lecture 2 motion in two or three dimensions
TRANSCRIPT
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Motion in Two or Three Dimensions
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Projectile motion
A projectile is any body that is given an initial velocity and then follows a path (its trajectory) determined entirely by the effects of gravitational acceleration and air resistance. Here, we consider an idealized model, representing the projectile as a particle with an acceleration due to gravity that is constant in both magnitude and direction, and neglecting the effects of air resistance, and the curvature and rotation of the earth.
0000 cosvv x
000 sinvv y
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Projectile motion (cont’)
Remark 1: Projectile motion is two-dimensional, always confined to a vertical plane determined by the direction of the initial velocity – the acceleration due to gravity being purely vertical can’t accelerate the projectile sideways.
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Projectile motion (cont’)
Remark 2: We can treat the x- and y-coordinates separately. The trajectory of a projectile is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration.
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Projectile motion (cont’)
0xa
gay
000 cos vvv xx
gtvgtvv yy 000 sin
At t = 0, x0 = y0 = 0, v0x = v0cos0, and v0y = v0sin0. For t > 0, tvx 00 cos
The trajectory is a parabola:
22
00
0cos2
tan xv
gxy
200 2
1sin gttvy
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Example 3.7 Height and range of a projectile
A batter hits a baseball so that it leaves the bat at speed v0 = 37.0 m/s at an angle of 0 = 53.1o. (a) Find the position of the ball and its velocity at t = 2.00 s. (b) Find the time when the ball reaches the highest point of its flight, and its height h at this time. (c) Find the horizontal range R – that is, the horizontal distance from the starting point to where the ball hits the ground.
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Example 3.7 Height and range of a projectile (cont’)Solution:
At t = 2.00 s,
m4.44cos 00 tvx m6.3921
sin 200 gttvy
sm2.22cos 00 vvx sm0.10sin 00 gtvvy
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Example 3.7 Height and range of a projectile (cont’)Solution (cont’):
At the highest point of its flight,0sin 100 gtvvy
s02.3sin 00
1 g
vt
g
vh
2
sin 022
0
m7.4421
sin 21100 gttvh
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Example 3.7 Height and range of a projectile (cont’)Solution (cont’):
When the ball hits the ground,
021
sin 200 gttvy
s04.6sin2 00 g
vt
m134cos 00 tvR
g
v
g
vR 0
2000
20 2sincossin2
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Example 3.10 The zookeeper and the monkey
A monkey escapes from the zoo and climbs a tree. After failing to entice the monkey down, the zookeeper fires a tranquilizer dart directly at the monkey. The monkey lets go at the instant the dart leaves the gun. Show that the dart will always hit the monkey, provided that the dart reaches the monkey before he hits the ground and runs away.
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Example 3.10 The zookeeper and the monkey (cont’) tvx 00dart cos
200dart 2
1sin gttvy
20monkey 21
tan gtdy 20dartdart 21
tan gtxy
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Position vector
To describe the motion of a particle in space, we must first be able to describe the particle’s position. Consider a particle that is at a point P at a certain instant.The Cartesian coordinates x, y, and z of point P are the x-, y-, and z-components of the position vector of the particle at this instant:
kzjyixr ˆˆˆ
– this vector goes from the origin of the coordinate system to the point P.
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Average velocity vector
Suppose during a time interval t = t2 – t1, the particle moves from P1 to P2. The displacement (change in position) during this interval is
kzzjyyixxrrr ˆˆˆ12121212
The average velocity during this interval,
12
12av tt
rrtr
v
Note ktz
jty
itx
v ˆˆˆav
kttzz
jttyy
ittxx ˆˆˆ
12
12
12
12
12
12
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Instantaneous velocity vector
The (instantaneous) velocity is the limit of the average velocity as the time interval approaches zero – it equals the instantaneous rate of change of position with time.
dtrd
tr
vt
0lim
Note
ktz
jty
itx
vttt
ˆlimˆlimˆlim000
kvjvivkdtdz
jdtdy
idtdx
zyxˆˆˆˆˆˆ
The magnitude of the (instantaneous) velocity vector, i.e., the speed, is given by 222
zyx vvvvv
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The acceleration vector
Suppose during a time interval t = t2 – t1, the particle moves from P1 to P2. The change in velocity vector during this interval is
12 vvv
The average acceleration during this interval,
12
12av tt
vvtv
a
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The acceleration vector (cont’)
Notek
ttvv
jtt
vvi
tt
vvktv
jt
vit
va zzyyxxzyx ˆˆˆˆˆˆ
12
12
12
12
12
12av
dtvd
tv
at
0lim
Note
ktv
jt
vit
va z
t
y
t
x
t
ˆlimˆlimˆlim000
kajaiakdtdv
jdt
dvi
dt
dvzyx
zyx ˆˆˆˆˆˆ
The (instantaneous) acceleration is the limit of the average acceleration as the time interval approaches zero – it equals the instantaneous rate of change of velocity with time.
kdtzd
jdtyd
idtxd ˆˆˆ
2
2
2
2
2
2
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The acceleration vector (cont’)
Remark 3: Parallel & perpendicular components of acceleration
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Examples 3.1, 3.2, & 3.3
A robotic vehicle, or rover, is exploring the surface of Mars. The stationary Mars lander is the origin of coordinates, and the surrounding Martian surface lies in the xy-plane. The rover, which we represent as a point, has x- and y-coordinates that vary with time:
22sm25.0m0.2 tx 33sm025.0sm0.1 tty (a) Find the rover’s coordinates and distance from the lander at t = 2.0 s.(b) Find the rover’s displacement and average velocity vectors for the interval t = 0.0 s to t = 2.0 s.(c) Find a general expression for the rover’s instantaneous velocity vector. Express its velocity at t = 2.0 s in component form and in terms of magnitude and direction.Solution: …
tvx2sm50.0 23sm075.0sm0.1 tvy
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Examples 3.1, 3.2, & 3.3 (cont’)
(d) Find the components of the average acceleration for the interval t = 0.0 s to t = 2.0 s.(e) Find the instantaneous acceleration at t = 2.0 s.(f) Find the parallel and perpendicular components of the acceleration at t = 2.0 s.Solution (cont’): …
2sm50.0xa tay3sm15.0
At t = 2.0 s, …sm0.1xv sm3.1yv2sm50.0xa
2sm30.0ya
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Uniform circular motion
When a particle moves in a circle of radius R with constant speed v, the motion is called uniform circular motion.
There is no component of acceleration parallel (tangent) to the path; otherwise, the speed would change – the acceleration vector is perpendicular (normal) to the path and hence directed inward (never outward!) toward the center of the circular path.
TR
v2
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Uniform circular motion (cont’)
Rv
a2
rad
Proof:v
v
Rs
Rv
ts
Rv
t
va
tt
2
00rad limlim
In uniform circular motion, the magnitude arad of the instantaneous acceleration is equal to the square of the speed v divided by the radius R of the circle. Its direction is perpendicular to v and inward along the radius – centripetal acceleration.
v
2
2
rad
4TR
a
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Nonuniform circular motion
When a particle moves in a circle of radius R with a varying speed v, the motion is called nonuniform circular motion.
Rv
a2
rad dtdv
a tan
2tan
2rad aaa
Remark 4:
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Relative velocity
In general, when two observers A and B measure the velocity of a moving body P, they get different results if one observer (B) is moving relative to the other (A).
ABBPAP rrr
ABBPAP rdtd
rdtd
rdtd
ABBPAP vvv
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Relative velocity (cont’)
ABBPAP vvv
Peter
Alice
Betty
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Example 3.14 and 3.15
An airplane’s compass indicates that it is headed due north, and its airspeed indicator shows that it is moving through the air at 240 km/h. If there is a 100-km/h wind from west to east,(a) what is the velocity of the airplane relative to the earth?(b) in what direction should the pilot head to travel due north? What will be her velocity relative to the earth?jv ˆ240airplane
The velocity of the air relative to the earth,iv ˆ100earthair
The velocity of the plane relative to the earth,jivvv ˆ240ˆ100earthairairplaneearthplane
…
Solution: The velocity of the plane relative to the air,