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Chapter-4

Chapter-4. Motion in Two and Three Dimensions. Ch 4-2 Position and Displacement in Two and Three Dimensions. Position vector r in three dimension r= x i + y j + z k x= -3 m, y= + 2m and z=+5m. Ch 4-2 Position and Displacement in Three Dimensions. Displacement r r = r 2 -r 1

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Chapter-4

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  1. Chapter-4 Motion in Two and Three Dimensions

  2. Ch 4-2 Position and Displacement in Two and Three Dimensions • Position vector r in three dimension • r= xi + yj+ zk • x= -3 m, y= + 2m and z=+5m

  3. Ch 4-2 Position and Displacement in Three Dimensions Displacement r r = r2-r1 where r1= x1i+y1j+z1k r2= x2i+y2j+z2k r= (x2- x1) i+(y2- y1) j+ (z2 – z1) k r = (x) i+(y) j+ (z) x = x2- x1;y = y2- y1 ; z = z2- z1

  4. Checkpoint 4-1 • r = (x)i+ (y)j + (z)k x= 6-(-2)=8 y= -2-4= -6 z= -3-(-3)=0 • r = 8i - 6j (b) The displacement vector r is parallel to xy plane • (a) If a wily bat flies from xyz coordinates ( -2m, 4 m, -3 m) to coordinates ( 6 m, -2 m, -3 m), what is its displacement r in unit vector notation? (b) Is r paralell to one of the three coordinates planes? If so, which one?

  5. Ch 4-3 Average and Instantaneous Velocity • Average velocity vavg vavg= displacement/time interval vavg = r /t r/t = (x/t) i+(y/t) j+ (z/t)k • Instantaneous velocity v v= dr/dt v= (dx/dt) i+(dy/dt) j+ (dz/dt) k v=vxi+vyj+vzk • Direction of v always tangent to the particle path at the particle position

  6. The figure shows a circular path taken by a particle. If the instantaneous velocity of the paticle is v= (2 m/s) i- (2 m/s) j , through which quadrant is the particle moving at that instant if it is traveling (a) clockwise (b) counterclockwise around the circle? Ans:v= (2 m/s) i- (2 m/s) j (a) clockwise motion vx= +ve and vy= -ve only in First quadrant (b) counterclockwise motion vx= +ve and vy= -ve only in Third quadrant Checkpoint 4-1(new book)

  7. Ch 4-4 Average and Instantaneous Acceleration • Average acceleration aavg aavg = change in velocity/time aavg = v/t =(v2-v1)/ t • Instantaneous acceleration a a= dv/dt a= (dvx/dt)i+ (dvy/dt)j+ (dvz/dt)k a=axi+ayj+azk

  8. Here are four description of the positions of a puck as it moves in the XY plane (1) X =-3t2+4t-2 and y= 6t2 – 4t (2) X = -3t3 -4t and Y = -5t2 +6 (3) r = 2t2 i –(4t +3) j (4) r = (4t3 -2)i +3 j (1) d2x/dt2 = -6 , d2y/dt2=12 (2) d2x/dt2 =-18 t ,d2y/dt2=-10 (3) d2r/dt2 = 4 i , (4) d2r/dt2 = 24 t i , ax is constant for 1 and 3,ay is constant for both ay is so a is constant ax is not constant for 2 and 4 while ay is constant. Hence a is not constant for case 2 and 4 Checkpoint 4-2

  9. If the position of a hobo’s marble is given by r=(4t3-2t)i+3j, with r in meters and t in seconds, what must be the units of coefficients 4, -2 and 3? r=(4t3-2t)i+3j =4t3i-2ti+3j Since units of 4t3-2t and 3 has to be meter then Unit of 4 is m/s2 Unit of -2 is m/s Unit of 3 is m Checkpoint 4-4

  10. Ch 4-5 Projectile Motion • Motion in two dimension: • Horizontal motion (along x-axis ) with constant velocity, ax=0 • Vertical motion (along y-axis) with constant acceleration ay= g • Horizontal motion and the vertical motion independent of each other • v0=v0xi+v0yj • v0x=v0 cos ; v0y=v0 sin

  11. Ch 4-5 Projectile Motion • Horizontal motion: x-x0=v0xt= v0cost • Horizontal Range R : R is the horizontal distance traveled by the projectile when it has returned to its initial launch height R= v0 cos t= (v02sin2)/g; Rmax=v02/g (=45)

  12. Ch 4-6 Projectile Motion Analyzed • Vertical motion : y-y0=v0yt – gt2/2 = v0 sint – gt2/2 vy= v0y t – gt = v0 sint – gt • Max. height h= v0y2 /2g  • Projectile path equation: Projectile path is a parabola given by y=(tan) x - gx2/2vx2

  13. Formule Summary for Projectile Motion

  14. A fly ball is hit to the outfield. During its flight (ignore the effect of the air), what happens to its (a) horizontal and (b)vertical components of velocity What are its (c) horizontal and (d) vertical components of its acceleration during ascent, during descent, and at the topmost point of its flight? vx=constant vy initially positive and then decreases to zero and finally it increases in negative value. ax= 0 ay= -g throughout the entire projectile path Checkpoint 4-4

  15. Ch 4-7 Uniform Circular Motion • Uniform Circular Motion: • Motion with constant speed v in a circle of radius r but changing speed direction • Change in direction of v causes radial or centripetal acceleration aR aR =v2/r • Period of motionT: T = (2r)/v

  16. Checkpoint 4-6 • Object has counterclockwise motion Then at y=2 m , its velocity is v= –(4m/s)i and centripetal acceleration magnitude aR=v2/R = (4)2/2 = 8 m/s2 aR=(-8m/s2)j An object moves at constant speed along a circular path in a horizontal xy plane, with the center at the origin. When the object is at x=-2m, its velocity is –(4m/s)j. Give the object (a) velocity and (b) acceleration when it is at y= 2m

  17. Ch 4-8 Relative Motion in One Dimension • Relative position xPA=xPB+xBA • Relative Velocities d/dt(xPA)=d/dt(xPB)+d/dt (xBA) vPA= vPB + vBA • Relative Acceleration d/dt(vPA)=d/dt(vPB)+d/dt (vBA) • aPA=aPB (vBA is constant)

  18. Ch 4-8 Relative Motion in Two Dimensions • Two observers watching particle P from origins of frames A and B, while B moves with constant velocity vbA with respect to A • Position vector of particle P relative to origins of frame A and B are rPA and rPB.If rBA is position vector of the origin of B relative to the origin of A then • rPA = rPB. + rBA and • vPA = vPB. + vBA • aPA = aPB. Because vBA = constant

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