copyright r. janow – spring 2012 physics 111 lecture 03 motion in two dimensions sj 8th ed.: ch....

Copyright R. Janow – Spring 2012

Physics 111 Lecture 03Motion in Two DimensionsSJ 8th Ed.: Ch. 4.1 – 4.4

• Position, velocity, acceleration vectors• Average & Instantaneous Velocity• Average & Instantaneous Acceleration • Two Dimensional Motion with Constant

Acceleration (Kinematics)• Projectile Motion (Free Fall)• Uniform Circular Motion• Tangential and Radial Acceleration• Relative Velocity and Relative

Acceleration

4.1 Position, Velocity and Acceleration Vectors4.2 Two Dimensional Motion with Constant Acceleration4.3 Projectile Motion4.4 A particle in Uniform Circular Motion4.5 Tangential and Radial Acceleration4.6 Relative Velocity and Relative Acceleration


Motion in two and Three Dimensions

Extend 1 dimensional kinematics to 2 D and 3 Dndescriptio z y,x, or yx, x

• Kinematic quantities become 3 dimensionalkz j y ix r x

kv jv iv v v zyx

ka ja ia a a zyx

• Motions in the 3 perpendicular directions can be analyzed independently

• Vectors needed to manipulate quantities

• Constant acceleration Kinematic Equations hold component-wise for each dimensiont a v v xx0x

2x2

1x00 t a t v x x

)x(x2a v v x2x0

2x 0

t a v v yy0y 2

y21

y00 t a t v y y

)y(y2a v v y2y0

2y 0

Same for z

• In vector notation each equation is 3 separate ones for x, y, z

t a v v 0

2

21

00 t a t v r r

z y, )x(x 2a v v for same &x2x0

2x 0


Position and Displacement

A particle moves along its path as time increases

r r r if

Displacement

Positions

Trajectory: y = f(x)Parameterized by time

j ]y[y i ]x[x

jy ix r

ifif

fiffii tt )(tr r )(tr r

always points from the origin to the particle’s location r

• Path does not show time dependence.• Slopes of (tangents to) x(t), y(t), z(t) graphs would show velocity components

Average Velocity

t

r

interval time

ntdisplaceme vavg

Same direction as r

Rate of change of position


Instantaneous Velocity

jv iv dt

rd

t

r Lim

0t )v(Lim

0t v yxavg

dt

dy v

dt

dx v yxwhere

x

y

v

parabolic path

Free Fall Example

is tangent to the path in x,yi.e, to a plot of y vs x v

slope vx

t

x

tv )t(x x

tyy(t) is

parabolic2tg

2

1-tv- )t(y y0

Components of are tangent to graphs of corresponding components of the motion, viz:

v

vx is tangent to x(t)

vy is tangent to y(t)


Displacement in 3 Dimensions

kz j y ix r iiii

kz jy ix r

r r r if

fr r i

r

kv jv iv t

r Lim

0t )v(Lim

0t v zyxavg


Average Acceleration

x

y2D Trajectory: y = f(x)Parameterized by time

Velocity vectors are tangent to the trajectory at A & B• Move them tail to tail

v v v if

jt

v i

t

v

t

v a

yxavg

is parallel to v

a avg

Instantaneous Acceleration

ja ia dt

vd

t

v Lim

0t )a(Lim

0t a yxavg

If the solution is known find: )t(r

• vx(t), vy(t) by differentiating x(t), y(t)

• ax(t), ay(t) by differentiating vx(t), vy(t)

is NOT tangent to the path a

• ax IS tangent to a graph of vx(t)

• ay IS tangent to a graph of vy(t)

a ax affects vx and x, but

not vy or ySimilarly for ay

Rate of change of velocity


Summary – 3D Kinematics Formulaskz j y ix r

k

t

z j

t

y i

t

x

t

r vavg

kv jv iv dt

rd )v(Lim

0t v zyxavg

dt

dz v

dt

dy v

dt

dx v zyxwhere

ka ja ia dt

vd )a(Lim

0t a zyxavg

$$

dt

dv a

dt

dv a

dt

dv a z

zy

yx

xwhere

Definitions:

Same for z

Easiest to use in Cartesian coordinates – The x, y, z, dimensions move independently- Choose ax, ay, az and initial conditions independentlyCartesian Scalar Form - Constant Acceleration

t a v v xx0xf 2

x21

x00f t a tv x x

)x(x2a v v fx2x0

2xf 0

t ]vv x x xfx0[21

0f

t a v v yy0yf 2

y21

y00f t a tv y y

)y(y2a v v fy2y0

2yf 0

t ]vv y y yfy0[21

0f

Vector Form - Motion with Constant Acceleration - 2D or 3D

t)v v( r r 0f21

0f

of0f t t t t a v v

221

00f t a tv r r

ra2 v vv v 20ff

2f

PARABOLAS


Kinematic Equations in 2D: Graphical Representation

t a v (t)v if

2

21

iif t a tv r (t)r

vx

vy

)(constants conditions initial are a r v

ally"parametric" motions yandx links t t t

ii

if


Projectile Motion: Motion of a particle under constant, downward gravitation only

Assume:• Free fall along y direction (up/down) with horizontal motion as well ay = constant = - g, ax, az = 0 Velocity in x-z direction is constant.• Trajectory (a parabola) lies in a plane. Can choose it to be x-y. Motion is 2D, not 3D. • Usually pick initial location at origin: x0 = 0, y0 = 0 at t=0• Initial conditions:

v0 has x and y components vx0 = v0cos(), vy0 = v0sin() vx is constant, no drag or non-gravitation forces usually If vx = 0, motion is strictly vertical, range = 0

x

ypath

0v

v0x

v0y

Equations:

v (t)v x0x

tv x x(t) x00

v (t)v 2x0

2x

No acceleration along x

gt v (t)v y0y

221

y00 gt tv y y(t)

]y[y(t) 2g v (t)v 2y0

2y 0

Acceleration = - g along y


Trajectory of a Projectile: y as a function of x

LaunchRange R

(return to launch altitude)

vy at E = -vyi

Maximum Height

tv x(t) xi 221

yi gt tv y(t)

xix v)t(v gt v (t)v yiy vR


Example: Show Projectile Trajectory is a Parabola (in x)

Initial Conditions:

x

ypath

iv

vix

viy

i

0ix

0iy

)cos(vv iixi )sin(vv iiyi

Eliminate time from Kinematics Equations:

tv x(t) xi v

x(t) t

xi

2xi

v

2x21

xiv x

yi2

21

yi g v gt tv y(t)

)

i(cos

iv

2x21

i g )tan(x y(x)

22Parabolic in x

Limiting Cases:

As i 0, tani 0, cosi 1,

iv

2x21 g y(x)

2

As i 90o, tani infinity, cosi 0, only vertical up, lowsb y(x)


Example: Find the Range of a Projectile (Level Ground)

)cos(

)sin()tan(

)

i(cos

iv

2R21

i g )tan(R y(x)

22

0

The range R is the horizontal distance traveled as a projectile returns to it’s launch height yi = 0.

y

x

Set x = R and y = 0 in the preceding trajectory formula:

iv

2R21

icos i g )()sin(R 2

0

Rearrange and put into standard quadratic form:

0 iii

2 ]g

v)cos( )[2sin(R R 2

R factors. Roots are:

0 t at 0 R n)informatio new (not

flight) of time ft time (atiii

g

v)cos()2sin( R 2

Trigonometric Identity: )sin(2 )cos()2sin( iii )sin(2g

v R i

i 2

“Range Formula”Extremes of Range versus launch angle i

] lyhorizontal launch [ i 0 R 0

] 0 R 90 launch vertical 0, )sin(2x90 :note [ oi

o

1 )sin(2 45 ] range maximum [i oi


Example: Time of Flight for a Projectile Find the time of flight tR for range R – the time to return to the original launch height yi = 0.

Set y = 0 in the y-displacement formula:

The roots are:

news) not - point launch (the R 0 t

2R2

1RyiR gt tv 0 ) y(t

This is a quadratic equation in tR – factor it: ]gt [v t 0 R21

yiR

R) range withlanding (the yi

R g

2v t and:

Find maximum height ymax reached at time tmax Set vy,max = 0 at tmax

gt v 0 v maxyimaxy, flight) of time total the (half Ryimax 2

t

g

v t

2

2yi

21yi

yi2max2

1maxyimax g

vg

g

vvgt tv y

2g

)(sinv y i

2i

max

2

Find speed vf at time tR - passing y = 0 on the way down

Same speed at the same altitude

)sin(vv iiyi

)sin(vv ffyf v v yiyf

g

2gv v gt v v yi

yiRyiyf

v v if vxf = vxi = vicos(i)

(constant) i = f


Projectile RangeLaunch with same initial speed, vary angle

Maximum range at = 45o

Complementary angles produce the same range, but different maximum heights and times of flight

2g

)(sinv y i

2i

max

2

)sin(2g

v R i

i 2


Problem Solving Strategy

Text (page 43) defines a general 4 step method:• Conceptualize• Categorize• Analyze• Finalize

Additional problem solving hints for mechanics:• Choose coordinate system(s). Try to make choices that simplify representing the problem. For example, where acceleration a is constant try choosing an axis along the direction of a.• Make a sketch showing axes, origin, particles.• Choose names for the important quantities that will help you remember what they mean. Be sure to distinguish quantities of the same type, such as v, vx, vy, vi, vf, ….• List the known quantities and the given initial conditions, like vxi, vyi, ax, ay,… Show these on your sketch.• Choose equations to use for describing the motion in the problem. Time connects the x, y, z dimensions and is sometimes to be eliminated via algebra.


path for g

= 0

xm

ym

s

What should monkey (a Physics 111 student) do? Where should hunter aim if monkey jumps?Assume: hunter fires at the instant monkey lets go and aims directly at falling monkey


What should monkey (a Physics 111 student) do? Where should hunter aim if monkey jumps?

When should hunter fire? When should monkey jump?

Coordinates, variables:Bullet: xb(t), yb(t) Monkey: s, ym(t)

Initial Conditions:Monkey at (s, h) Bullet at (0,0)

x

y

0v

h

s

)sin(vv

)cos(vv

y

x

00

00

To hit monkey:yb(t) = ym(t) at same time t1 whenbullet is at s

For bullet at s:

stv )t(x xb 101

21

0

021101 2

1

2

1t gs

v

vgttv )t(y

x

yyb

v

st

x01

For monkey: 211 2

1gth )t(ym

Monkey should wait until hunter fires before jumping

Hunter should aim directly at monkey after he jumpsBoth know Physics, so everybody waits forever

s

h)tan(

v

v

x

y 0

0)t(y )t(y mb 11

Equate positions:


Example: Find the arrow’s speedAn arrow is shot horizontally by a person whose height is not known.It strikes the ground 33 m away horizontally, making an angle

of 86o with the vertical. What was the arrow’s speed as it was released?

86oR = 33 m

v0

vf

R = v0t t = R / v0

t = flight time = time to hit ground

Final velocity components:

)sin(vvv offx 860 No acceleration along x

)cos(vgtvv ofyfy 860 Vy0 = 0, t as above

Divide equations to eliminate vf:

gR

v

gt

v)tan(

)cos(

)sin(

v

v oo

o

fy

fx20086

86

86

Rearrange:

14.3 33 9.8 )tan(gR v o 8620

m/s 68 v 0


Uniform Circular Motion• A particle in Uniform Circular Motion moves in a circular path with a constant speed The radius vector is rotating.• The velocity vector is always tangent to the particle’s path (as usual). It is perpendicular to the (rotating) radius vector.• The particle is accelerating, since the direction of the velocity is constantly changing.

r

v

a

a

a

v

v

r

r

• The centripetal acceleration vector always points toward the center of the circle.• The magnitudes of all three vectors are constant, but their directions are changing • All three vectors are rotating with the same constant angular velocity and period.• The period of revolution T is:

a ,v ,r

v

rT

2

a ,v ,r

• The frequency f is the reciprocal of the period f = 1/T


Centripetal Acceleration Magnitude Formula

The position vectors and velocity vectors both change due to changes in their directions:

The triangles for r and v are both right isosceles, with the same angle at their apexes. They are similar:

tail to tail

v

|v|

r

|r|

|r|

r

v |v|

The definition of the (magnitude of) the average acceleration is: t

|r|

r

v

t

|v| |a| avg

In the limit t 0: v t

|r|

and a |a| avg

r

v a

2

c The magnitude of the centripetal acceleration is

It always points to the center of the circular motion


Example: Find the Earth’s centripetal acceleration in its orbit about the Sun

Assume uniform circular motionThe radius of the Earth’s orbit is r = 1.5 x 1011 m

r

v a

2

c Need to find the speed v

T

r 2

period

ncecircumfere v

T = 1 year = 365X24x3600 s = 3.156 x

107 s

m/s ..

. v 4

7

1110992

101563

10512

m/s .

.

]10[2.99 a 2

4

c3

11

210945

1051

Example: Passengers on an amusement park ride move in circles whose radius

is 5 m. They complete a full circle every 4 seconds. What acceleration do they feel, in “g”s”?

Need to find the speed v

22

c m/s .rT

r)(2

r

v a 3412

4

542

2

2

22

.

m/s .

m/s .

g

a

2

2c 261

89

3412

dimensionless


Uniform Circular Motion Formulas via Calculus

] j)sin( i)cos( [ r j)rsin( i)rcos( j y ix r

Calculate time derivative. Constant r means dr/dt = 0

] j d

)dsin( i

d

)dcos()[

dt

d( r

dt

rd v

] j )os(c i)sin([ v v

Note: r d/dt = v = magnitude of the speed = 2x r / T for constant v

Unit vector tangent to circle

Unit vector along r

] j d

)dcos( i

d

)dsin()[

dt

d( v

] j )os(c i)sin([ dt

dv

dt

vd a

v is constant

] j )in(s i)(osc[ r

va

2

r

v

dt

d

Note:

Unit vector pointing inward along r

r

v a

2

c

copyright r. janow – spring 2012 physics 111 lecture 03 motion in two dimensions sj 8th ed.: ch....

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