Mechanics for systems of particles and extended bodies

Mechanics for systems of particles and extended bodies PHYS2-3 UNSW. Session Left: trajectory of end of rod. Right: parabola is the trajectory followed by the Centre of mass In a finite body, not all parts have the same acceleration. Not even if it is rigid. How to apply F_ = m a_? Total mass M = Σ m i m j Define the centre of mass as the point with displacement m i r i r cm r j r_ cm = Σ m i r_ i M origin Why this definition? Consider n particles, m i at positions r_ i, F_ i acts on each. For each particle, N2 gives F_ i = m i a_ i Add these to get total force acting on all particles: m i F i r i origin F j r j m j Σ F_ i = Σ m i a_ i = Σ m i d 2 dt 2 r_ i definition of acceleration if masses constant, can change order of d/dt and multiply: = Σ d2 dt 2 m i r_ i = d2 dt 2 Σ m i r_ i = M d2 dt 2 Σ m i r_ i M multiply top and bottom by M But we defined r_ cm = Σ m i r_ i M Then Σ F_ i = M d2 dt 2 r_ cm = M a_ cm (total force) = (total mass)*(acceleration of centre of mass)

2 Look at forces in detail: F F j j,ext F i F m i,ext j Each F_ i is the sum of internal forces (from other particles in the body/ system) and external forces (from outside the system) m i F i,int F j,int Σ F_ i = Σ F_ i,internal + Σ F_ i, external Newton 3: All internal forces F_ ij between i th and j th particles are Newton pairs: F_ ji = F_ ij Σ internal forces = 0 Σ F_ i = Σ F_ i, external = F_ external F_ external = M a_ cm total external force = mass total * acceleration of centre of mass

3 For n discrete particles, centre of mass at r_ cm = Σ m i r_ i Σ m i = Σ m i r_ i M (i) For a continuous body, elements of mass dm at r_ r_ cm = r_ dm body body Can rearrange (i): dm = r_ dm body M (ii) This is the same equation. Really. 0 = Σ m i r_ i m i r_ cm M > Σ m i (r_ i r_ cm) = 0 law of the see-saw (ii) > (r_ i r_ cm)dm = 0 body cm N Later, when doing rotation, we'll consider which is a useful way to find c.m. experimentally. Three boys with equal mass m: W

4 Example. Where is the c.m. of the earth moon system? r_ cm = Σ m i r_ i Σ m i Take origin at centre of earth. x cm = m ex e + m m x m m e + m m = m m d m e + m m = 4,600 km i.e. inside the earth. recall: when doing Newton's 3rd, we derived the centre of rotation of this system Example m m 2 r plate On a square plate (mass m p ), we place m and m 2 as indicated. m p = 35 g, m = 00 g and m 2 = 50 g Where is the cm of the system? O r_ cm = Σ m i r_ i Σ m i = m p(.5i +.5j) + m (2.0j) + m 2 (2.0i + 2.0j) m p + m + m 2 = (303g)i + (503g)j 285g =. i +.8j Check? check that Σ m i (r_ i r_ cm) = 0

5 Example. Rod, cross-section A, made of length a of material with density ρ 2 and length b of material with density ρ. Where is c.m.? If ρ / = 2ρ 2, and a = 2b, where is cm? r_ cm = dm x cm = " r i dm " dm = ρdv = ρadx " xdm " dm How am I going to integrate dm? and over the whole body? Put origin at join and ρ is constant for the integrations = 0 ρ Ax dx + a -b 0 0 ρ A dx + a -b 0 ρ 2 Ax dx ρ 2 A dx messy? But it's only four easy integrations = = 2 ρ b 2 + 2 ρ 2 a 2 ρ b + ρ 2 a a 2 rb 2 2(a + rb) where we define r = ρ ρ 2 Internal vs external work. Problem. Skateboarder pushes away from a wall F F v Point of application of F_ does not move, normal force does no work, but K changes. Where does energy come from? Obvious: arms! F ext = Ma cm F ext dx = Ma cm dx cm = M dv cm dx cm dt = Mv cm dv cm "Centre of mass work" f F ext dx = ( ) 2 Mv 2 cm f ( ) W cm = 2 Mv 2 cm i i Work done = that which would have been done if F ext had acted on cm.

6 Momentum Definition: p_ mv_ Generalised form of In relativity, we ll find that this is a low v approximation to mv_ p_ = - v 2 /c 2 and also that K = (γ )mc 2 Newton 2: Σ F_ = d dt p_ Σ F_ = m d dt v_ + v_ d dt m If m constant, Σ F_ = m a_ System of particles: but for the general case, use the general expression What is system? - you choose: draw a boundary around it. P_ = Σ p_ i and M = Σ m i P_ P_ = Σ m i v_ i = Σ m d i dt r_ i d = dt Σ m i r_ i = M d Σ m i r_ dt i M = Mv_ cm If M constant: d dt P_ = Ma_ cm Σ F_ i = Σ d dt p_ i = d dt P_ All internal forces are in pairs F_ ji = F_ ij F_ ext = d dt P_ two very important conclusions: i) Motion of cm is like that of particle mass M at r_ cm subjected to F_ ext. ii) If F_ ext = 0, momentum of whole system is conserved Note that momentum is a vector, so we have a conservation law that can apply in one or more directions.

7 Example 90 kg man jumps (v j = 5 ms - ) into a (stationary) 30 kg dinghy. What is their final speed? (Neglect friction.) v i m m m d v f No external forces act in horizontal direction so P x is conserved. P i = P f man dinghy man dinghy m m v j + 0 = (m m + m d )v f v f = m m m m + m d v j Example Rain falls into an open trailer (area 0 m 2 ) at 0 litres.min -.m -2. Neglecting friction, what F required to maintain constant speed of 0 ms -? F x = d dt (mv x) = m d dt v x + v x d dt m = 0 ms - x 0 kg.m -2 60 s 0 m 2 = 7 N. 0 litres has mass 0 kg

8 Example. Rocket has mass m = m(t), which decreases as it ejects exhaust at rate r = dm dt and at relative velocity u. What is the acceleration of the rocket? dm rate of increase of dt = mass of rocket < 0 No external forces act so momentum conserved. In the frame of the rocket, forwards direction: dp rocket + dp exhaust = 0 m.dv + ( dm).( u) = 0 dv = u dm m a a = dv dt = u m. dm dt = ur m st rocket equation dv = u dm m = dv = u d(ln m) f i dv = v f - v i = u ln m i m f 2nd rocket equation need high exhaust velocity u (c?), else require m i >> m f

9 Collisions time. Definition: in a collision, "large" forces act between bodies over a "short" In comparison, we shall often neglect the momentum change due to external forces. Example : v a v b F ab = F ba = 0 F ab F ba F ab = F ba = large! F ab = F ba = 0 forces that crumple cars during (brief) collision are much larger than friction force (tires - road), neglect F ext. Be quantitative: suppose car decelerates from 30 kph to rest in a 20 cm 'crumple zone'. Approximate as constant acceleration a = (v f 2 vi 2 )/2Δx = 70 ms -2, so force on car during collision ~ m a ~ 200 kn, compared with friction at ~ 0 kn. Other safety questions while we are here: What is the force on occupants? If they have the same acceleration as the car, then F = ma ~ 0 kn: seat belts, air bags. What about reducing crumple zones? Forces on pedestrians? How big are crumple zones for pedestrians?

0 Example 2 spacecraft Jupiter doesn't "hit" Examples: deep space probes Here, start and finish of collision not well defined At large separation before and after, Fab = Fba 0 During collision (fly-by), forces are considerable. However, Fgrav /r2, so much smaller at large distances. Impulse (J _ ) and momentum Newton 2 dp _ dt _ = F f dp _ = i Definition: I f F _ dt i so f p _ dt _f p _i = F i In collisions, Impulse is integral of large internal force over short time Ball is inflated to normal pressure. Can get an underestimate of force: pressure F > in ball * deformed area ~ 70 kpa* 0.02 m2 ~ kn Camera flashes at equal times When is head of club travelling fastest? Speed of ball compared to speed of club?

Usual case: external forces small, act for small time, therefore i f F_ ext dt is small. F F 2 m m 2 Δ p _ = i f F_ dt - F_ Δt Δ p _ 2 = i i f F_2 dt = f F_ dt Δ p _ = Δ p _ 2 Δ P_ = Δ p _ Δ p _ 2 = 0 If external forces are negligible (in any direction), then the momentum of the system is conserved (in that direction). Example. Cricket ball, m = 56 g, travels at 45 ms -. What impulse is required to catch it? If the force applied were constant, what average force would be required to stop it in ms? in 0 ms? What stopping distances in these cases? v i v f = 0 m = 0.56 kg, v i = 45 m.s - v f = 0. I p _ f p _ i I = m(v f - v i ) to right =... = 7.0 kgms - to left. = f F_ dt if F_ constant, I = FΔt. i F av = I/Δt. If const F const a. s = v av Δt. v av = 23 m.s - Δt.0 ms 0 ms F 7 kn 0.7 kn ouch! s 2 cm 20 cm

2 Example. (A common method to measure speed of bullet.) Bullet (m) with v b fired into stationary block (M) on string. (i) What is their (combined) velocity after the collision? (ii) What is the kinetic energy of the bullet? (iii) of the combination? (iv) How high does the block then swing? a) b) c) v b m M M+m v t h v=0 Note the different stages and three diagrams: a-b):collision, no horizontal external forces momentum conserved. Friction does work, so mechanical energy is lost, not conserved b-c): during this phase, external forces do act, so momentum is lost, not conserved. However, there are no non-conservative forces, so mechanical energy conserved. a) b) Analyse a) to b) v b m M M+m v t No horizontal ext forces during collision momentum conserved i) P xi = P xf mv b = (m + M)v t v t = m m + M v b ii) K b = 2 mv b 2 iii) K t = 2 (m + M) v t 2 = 2 (m + M) m m + M v b 2 = m 2 2 m + M v b 2 < K b. Conclusion: U i = U f, K i =/ K f. Mechanical energy is not conserved - deformation of block is not elastic; heat is produced. Let s look in more detail: little digression about elastic and inelastic collisions

3 During a collision with negligible external forces, P_ = (Σm) v_ cm is conserved Σm constant v_ cm is constant 2 Mv_ cm 2 constant K of c.m. is not lost. But the K of components with respect to c.m. can be lost. Greatest possible loss of K: if all final velocities = v_ cm, i.e. if all objects stick together after collision. Called completely inelastic collision. Completely inelastic collision. v i v 2i v m m 2 m + m 2 Contrast: Completely elastic collision is one in which non-conservative forces do no work, so mechanical energy is conserved. Inelastic collision is one in which non-conservative forces do some work, so mechanical energy is not conserved. Completely inelastic collision is one in which all kinetic energy with respect to the centre of mass is lost: Non-conservative forces do as much work as possible, so as much mechanical energy as possible is lost Part (iv) of previous example (b-c): b) c) M+m v t h v=0 here the external forces (gravity and tension) do do work and change momentum. But there is no non-conservative force and so in this part of the process conservation of mechanical energy applies: ΔU + ΔK = 0 (M + m)g (Δh 0) + (0 K t ) = 0 Δh =... = 2 m 2 g(m + M) 2 v b 2 so we can rearrange and get v b from Δh

4 Example Elastic collision in one dimension v i v 2i v f v 2f (bang) m m 2 m m 2 note the before and after diagrams again Collision: neglect external forces p i = p f m v i + m 2 v 2i = m v f + m 2 v 2f (i) elastic K i = K f 2 m v i 2 + 2 m 2 v 2i 2 = 2 m v f 2 + 2 m 2 v 2f 2 (ii) usually know m, m 2, v i, v 2i. Two unknowns (v f, v 2f ), we can always solve. Or: transform to frame where (e.g.) v = 0 can simplify the algebra Or: transform to centre of mass frame. Example. Take m = m 2, v 2i = 0, v i = v. v v 2i = 0 v f v 2f m m m neglect external forces mv + 0 = mv f + mv 2f m p i = p f (i) 2 mv 2 + 0 = 2 mv f 2 + 2 mv 2f 2 (ii) (i) > v 2f = v v f (iii) substitute in (ii) > 2 mv 2 + 0 = 2 mv f 2 + 2 m(v 2 + v f 2-2vv f ) 0 = v f 2 vv f 0 = v f (v f v) 2 solutions Either: v f = 0 and (iii) > v 2f = v i.e. st stops dead, all p and K transferred to m 2 or: v f = v and (iii) > v 2f = 0 i.e. missed it.

5 Example Show that, for an elastic collision in one dimension, the relative velocity is unchanged. p and K conservation gave: (i) m (v f v i ) = m 2 (v 2f v 2i ) i.e. show v i - v 2i = v 2f - v f (ii) 2 m (v f 2 v i 2) = 2 m 2 (v 2f 2 v 2i 2) If they hit, (v f v i ) /= 0, (v 2f v 2i ) /= 0 use a 2 b 2 = (a b)(a + b) (ii)/(i) v i + v f = v 2i + v 2f v i v 2i = v 2f v f i.e. relative velocity the same before and after Solve > v 2f = 2m m + m 2 v i + m 2 - m m 2 + m v 2i Example Two similar objects, mass m, collide completely inelastically. case : v i = v, v 2i = 0. case 2: v i = v, v 2i = v. What energy is lost in each case? p conserved > v f = v i + v 2i 2 mv i + mv 2i = 2mv f ΔΚ = K f K i = 2 (2m) v f 2-2 mv i 2-2 mv 2i 2 case : ΔK = 2 (2m) v + 0 2 2-2 mv 2 = 4 mv 2 case 2: ΔK = 2 (2m) 0 + 0 2 2-2 mv 2-2 mv 2 = mv 2 4 times as much energy lost Remember this if you have the choice in traffic, rugby etc

6 Elastic collisions in 2 (& 3) dimensions b y v 2f! 2 x v i v f! Choose frame in which m 2 stationary, v i in x dir n b is called impact parameter (distance "off centre") p x conserved m v i = mv f cos θ + mv 2f cos θ 2 p y conserved 0 = mv 2f sin θ 2 mv f sin θ K conserved 2 m v i 2 = 2 mv f 2 + 2 mv 2f 2 + ΔK (iii) where ΔK = 0 for elastic case 3 equations in v f, v 2f, θ and θ 2 : need more info (often given θ or θ 2 ) Incidentally: for hard spheres, neglecting rotation and friction (reasonable during collision, but not after) b! 2 (R + R) sin θ 2 = b θ 2 = sin - b 2R i) Note that as θ > 90, small error in b gives large error in θ 2. ii) Experiment on billiard table: Does b = R give θ 2 = 30? friction, rotation ignored

7 v v! s m m C v 2 " s 2 m 2 Example. Police report of road accident. Car, mass m strikes stationary car m 2 at point C. They then slide to rest in positions shown. Given µ k = µ (assumed same for both) find the initial speed v of m. Can you check assumption? (real example) After collision, a for both = F f m = µ W m = µg v f 2 v i 2 = 2as = 2µgs 0 v 2 = 2µgs v = 2µgs v 2 = 2µgs 2 Neglect external forces during collision: ΔP = 0 P x : m v = m v cos θ + m 2 v 2 cos φ P y : 0 = m v sin θ m 2 v 2 sin φ (ii) (i) v = (i) 2µgs cos θ +(m 2 /m ) 2µ 2 gs 2 cos φ Note the "spare" equation we can use it to check the model or assumptions: (ii) m 2µ gs sin θ = m 2 2µ 2 gs 2 sin φ µ 2 µ = s m 2 sin 2 θ s 2 m 2 2 sin 2 φ (The µ may not be the same for the two: surfaces different, orientation of wheels etc)

8 y x v z v b v f! Balloonist Albert writes message on a bottle ( kg) and drops it over the side. It is falling vertically at 40 m.s - when caught by parachutist Zelda (m = 50 kg), travelling at ms - at 45 to vertical. Collision (bottle Zelda's hand) lasts 0 ms. i) If only gravity acted, what is Δp for Zelda over 0 ms? ii) Neglecting ext forces during collision, what is the velocity of (Zelda+bottle) after collision? iii) What impulse is applied to bottle during collision? iv) What is the impulse applied to Zelda? v) What is the average force during collision? vi) Will Albert and Zelda live happily ever after? i) due to W_, Δp_ = W_ Δt =.. = 5 kgm.s - down ii) Neglect ext forces momentum conserved. m b v_ bi + m Z v_ Zi = m (Z+b) v_ (Z+b) f (-40 j _ ) + 50( cos 45 i _ cos 45 j _ ) = 5(v x i _ + v y j _ ) i_ dirn : v x = cos 45. 50 5 j_ dirn : v y = x = 0.7 ms- x40 50 cos 45 5 =.5 ms - v f = v x 2 + v y 2 =.6 ms - θ f = tan - v y v x 67 to horizontal iii) I b p _ bf p _ bi = x(v x i _ + v y j _ ) - (-40 j _ ) = (.6 i _ + 38 j _ ) kgm.s - iv) I Z = I b = (.6 i _ + 38 j _ ) kgm.s - I Z =.6 2 + 38 2 = 38 kgm.s - v) - F_ Z = Δ p _ Z Δt = I Z Δt =.. = 380 N

9 S h n 2 G Example Controlled demolition. A building has S stories, each of height h. Explosions destroy the strength of the n th floor. How long before (n+) th floor hits ground, falling vertically? Assume that the building remains intact above the explosion and inelastic collisions with the lower floors. To obtain a lower bound, assume negligible strength between lower floors. (S n) floors have mass (S n)m. To get the lower estimate on falling time, assume no strength in the demolished floor, s the upper floors are in free fall (as a rigid body) for a distance h with acceleration g. So they strike the next floor with speed v n = 2gh. Inelastic collision with next floor gives speed v where: (S n)mv n = (S n+)mv Let the falling mass after any collision have initial speed v 0 and speed before the next collision be v c. v c 2 v 0 2 = 2gh v c = 2gh + v 0 2 = v 0 + 2gh v 0 2 /2 For i th collision (S n+i )mv c (i) = (S n+i)mv 0 (i+) v c (i) = 2gh + (v 0 (i)) 2 S n+i S n+i 2gh + (v 0 (i)) 2 = v 0 (i+) v 0 (i+) = S n+i 2gh + (v 0 (i)) 2 h = v 0 t + 2 gt2 0 = 2 gt2 + v 0 t h t = v 0 ± v 0 2 + 2gh g