3 motion in two & three dimensions displacement, velocity, acceleration case 1: projectile...

3 Motion in Two & Three Dimensions

• Displacement, velocity, acceleration

• Case 1: Projectile Motion• Case 2: Circular Motion

• Hk: 51, 55, 69, 77, 85, 91.

Position & Displacement Vectors

jyixr ˆˆ

12 rrr

Velocity Vectors

t

rvav

dt

rdv

222 )()()( zyx vvvv

Relative Velocity

• Examples: • people-mover at airport• airplane flying in wind• passing velocity (difference in velocities)• notation used:

velocity “BA” = velocity of B with respect to A

Example:

Acceleration Vectors

t

vaav

dt

vda

Direction of Acceleration

• Direction of a = direction of velocity change (by definition)

• Examples: rounding a corner, bungee jumper, cannonball (Tipler), Projectile (29, 30 below)

Projectile Motion

• begins when projecting force ends

• ends when object hits something

• gravity alone acts on object

Horizontal V Constant

Two Dimensional Motion (constant acceleration)

tavv xoxx

221 tatvx xox

tavv yoyy

221 tatvy yoy

Range vs. Angle

Example 1: Calculate Range (R)

vo = 6.00m/s o = 30°

xo = 0, yo = 1.6m; x = R, y = 0

smvv ooox /20.530cos00.6cos

smvv oooy /00.330sin00.6sin

Example 1 (cont.)

2

221

221

9.436.1

)8.9(30sin66.1

tt

tt

tatvy yoy

06.139.4 2 tt

Step 1

Quadratic Equation

02 cbxaxa

acbbx

2

42

06.139.4 2 tt

6.1

3

9.4

c

b

a

a

acbbx

2

42

Example 1 (cont.)

6.1

3

9.4

c

b

a

)9.4(2

)6.1)(9.4(4)3(3 2 t

)9.4(2

353.63t

954.0

342.0

t

t

End of Step 1

Example 1 (cont.)

tvtatvx oxxox 221Step 2

(ax = 0)

mtvx oo 96.4)954.0(30cos6cos

“Range” = 4.96m

End of Example

Circular Motion

• Uniform

• Non-uniform

• Acceleration of Circular Motion

18

Centripetal Acceleration

• Turning is an acceleration toward center of turn-radius and is called Centripetal Acceleration

• Centripetal is left/right direction

• a(centripetal) = v2/r

• (v = speed, r = radius of turn)

• Ex. V = 6m/s, r = 4m. a(centripetal) = 6^2/4 = 9 m/s/s

Tangential Acceleration

• Direction = forward along path (speed increasing)

• Direction = backward along path (speed decreasing)

t

dvat

Total Acceleration

• Total acceleration = tangential + centripetal

• = forward/backward + left/right

• a(total) = dv/dt (F/B) + v2/r (L/R)

• Ex. Accelerating out of a turn; 4.0 m/s/s (F) + 3.0 m/s/s (L)

• a(total) = 5.0 m/s/s

Summary

• Two dimensional velocity, acceleration

• Projectile motion (downward pointing acceleration)

• Circular Motion (acceleration in any direction within plane of motion)

Ex. A Plane has an air speed vpa = 75m/s. The wind has a velocity with respect to the ground of vag = 8 m/s @ 330°. The plane’s path is due North relative to ground. a) Draw a vector diagram showing the relationship between the air speed and the ground speed. b) Find the ground speed and the compass heading of the plane.

(similar situation)

)v,0()330sin8,330cos8()sin75,cos75(

vvv

pg

agpapg

0330cos8cos75 3.95

pgvm/s 70.7330sin83.95sin75

PM Example 2:

vo = 6.00m/s o = 0°

xo = 0, yo = 1.6m; x = R, y = 0

smvv ooox /00.60cos00.6cos

smvv oooy /00sin00.6sin

PM Example 2 (cont.)

2

221

221

9.406.1

)8.9(0sin66.1

t

tt

tatvy yoy

571.09.4

6.1

6.19.4 2

t

t

Step 1

PM Example 2 (cont.)

tvtatvx oxxox 221Step 2

(ax = 0)

mtvx oo 43.3)571.0(0cos6cos

“Range” = 3.43m

End of Step 2

PM Example 2: Speed at Impact

st 571.0

tavv xoxx tavv yoyy

smtvx /6)0(6 sm

vy

/59.5

571.0)8.9()0(

smvvv yx /20.8)59.5()6( 2222

v1

0

1

2

3

4

5

6

0 2 4 6 8 10 12 14

x(m)

y(m

)

1. v1 and v2 are located on trajectory.

1v

2v

va

Q1. Given 1v2v

v

locate these on the trajectory and form v.

0

1

2

3

4

5

6

0 2 4 6 8 10 12 14

x(m)

y(m

)

1v

2v

Velocity in Two Dimensions

• vavg // r

• instantaneous “v” is limit of “vavg” as t 0

t

rvavg

Acceleration in Two Dimensions

t

vaavg

• aavg // v

• instantaneous “a” is limit of “aavg” as t 0

Displacement in Two Dimensions

ro

r

r

orrr

rrr o

v1

0

1

2

3

4

5

6

0 2 4 6 8 10 12 14

x(m)

y(m

)

1. v1 and v2 are located on trajectory.

1v

2v

va

Ex. If v1(0.00s) = 12m/s, +60° and v2(0.65s) = 7.223 @ +33.83°, find aave.

)39.10,00.6())60sin(0.12),60cos(0.12(1 v

)02.4,00.6())83.33sin(223.7),83.33cos(223.7(2 v

smvvv /)37.6,0()39.10,00.6()02.4,00.6(12

ssms

sm

t

va //)8.9,0(

00.065.0

/)37.6,0(

Q1. Given 1v2v

v

locate these on the trajectory and form v.

0

1

2

3

4

5

6

0 2 4 6 8 10 12 14

x(m)

y(m

)

1v

2v

Q2. If v3(1.15s) = 6.06m/s, -8.32° and v4(1.60s) = 7.997, -41.389°, write the coordinate-forms of these vectors and calculate the average acceleration.

)8777.0,00.6())32.8sin(06.6),32.8cos(06.6(3 v

)2877.5,00.6())39.41sin(997.7),39.41cos(997.7(4 v

smvvv /)41.4,0()8777.0,00.6()2877.5,00.6(12

ssms

sm

t

va //)8.9,0(

15.160.1

/)41.4,0(

v3

v4v a