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Motion in Two and Three Dimensions

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(a) If a wily bat flies from xyz coordinates ( -2m, 4 m, -3 m) to coordinates ( 6 ... Ch 4-8 Relative Motion in Two Dimensions ... – PowerPoint PPT presentation

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Title: Motion in Two and Three Dimensions


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 z5m

3
Ch 4-2 Position and Displacement in Three
Dimensions
  • Displacement ?r
  • ?r r2-r1
  • where r1 x1iy1jz1k
  • r2 x2iy2jz2k
  • ?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
  • vvxivyjvzk
  • Direction of v always tangent to the particle
    path at the particle position

6
Checkpoint 4-1(new book)
  • 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

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
  • aaxiayjazk

8
Checkpoint 4-2
  • (1) d2x/dt2 -6 , d2y/dt212
  • (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
  • Here are four description of the positions of a
    puck as it moves in the XY plane
  • (1) X -3t24t-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

9
Checkpoint 4-4
  • r(4t3-2t)i3j
  • 4t3i-2ti3j
  • 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
  • If the position of a hobos marble is given by
  • r(4t3-2t)i3j, with r in meters and t in
    seconds, what must be the units of coefficients
    4, -2 and 3?

10
Ch 4-5 Projectile Motion
  • Motion in two dimension
  • Horizontal motion (along x-axis ) with constant
    velocity, ax0
  • Vertical motion (along y-axis) with constant
    acceleration ay g
  • Horizontal motion and the vertical motion
    independent of each other
  • v0v0xiv0yj
  • v0xv0 cos? v0yv0 sin?

11
Ch 4-5 Projectile Motion
  • Horizontal motion
  • x-x0v0xt 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
  • Rmaxv02/g (?45?)

12
Ch 4-6 Projectile Motion Analyzed
  • Vertical motion
  • y-y0v0yt 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
Equation of motion Horizontal Motion (ax0) Vertical Motion (ay g)
vavg (vivf)/2 vavg v0x v0 cos? Vavg (v0yvy)/2 v0y v0 sin?
vf vi at vx v0x v0 cos? vy v0 sin? -gt
(vf)2 (vi)2 2ax (vx)2 (v0 cos?)2 (vy)2 (v0 sin?)2 -2gy
x vitat2/2 x v0 cos? t y v0 sin? t- (gt2)/2
Max. Distance R 2 v0 cos? tup (v0)2 sin2? /2g Rmax (v0)2 /2g h (v0 sin? )2 / 2g
14
Checkpoint 4-4
  • 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?
  • vxconstant
  • vy initially positive and then decreases to zero
    and finally it increases in negative value.
  • ax 0
  • ay -g
  • throughout the entire projectile path

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 motion T
  • T (2?r)/v

16
Checkpoint 4-6
  • Object has counterclockwise motion
  • Then at y2 m , its velocity is v (4m/s)i
  • and centripetal acceleration magnitude aRv2/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
  • xPAxPBxBA
  • 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)
  • aPAaPB (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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