lecture 2 motion in two or three dimensions

1

Motion in Two or Three Dimensions

2

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,