Lecture3 - From Lecture 2: vxf = vxi + ax t 12 x f = xi...

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Unformatted text preview: From Lecture 2: vxf = vxi + ax t 12 x f = xi + vxi t + ax t 2 v 2f = v 2i + 2 ax ( x f − xi ) x x y y Ay O Ax = Acosθ A = A 2 + A 2 x y Ay = Asinθ θ = tan −1 ⎛ Ay ⎞ ⎜A ⎟ ⎝ x⎠ Ax x Once have Rx and Ry , you get R In terms of unit vectors: ˆ ˆ R = Rx i + Ry ˆ = ( Ax + Bx ) i + Ay + By ˆ j j A θ R B O A R= A+ B Rx = Ax + Bx Ry = Ay + By x ( ) Ques%on 1: If velocity of a par;cle changes from m/s in 5 seconds, what is its average accelera;on in that interval? ˆ j ˆ j vi = 3i + ˆ m/s to v f = i + 2 ˆ Ques%on 2: Using unit vectors, redo the problem (the one we did on board in the last class) of finding the average accelera;on of a par;cle in uniform circular mo;on when the par;cle covers 90o in the circle. From Lecture 2: Ai B = AB cosθ A × B = AB sin θ z A× B Direc1on of cross product is given by the right ­hand rule y x A B A liEle bit more on vectors Ai B = Bi A A × B = −B × A (Cross product is not commuta1ve) If A and B are given by A = Ax iˆ + A ˆ + Az k ˆ yj then, A ⋅ B = Ax Bx + Ay By + Az Bz A× B= ˆ i Ax Bx ˆ j Ay By ˆ k Az Bz ˆ ˆ B = Bx i + By ˆ + Bz k j ˆj iiˆ = ˆˆ i ii = ˆ i k = k ii = 0 j ˆ ˆˆ ˆi ˆ = k i k = 1 jj ˆˆ ˆ ˆ = i ( Ay Bz − Az By ) − ˆ( Ax Bz − Az Bx ) + k ( Ax By − Ay Bx ) j Mo1on in two dimension Posi1on, displacement and velocity •  Posi1on vector r = xiˆ + yˆ j y vi ri rf vf •  Displacement: ˆ Δ r = rf − ri = ( x f − xi )i + ( y f − yi ) ˆ = i Δx + ˆΔy jˆ j •  A par1cle has posi1on vector r i and aGer 1me Δt its new posi1on vector is r . The average velocity of the par1cle is f Δr ˆ Δx ˆ Δy vavg = =i +j Δt Δt Δt O x In the limit of Δt approached 0, we get the instantaneous velocity dr ˆ dx ˆ dy ˆ v= =i +j = ivx + ˆvy j . dt dt dt Accelera1on: Δv ˆ Δvx ˆ Δvy •  The average accelera1on is given by aavg = =i +j Δt Δt Δt in the limit Δ → 0 , this becomes instantaneous accelera1on: t dv ˆ dvx ˆ dvy ˆ d 2 x ˆ d 2 y ˆ a= =i +j = i 2 + j 2 = iax + ˆay j dt dt dt dt dt •  Note that a par1cle may accelerate because  ­ the magnitude of the velocity changes but direc1on remains the same  ­ magnitude remains the same but direc1on changes  ­ both magnitude and direc1on of velocity change Ques;on: examples? By now, you may have started to realized that … Mo1on in two dimensions can be modeled as two independent mo1ons in each of the two perpendicular direc1ons associated with the x and y axes. Any influence in the y direc1on does not affect the mo1on in the x direc1on and vice versa. Or, in other words, when you are solving a 2D mo;on problem, you can, for example, solve the x ­mo;on without caring at all about y ­mo;on at first, and then do y ­mo;on without caring about x ­mo;on. This is a sort of divide and conquer: a 2D mo;on is in fact broken up into a set of two independent 1D mo;ons. We can do the same thing with 3D mo;ons. Example The coordinates of a dog running across a field can be described as a func1on of 1me as x = 0.3t2 ­5.6t+3, y =  ­0.2t2+4t+2 (in meters)   What is the posi1on vector at 1me t=0.4 s? Answer: At t=0.4 s: x = 0.8, y = 3.6, therefore:   Find posi1on at t = 1.0 s: x =  ­2.3 and y = 5.8.   What is the average velocity of the dog between ti=0.4 s and tf=1 s and what is its instantaneous velocity at these points in 1me ?   Ques1on: What’s the dog’s accelera1on at t = 0.4s? Projec1le mo1on Mo1on of baseball in air is an example of projec1le mo1on. The ball moves in a curved path and returns to the ground. The 2D mo1on shown in the figure is a mo1on of a projec1le fired with a velocity vi at an angle θ . In this kind of mo1on we assume •  the free ­fall accelera1on is constant in the range of mo1on and is directed down •  the effect of air resistance is negligible. y vx vy vx x θi Time to reach the maximum height Assume that a projec1le is launched vy vx viy = visinθ vi at t = 0, as indicated in the figure. θi Take the ver1cal mo1on (y mo1on). vix = vicosθ At maximum height vy = 0,. We can use this formula: vyf = vyi + ayt Here, vyi vi s in θ i , and ay = − g . Now get 0 = vi sin θ i + (− g )t = vy t= vi sin θ i g Projec1le mo1on (contd …) Maximum height We can use the following kinema1c formula to get the expression for maximum height, say h: 2 2 Here y f − yi = h and we get v yf = v yi + 2 ay ( y f − yi ) 0 = vi2 sin 2 θ + 2(− g )h vi2 sin 2 θ i h= 2g Total 1me in air 1 y f = yi + vyi t + a yt 2 and To find the total 1me in air (for level surface), we can use 2 note that yi = yf = 0: 1 2 v sin θ i 0 = 0 + (vi sin θ )t + (− g )t 2 t= i 2 g Note that this is exactly twice the 1me to reach the maximum height. Thus (for level surface) it takes equal 1me to ascend to maximum height and to descend from there. Note also that the we did not men1on the other other solu1on t = 0 of the quadra1c equa1on. It is trivial and simply means that the par1cle might not have moved at all (a possibility since yi = yf = 0). Projec1le mo1on (contd …) Horizontal range This is the horizontal distance between the point of projec1on and the point of return of the projec1le at the same horizontal level and we denote it by R. We can easily get an expression for it by solving the horizontal mo1on and use the total 1me of flight for level surface. Note that there is no accelera1on in the x ­mo1on. x f = xi + vxi t = 0 + (vi cosθ i )t ⎛ 2 v sin θ i ⎞ R = (vi cosθ i ) ⎜ i ⎟ ⎝ ⎠ g vi2 sin 2θ i R= g Ques1on: At what angle you should project in order to achieve maximum range? Ques1on: Use x mo1on and y mo1on to eliminate t in the kinema1c rela1ons for a projec1le to get the rela1on between x and y. The resul1ng equa1on given below tell us about the shape of the trajectory of a projec1le. Can you tell what is the shape ? 2 g ( x − xi ) y − yi = (tan θ i )( x − xi ) − 2 2 ( vi cosθ i ) Posi1on and velocity at any 1me t We can write velocity and posi1on vectors of a projec1le as ˆ ˆ v (t ) = vi + gt = vix i + (viy − gt ) ˆ = vi cosθ i + (vi sin θ − gt ) ˆ j j 1 rf = ri + vi t + gt 2 2 Example How fast does a motorcycle have to go to jump a gap between two ramps placed D=77 m apart, when he lands half ­way down on the landing ramp. Ramps: H=3 m high, angled at θi =120. y H 2 2 Landing x y − yi = (tan θ i )( x − xi ) − 2 ( vi cosθ i ) g ( x − xi ) D − − − (1) yi = H and y = H/2 (Ini1al & final height) xi = 0 and with tan θ = H/L we find x = D + H/(2 tanθi )=84.1 m 1 g( x − xi )2 2(cosθ i )2 ( x − xi ) tan θ i − ( y − yi ) Solve (1) for vi: vi = Plug in numbers: vi = 42.3 m/s = 96.7 mph = 155.6 km/h The par1cle in uniform circular mo1on Δv = v f − vi = v f + (− vi ) Even though the magnitude of the velocity has not changed, The direc1on has; which leads to an accelera1on + vi In the limit we get instantaneous accelera1on: dv a= dt ⇒ dv = a dt Note that the direc1on of accelera1on is in the direc1on of the increment in velocity. Since magnitude of the velocity (speed) is not changing, accelera1on should have no component in the direc1on of the velocity at any 1me. This means that the direc1on of the accelera1on is perpendicular to the velocity all the 1me. accelera1on is always toward the center to the circle. centripetal accelera1on Circular mo1on analyzed Note: v and r are constant (constant speed and radius (circle)): dv ⎛ v dy p ⎞ ˆ ⎛ v dx p ⎞ ˆ a= = ⎜− i +⎜ j ....( 3) ⎝ r dt ⎟ ⎠ ⎝ r dt ⎟ ⎠ dt v ˆ a= − vyi + vx ˆ ...( 4 ) j r ( ) What is the angle between a and v? Use (1) and (4) Magnitude of centripetal accelera1on v2 From (4), a = − cosθ iˆ + sin θ ˆ j r ( ) v2 a= r Period of rota1on T (the 1me to complete one cycle) of uniform circular mo1on is given by 2π r T= v Example A jet pilot can withstand about 4g accelera;on before there is a chance that she might pass out. If her jet plane is traveling at 1500 km/h, what radius arc can she fly without having the risk of her death? v = 1500 km/h = 1500 km/h x 1000 m/km x 1 h/3600 s = 416.7 m/s Rela1ve Mo1on What is the velocity of car P ? A B P Depends on the observer (A or B) ! Reference Frames In Physics we have to use some “frame” to which we reference our measurements and “aEach” our coordinate system (in our mind). Typically this frame is “aEached” to the observer. Reference Frame: The physical object (observer) to which we “aEach” our coordinate system. Certain frames are equivalent and give the same physical laws. They are called “Iner1al Reference Frames” Assume the observers B moves with constant velocity vBA rela1ve to the observer A. They both observe a moving object P. What is its velocity and accelera1on rela1ve to each observer ? We no1ce that Take 1me deriva1ve: Take deriva1ve again: r = r + rB PA PB A vPA = vPB + vBA aPA = aPB The second equa1on is usually called Galilean velocity addi1on and the frames in which this is true are called Galilean reference frames. If the one of the reference frames were accelerated, the result aPA = aPB would not hold. Then in the accelerated frame, we would observe addi1onal accelera1ons which would not appear in the non ­accelerated frame. Such an accelerated frame is called a “non ­iner1al reference frame” Rela1ve velocity We can use Galilean velocity addi1on formula to calculate rela1ve velocity of an object with respect to another object. For example, if the driver B knows the the v P B ) and his own velocity with velocity of velocity of P with respect to him ( respect to the road ( v B A ), then he can calculate the velocity of P with respect to the the standing observer ( v PA ), that is, P’s velocity with respect to the road. All he has to do is the vector addi1on. As an example of this kind of problems solving let’s see the worked out problem, Example 4.8 (A), from the text (chalk board stuff) ...
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