Physics 111: Mechanics Lecture 3

Dale Gary

NJIT Physics Department

February 5-8, 2013

Motion in Two Dimensions Reminder of vectors and vector algebra Displacement and position in 2-D Average and instantaneous velocity in 2-D Average and instantaneous acceleration in

2-D Projectile motion Uniform circular motion Relative velocity*

February 5-8, 2013

Vector and its components The components are

the legs of the right triangle whose hypotenuse is A

yx AAA

2 2 1tan yx y

x

AA A A and

A

)sin(

)cos(

AA

AA

y

x

Or,

x

y

x

y

yx

A

A

A

A

AAA

1

22

tanor tan

February 5-8, 2013

Which diagram can represent ?

A) B)

C) D)

Vector Algebra

12 rrr

r

2r1r

r

2r1r

r

2r1r

r

2r1r

1r

February 5-8, 2013

Kinematic variables in one dimension Position: x(t) m Velocity: v(t) m/s Acceleration: a(t) m/s2

Kinematic variables in three dimensions Position: m Velocity:

m/s Acceleration: m/s2

All are vectors: have direction and magnitudes

Motion in two dimensions

kvjvivtv zyxˆˆˆ)(

y

x

z

ij

k

x

kzjyixtr ˆˆˆ)(

kajaiata zyxˆˆˆ)(

February 5-8, 2013

In one dimension

In two dimensions Position: the position of an object

is described by its position vector --always points to particle from origin.

Displacement:

x1 (t1) = - 3.0 m, x2 (t2) = + 1.0 mΔx = +1.0 m + 3.0 m = +4.0 m

Position and Displacement

)(tr

12 rrr

jyix

jyyixx

jyixjyixr

ˆˆ

ˆ)(ˆ)(

)ˆˆ()ˆˆ(

1212

1122

)()( 1122 txtxx

12 rrr

February 5-8, 2013

Average velocity

Instantaneous velocity

v is tangent to the path in x-y graph;

Average & Instantaneous Velocity

dt

rd

t

rvv

tavg

00t

limlim

jvivjt

yi

t

xv yavgxavgavg

ˆˆˆˆ,,

t

rvavg

jvivjdt

dyi

dt

dx

dt

rdv yx

ˆˆˆˆ

February 5-8, 2013

Motion of a Turtle

A turtle starts at the origin and moves with the speed of v0=10 cm/s in the direction of 25° to the horizontal.

(a) Find the coordinates of a turtle 10 seconds later.

(b) How far did the turtle walk in 10 seconds?

February 5-8, 2013

Motion of a TurtleNotice, you can solve the equations independently for the horizontal (x) and vertical (y) components of motion and then combine them!

yx vvv

0

0 0 cos 25 9.06 cm/sxv v

X components:

Y components:

Distance from the origin:

0 90.6 cmxx v t

0 0 sin 25 4.23 cm/syv v 0 42.3 cmyy v t

cm 0.10022 yxd

February 5-8, 2013

Average acceleration

Instantaneous acceleration

The magnitude of the velocity (the speed) can change The direction of the velocity can change, even though

the magnitude is constant Both the magnitude and the direction can change

Average & Instantaneous Acceleration

dt

vd

t

vaa

tavg

00t

limlim

jaiajt

vi

t

va yavgxavg

yxavg

ˆˆˆˆ,,

t

vaavg

jaiajdt

dvi

dt

dv

dt

vda yx

yx ˆˆˆˆ

February 5-8, 2013

Position

Average velocity

Instantaneous velocity

Acceleration

are not necessarily same direction.

Summary in two dimensionjyixtr ˆˆ)(

jaiajdt

dvi

dt

dv

dt

vd

t

vta yx

yx

tˆˆˆˆlim)(

0

jvivjt

yi

t

x

t

rv yavgxavgavg

ˆˆˆˆ,,

jvivjdt

dyi

dt

dx

dt

rd

t

rtv yx

tˆˆˆˆlim)(

0

dt

dxvx

dt

dyvy

2

2

dt

xd

dt

dva x

x 2

2

dt

yd

dt

dva y

y

)( and ),( , tatv(t)r

February 5-8, 2013

Motion in two dimensions

tavv

0

Motions in each dimension are independent components Constant acceleration equations

Constant acceleration equations hold in each dimension

t = 0 beginning of the process; where ax and ay are constant; Initial velocity initial displacement ;

221

0 tatvrr

tavv yyy 0

221

00 tatvyy yy

)(2 02

02 yyavv yyy

tavv xxx 0

221

00 tatvxx xx

)(2 02

02 xxavv xxx

jaiaa yxˆˆ

jvivv yxˆˆ

000 jyixr ˆˆ

000

February 5-8, 2013

Define coordinate system. Make sketch showing axes, origin.

List known quantities. Find v0x , v0y , ax , ay , etc. Show initial conditions on sketch.

List equations of motion to see which ones to use. Time t is the same for x and y directions. x0 = x(t = 0), y0 = y(t = 0), v0x = vx(t = 0), v0y = vy(t = 0).

Have an axis point along the direction of a if it is constant.

Hints for solving problems

tavv yyy 0

221

00 tatvyy yy

)(2 02

02 yyavv yyy

tavv xxx 0

221

00 tatvxx xx

)(2 02

02 xxavv xxx

February 5-8, 2013

2-D problem and define a coordinate system: x- horizontal, y- vertical (up +)

Try to pick x0 = 0, y0 = 0 at t = 0

Horizontal motion + Vertical motion Horizontal: ax = 0 , constant velocity

motion Vertical: ay = -g = -9.8 m/s2, v0y = 0

Equations:

Projectile Motion

221 gttvyy iyif

tavv yyy 0

221

00 tatvyy yy

)(2 02

02 yyavv yyy

tavv xxx 0

221

00 tatvxx xx

)(2 02

02 xxavv xxx

Horizontal Vertical

February 5-8, 2013

X and Y motions happen independently, so we can treat them separately

Try to pick x0 = 0, y0 = 0 at t = 0

Horizontal motion + Vertical motion Horizontal: ax = 0 , constant velocity

motion Vertical: ay = -g = -9.8 m/s2

x and y are connected by time t y(x) is a parabola

Projectile Motion

gtvv yy 0

221

00 gttvyy y xx vv 0

tvxx x00

Horizontal Vertical

February 5-8, 2013

2-D problem and define a coordinate system.

Horizontal: ax = 0 and vertical: ay = -g.

Try to pick x0 = 0, y0 = 0 at t = 0.

Velocity initial conditions: v0 can have x, y components. v0x is constant usually. v0y changes continuously.

Equations:

Projectile Motion

000 cosvv x

Horizontal Vertical

000 sinvv x

gtvv yy 0

221

00 gttvyy y xx vv 0

tvxx x00

February 5-8, 2013

Initial conditions (t = 0): x0 = 0, y0 = 0 v0x = v0 cosθ0 and v0y = v0 sinθ0

Horizontal motion:

Vertical motion:

Parabola; θ0 = 0 and θ0 = 90 ?

Trajectory of Projectile Motion

221

00 gttvy y

xx v

xttvx

00 0

2

000 2

xxy v

xg

v

xvy

2

022

0

0cos2

tan xv

gxy

February 5-8, 2013

Initial conditions (t = 0): x0 = 0, y0 = 0 v0x = v0 cosθ0 and v0x = v0 sinθ0, then

What is R and h ?

Horizontal Vertical

221

000 gttv y tvx x00

g

v

g

vvtvxxR x

02

0000000

2sinsincos2

g

v

g

vt y 000 sin22

2

02

21

00 222

tgtvgttvyyh yhhy

g

vh

2

sin 022

0

yy

yyy vg

vgvgtvv 0

000

2

h

gtvv yy 0

221

00 gttvyy y

xx vv 0

tvxx x00

February 5-8, 2013

Projectile Motion at Various Initial Angles

Complementary values of the initial angle result in the same range The heights will be

different The maximum

range occurs at a projection angle of 45o

g

vR

2sin20

February 5-8, 2013

Uniform circular motion

Constant speed, or,constant magnitude of velocity

Motion along a circle:Changing direction of velocity

February 5-8, 2013

Circular Motion: Observations Object moving along a

curved path with constant speed Magnitude of velocity:

same Direction of velocity:

changing Velocity: changing Acceleration is NOT zero! Net force acting on the

object is NOT zero “Centripetal force”

amFnet

February 5-8, 2013

Centripetal acceleration

Direction: Centripetal

Uniform Circular Motion

r

v

t

va

r

v

r

v

t

r

t

v

r

rvv

r

r

v

v

r

2

2

so,

Ox

y

ri

R

A Bvi

rf

vf

Δr

vi

vf

Δv = vf - vi

February 5-8, 2013

Uniform Circular Motion Velocity:

Magnitude: constant v The direction of the velocity

is tangent to the circle Acceleration:

Magnitude: directed toward the center

of the circle of motion Period:

time interval required for one complete revolution of the particle

r

vac

2

r

vac

2

v

rT

2

vac

February 5-8, 2013

Position

Average velocity

Instantaneous velocity

Acceleration

are not necessarily in the same direction.

Summaryjyixtr ˆˆ)(

jaiajdt

dvi

dt

dv

dt

vd

t

vta yx

yx

tˆˆˆˆlim)(

0

jvivjt

yi

t

x

t

rv yavgxavgavg

ˆˆˆˆ,,

jvivjdt

dyi

dt

dx

dt

rd

t

rtv yx

tˆˆˆˆlim)(

0

dt

dxvx

dt

dyvy

2

2

dt

xd

dt

dva x

x 2

2

dt

yd

dt

dva y

y

)( and ),( , tatv(t)r

February 5-8, 2013

If a particle moves with constant acceleration a, motion equations are

Projectile motion is one type of 2-D motion under constant acceleration, where ax = 0, ay = -g.

Summary

jtatvyitatvxjyixr yiyiixixiifffˆ)(ˆ)(ˆˆ 2

212

21

jtavitavjvivtv yiyxixfyfxfˆ)(ˆ)(ˆˆ)(

tavv i

221 tatvrr iif