Physics 201: Lecture 5, Pg 1 Lecture 5 l Goals: (Chapter 4.1-3) Introduce position, displacement, velocity and acceleration in 2D Address 2D motion.

Physics 201: Lecture 5, Pg 1

Lecture 5 Goals: (Chapter 4.1-3)

Introduce position, displacement, velocity and acceleration in 2D Address 2D motion in systems with constant acceleration (i.e. both magnitude and acceleration) Discuss horizontal range (special)

Key point 1: In many case motion occurs 2D with constant acceleration (typically on the surface of a planet)

Key point 2: The “superposition principle” allows us to discuss the x & y motion individually


Decomposing vectors Any vector can be resolved into components along the x and y axes

tan-1 ( y / x )

22 yxr

y

x

(x,y)

rry

rx

j i

sin

cos

yxr

ryr

rxr

y

x

vy

vx

(vx,vy)

vvy

vx

ay

ax

(ax,ay)

aay

ax

j i

sin

cos

yx

y

x

vvv

vv

vv

tan-1 ( vy / vx )

22yx vvv

j i

sin

cos

yx

y

x

aaa

aa

aa

22yx aaa

tan-1 ( ay / ax )


Dynamics: Motion along a line but with a twist(2D dimensional motion, magnitude and directions)

Particle motions involve a path or trajectory

In 2D the position of a particle is r = x i + y j

and this vector is dependent on the origin. (i , j unit vectors )

ri’

rf’

O’Displacement, r, is independent of origin


Motion in 2DPosition

Displacement

Velocity (avg)

ffii trtr , and ,

if rrr

t

rv

avg


Instantaneous Velocity

As t 0 r shrinks and becomes tangent to the path The direction of the instantaneous velocity is along a line

that is tangent to the path of the particle’s direction of motion.

v

dt

rd

t

rv

t

0lim


Average Acceleration The average acceleration of particle motion reflects

changes in the instantaneous velocity vector (divided by the time interval during which that change occurs).

Instantaneous

acceleration

t

vv

tv if

avga

avga

t

va

t

0

lim


Instantaneous Acceleration

The instantaneous acceleration is a vector with components parallel (tangential) and/or perpendicular (radial) to the tangent of the path

Changes in a particle’s path may produce an acceleration The magnitude of the velocity vector may change The direction of the velocity vector may change Both may change simultaneously (depends: path vs time)

t

va

t

0

lim


2 D Kinematics Position, velocity, and acceleration of a particle:

r = x i + y j

v = vx i + vy j (i , j unit vectors )

a = ax i + ay j

2

2

dtxd

ax 2

2

dtyd

ay

dt

dxvx

dt

dyvy

)( txx )( tyy

ji

ji

)jˆ(

yx vv

dtdy

dtdx

dtyixd

dtrd

v


2 D Kinematics (special case) If ax and ay are constant then

tavtv

tatvyty

tavtv

tatvxtx

t

tatyty

tatxtx

yyiy

yyii

xxix

xxii

i

yyii

xxii

)(

)(

)(

)(

0 if and

v)(

v)(

221

221

221

221

In many case one of the a’s is zero


Trajectory with constant acceleration along the vertical

What do the velocity and acceleration

vectors look like? x

t = 0

4y

x vs y

Position

Special Case:

ax=0 & ay= -g

vx(t=0) = v0 vy(t=0) = 0

vy(t) = – g t

x(t) = x0 v0 t

y(t) = y0 - ½ g t2




vectors look like?

Velocity vector is always tangent to the curve!

x

t = 0

4y

x vs yVelocity




vectors look like?

Velocity vector is always tangent to the curve!

Acceleration may or may not be!

x

t = 0

4y

x vs yAcceleration

PositionVelocity


Another trajectory

x vs yt = 0

t =10

Can you identify the dynamics in this picture?

How many distinct regimes are there?

0 < t < 3 3 < t < 7 7 < t < 10

I. vx = constant = v0 ; vy = 0

II. vx = vy = v0

III. vx = 0 ; vy = constant < v0

y

x

What can you say about the acceleration?


Exercises 1 & 2 Trajectories with acceleration

A rocket is drifting sideways (from left to right) in deep space, with its engine off, from A to B. It is not near any stars or planets or other outside forces.

Its “constant thrust” engine (i.e., acceleration is constant) is fired at point B and left on for 2 seconds in which time the rocket travels from point B to some point C Sketch the shape of the path

from B to C. At point C the engine is turned off.

Sketch the shape of the path

after point C (Note: a = 0)


Exercise 1Trajectories with acceleration

A. A

B. B

C. C

D. D

E. None of these

B

C

B

C

B

C

B

C

A

C

B

D

From B to C ?


Exercise 2Trajectories with acceleration

A. A

B. B

C. C

D. D

E. None of these

C

C

C

C

A

C

B

D

After C ?


Kinematics in 2 D; Projectile Motion The position, velocity, and acceleration of a particle

moving in 2-dimensions can be expressed as:

r = x i + y j v = vx i + vy j a = ax i + ay j

Special Case: ax=0 & ay= -g

vx(t) = v0 cos

vy(t) = v0 sin – g t

x(t) = x0 v0 cos t

y(t) = y0 v0 sin t - ½ g t2

cos00 vvx

0yv0v


Kinematics in 2 D; Horizontal Range

Given v0 and how far will and object travel horizontally?Let y0 = 0 = yinitial = yfinal x0 =0

Again: ax=0 & ay= -g

1. vx(t) = v0 cos

2. vy(t) = v0 sin – g t

3. x(t) = 0 v0 cos t = R (range)

4. y(t) = 0 = 0v0 sin t - ½ g t2

4 gives: 0 = t (v0 sin - ½ g t) t = 0, 2 v0 sin g

R = v0 cos 2 v0 sin g = v02 sin 2 / g

Maximum when dR/d = 0 cos 2 = 0 or 45°

cos00 vvx

0yv0v


Parabolic trajectories ( v=10 m/s , g = - 10 m/s2)

90° R:0.0m H:5.0m t=2.00s

75° R:5.0m H:4.7m t=1.93s

60° R:8.7m H:3.7m t=1.73s

45° R:10.0m H:2.5m t=1.41s

30° R:8.7m H:1.2m t=1.00s

15° R:5.0m H:0.3m t=0.52s


Example Problem

A medieval soldier is at the top of a castle wall. There are two cannon balls. The 1st one he fires from the cannon and it lands 200 m away. Simultaneously the 2nd cannon ball is dislodged and falls directly down. The 2nd cannon ball lands after 2.0 seconds and the fired cannon ball lands 2.0 seconds later. The ground is completely level around the castle. At what angle from horizontal did he fire the cannon (if g= -10 m/s2 & no air resistance)?


Example Problem

A medieval soldier is at the top of a castle wall. There are two cannon balls. The 1st one he fires from the cannon and it lands 200 m away. Simultaneously the 2nd cannon ball is dislodged and falls directly down. The 2nd cannon ball lands after 2.0 seconds and the fired cannon ball lands 2.0 seconds later. The ground is completely level around the castle. At what angle from horizontal did he fire the cannon (if g= -10 m/s2 & no air resistance)?

Find height of wall from 2nd cannon ball

0 = h + 0 – ½ 10 t2 h = 5(4) m = 20 m

Find angle

0 = h + v sin (t+2) – ½ 10 (t+2)2 0 = 20 m + 4v sin - 80 m

R = 200 m = v cos (t+2) 4v = 200 / cos

Combining 60 m = 200 m tan 16.7 degrees


Example Problem

Example Problem: If the knife is thrown horizontally at 10 m/s second and the knife starts out at 1.25 m from the ground, then how far does the knife travel be for it hit the level ground (if g= -10 m/s2 & no air resistance)?


Example Problem

If the knife is thrown horizontally at 10 m/s second and the knife starts out at 1.25 m from the ground, then how far does the knife travel before it hits the level ground

(assume g= -10 m/s2 & no air resistance)? at t=0 the knife is at (0.0 m, 1.0 m) with vy=0 after t the kinfe is at (x m, 0.0 m) x = x0 + vx t and y = y0 – ½ g t2

So x = 10 m/s t and

0.0 m = 1.25 m – ½ g t2 2.5/10 s2 = t2 t = 0.50 sx = 10 m/s 0.50 s = 5 .0 m


Fini

For Thursday, Read all of Chapter 4

Physics 201: Lecture 5, Pg 1 Lecture 5 l Goals: (Chapter 4.1-3) Introduce position, displacement, velocity and acceleration in 2D Address 2D motion.

Documents