensions. Reasonable? Suggestions? Probles? which we rearrange to give
6 Proble. Horse and cart. Wheels roll freely. Let's go! Why should I pull? The force of the cart on e equals y force on it, but opposes it. Σ F = 0. We'll never accelerate. What would you say to the horse? F c F c F c F c F g F g This is a good exercise in looking at the external forces Horizontal forces on cart (ass c ) F c Horizontal forces on horse (ass h ) F c = c a c = c a F c Horizontal forces on Earth (ass E ) F g F g F c = h a F g E >> h + c In principle, the earth accelerates to the left. But E >> h so a E << a h
7 "light" ropes etc. Here, light eans << other asses Truck ( t ) pulls wagon ( w ) with rope ( r ). All have sae a_. i) wagon: F 2 = w a. Let's apply Newton's second to each eleent: ii) iii) rope: F 1 - F 2 = r a truck: F 1 + F ext = t a (ii)/(i) > F 1 F 2 F 2 = ra w a if r << w, F 1 = F 2. iportant result: Forces at opposite ends of light ropes etc are equal and opposite. Again, see Physclips if this isn't clear Tension. When the ass of a string, coupling etc is negligible, forces at opposite ends are equal and opposite and we call this the tension in the string.
8 Exaple. (one diensional otion only) Consider a train. Wheels roll freely. Locootive exerts horizontal force F on the track. What are the tensions T 1 and T 2 in the two couplings? car 2 T car 1 2 T 1 Whole train accelerates together with a. F loco Look at the external forces acting on the train (horiz. only). F F F = (++)a > a = F/3 Look at external horiz forces on car 2: T 2 T 2 = a = F/3 and on cars 2 and 1 together T 1 T 1 = 2a Newton's 3rd: A question for you: How does floor "know" to exert N = W = g?
9 Hookes law Fro Physclips Ch 6.3 No applied force (x = 0) Under tension x > 0 Note that _F spring is in the opposite direction to x. Experientally (see Physclips exaple above), _F s x over sall range of x F = kx Hooke's Law. linear elastic behaviour - ore in S2. Linear approxiations are v. useful! www.aniations.physics.unsw.edu.au/jw/elasticity.ht
10 Why is linear elasticity so coon? Consider the interolecular forces F and energies U and separate into attractive and repulsive coponents: See hoework proble on interatoic forces We'll do energy later, and we'll see:
11 Mass and weight (inertial) ass defined by F = a observation: near earth's surface and without air, all bodies fall with sae a ( = g) weight W = g What is your weight? (so far. This is still subject to experiental test!) Warning: do not confuse ass and weight, or their units kg > ass N > force (kg..s -2 ) kg wt = weight of 1 kg = g = 9.8 N Why is W? Why is g i? a = F = W = (Grav field).(grav. property of body) Mach's Principle not in our syllabus Principle of General Relativity, but v. interesting questions Interactions with vacuu field,
Exaple Grav. field on oon g = 1.7 s -2. An astronaut weighs 800 N on Earth, and, while juping, exerts 2kN while body oves 0.3. What is his weight on oon? How high does he jup on earth and on oon? g E = W E > = 800 N 9.8 s-2 = 82 kg W = g = 82kg 1.7s -2 = 140 N Vertical (y) otion with constant acceleration. While feet are on ground, Σ F = 2 kn - W E = 1.2 kn (Earth) Moon: Σ F = 2 kn - g = 1.9 kn 12 Jup has two parts: feet on ground a = Σ F v i = 0, v f = v j feet off ground a = - g v i = v j, v f = 0 v y v j y h Δ y feet leave ground y easure y of centre of ass. While on ground: v j 2 v o 2 = 2a j Δy = 2 Σ F Earth > v j = 3.0 s 1 Δy Moon > v j = 3.7 s 1 While above ground: v 2 v j 2 = 2gh > h = v j 2 2g h E = 0.5. h = 4
13 Exaple (an iportant proble) Light pulley, light inextensible string. What are the accelerations of the asses? A nuber of different conventions are possible. The iportant thing is to define yours carefully and use it consistently. Let a be acceleration (down) of 1 = acceleration (up) of 2. Take up as positive, look at the free body diagra for 1 : we are only interested in the vertical direction. By assuption, the acceleration is down. T is up. 1 g is down. So Newton 2 for 1 gives: And Newton 2 for 2 : subtract: T 1 g = 1 a T 2 g = + 2 a 1 g + 2 g = 1 a 2 a a = 1 2 1 + 2 g (Check: if 1 = 2, a = 0. If 2 = 0, a = g.) Alternatively, we ight have said a 1 be the acceleration up of 1 and a 2 be the acceleration of 2. Inextensible string so a 1 = a 2. etc.
14 Contact forces N F contact The noral coponent of a contact force is called the noral force N. The coponent in the plane of contact is called the friction force F f. F friction Noral force: at right angles to surface, is provided by deforation. If there is relative otion, kinetic friction (whose direction opposes relative otion) If there is no relative otion, static friction (whose direction opposes applied force) See the experient on Physclips Ch 6.5 Define coefficients of kinetic (k) and static (s) friction: F f = µ k N F f µ s N n.b.: Friction follows this approxiate epirical law µ and µ are approx. independent of N and of contact area. s k Often µ < µ. k s (It takes less force to keep sliding than to start sliding.)
15. Exaple. θ is gradually increased to θ c when sliding begins. What is θ c? What is a at θ c? (as before, free body diagra, N2:) Newton 2 in noral direction: N - g cos θ = 0 Newton 2 in direction down plane: g sin θ F f = a. (i) (ii) No sliding: a = 0 so total force is zero. (ii) g sin θ = F f µ s N (i) = µ s g cosθ g sin θ µ s g cosθ tan θ µ s, θ c = tan 1 µ s useful technique for µ s Sliding at θ = θ c : a > 0 (ii) a = g sin θ c F f = g sin θ c µ kn (i) = g sin θ c µ k g cosθ c we had θ c = tan 1 µ s so a = g cosθ c (µ s µ k )
16 Exaple. Rear wheel drive car, 3 kn weight on each front wheel, 2 kn on rear. Rubber-road: µ s = 1.5, µ k = 1.1 (ass of car)*g = weight = 2(3 kn + 2 kn) = 10 kn = 1 tonne Neglect rotation of car during accelerations. Assue that brakes produce 1.8 ties as uch force on front wheels as on back (why? Look at brake discs and pads). (i) What is ax forward acceleration without skidding? What is axiu deceleration for (ii) not skidding? (iii) 4 wheel skid? (i) F frs µ s N R = µ s W R on each rear wheel a ax = F = 1.5x2 kn = 2.9 kn on each rear wheel = 2*2.9kN = 5.8 s 2 1000kg Stopping. For all wheels, F fs µ s N = µ s W F ff = 1.8 F fr. µ s W F = 1.5 µ s W R, front wheels skid first, when F ff > µ s W F. 2 front and 2 rear wheels. ax total friction = (front + rear) = 2 + 2 1.8.µ sw F (brakes -> 1.8 ties as uch force on front wheels) ii) iii) = 1400 kg.wt = 14 kn a ax = F ax a = Σ F k = Σ µ kw = 14 kn 1000 kg = 14 s-2 =... = 11 s -2 Questions: Does area of rubber-road contact ake a difference? Does the size of the tire ake a difference?
17 Centripetal acceleration and force Circular otion with ω = const. and v const. eg bus going round a corner F = a centrip Or consider a haer thrower T resultant W Resultant force produces acceleration in the horizontal direction, towards the centre of the otion Centripetal force, centripetal acceleration
18 Exaple Conical pendulu. (Unifor circular otion.) What is the frequency? Free body diagra Apply Newton 2 in two directions: Vertical: a y = 0 Σ F y = 0 T cos θ W = 0 T = Horizontal: v 2 r v 2 r g cos θ = a c = T sin θ = g tan θ v = rg tan θ 2π r period = = rg tan θ g sin θ cos θ f = 1 period = 1 2π g tan θ r
19 Exaple. Foolhardy lecturer swings a bucket of bricks in a vertical circle. How fast should he swing so that the bricks stay in contact with the bucket at the top of the trajectory? r Draw diagra & identify iportant variables pose question atheatically. N W a centrip W and N provide centripetal force. g + N = a c For contact, we need N 0 so a c g how to express a c? a c = v2 r = rω 2 = r 2π T 2 T is easy to easure T = 2π r a c 2π r g r ~ 1 > T 2 s.
20 Exaple. (1,3) > stationary if F cos θ µ s N F cos θ µ s (g + F sin θ) (using (2)) F (cos θ µ s sin θ) µ s g (*) note iportance (cos θ µ s sin θ) Apply force F at θ to horizontal. Mass on floor, coefficients µ s and µ k. For any given θ, what F is required to ake the ass ove? Eliinate 2 unknowns N and F f > F(θ, µ s,,g) Stationary if F f µ s N (1) Newton 2 vertical: N = g + F sin θ (2) Newton 2 horizontal: F cos θ = F f (3) if (cos θ µ s sin θ) = 0, (F very large) θ = θ crit = tan 1 (1/µ s ). If θ < θ c, then (cos θ µ s sin θ) > 0 stationary if F i.e. oves when F > F crit = What if (cos θ µ s sin θ) = 0? θ = θ c,? µ s g cos θ µ s sin θ µ s g cos θ µ s sin θ (*) stationary if F µ s g cos θ µ s sin θ i.e. stationary no atter how large F becoes.
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22 Exaple Plane travels in horizontal circle, speed v, radius r. For given v, what is the r for which the noral force exerted by the plane on the pilot = twice her weight? What is the direction of this force? Centripetal force F = v2 r = N cos θ Vertical forces: N sin θ = g eliinate θ: N 2 = 2 v4 r 2 + g2 N 2 2 g2 = v4 r 2 > r = v 2 N 2 2 g2 sin θ = g N = 1 2 30 above horizontal, towards axis of rotation Question. Three identical bricks. What is the iniu force you ust apply to hold the still like this? N F f F f N g Vertical forces on iddle brick add to zero: 2 Ff = g Definition of µs Ff µs N N F f µs = g 2µs Bricks not accelerating horizontally, so noral force fro hands = noral force between bricks. (each) hand ust provide g 2µs horizontally. Vertically, two hands together provide 3g. 3g/2 iniu force fro hands g/2µ s