4.5projectile motion 4.6range of projectile dynamics 4.8relative motion in 1d 4.9relative motion in...

4.54.5 Projectile motionProjectile motion

4.64.6 Range of projectileRange of projectile

Dynamics

4.84.8 Relative motion in 1DRelative motion in 1D

4.94.9 Relative motion in 3-DRelative motion in 3-D

4.74.7 Circular motion (study at home) Circular motion (study at home)

5.25.2 Newton 1Newton 1

5.3-5.55.3-5.5 Force, mass, and Newton 2Force, mass, and Newton 2

4.54.5 Projectile motionProjectile motion

4.64.6 Range of projectileRange of projectile

Dynamics

4.84.8 Relative motion in 1DRelative motion in 1D

4.94.9 Relative motion in 3-DRelative motion in 3-D

4.74.7 Circular motion (study at home) Circular motion (study at home)

5.25.2 Newton 1Newton 1

5.3-5.55.3-5.5 Force, mass, and Newton 2Force, mass, and Newton 2

Summary for Lecture 3Summary for Lecture 3

Problems Chap. 4 24, 45, 53, Problems Chap. 4 24, 45, 53,

First-year Learning Centre

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In the First-year Learning Centre

Notices

NoticesRepresentative for

SSLC(Staff-Student Liaison Cttee)

Representative forSSLC

(Staff-Student Liaison Cttee)

Physics Phrequent PhlyersClarify the physics

Extend your knowledge Ask the lecturer

Physics Phrequent PhlyersClarify the physics

Extend your knowledge Ask the lecturer

Room pp211Physics Podium

Thursdays 12 – 2 pm

Room pp211Physics Podium

Thursdays 12 – 2 pm

Gallileo

Parabola

Max range 45o

2 angles for any range

Parabola

Max range 45o

2 angles for any range

Running out of “impetus”

What do we know from experience?

The trajectory depends on:

Initial velocity

Projection angle

Anything else?Air resistance Ignore for now

Projectile Motion

To specify the trajectory we need to specify every point (x,y) on the curve.

That is, we need to specify the displacement vector r at any time (t).

x

y

r(t)

r(t) = i x(t) + j y(t)

x(t)

y(t)

v0 = ivox + j voy

-g

v0sin

v0cos

Usually we know the

initial vector velocity vo.

We know acceleration is constant = -g.

accel is a vector a = iax + jay

= i 0 - j g

= i v0cosjv0sin

What do we know?

vo

x

y

The vertical component of the projectile motion is the same as for a falling object

The horizontal component is motion at constant velocity

Finding an expression for Projectile motion

To define the trajectory, we need r(t)

That is:- we need x(t) and y(t)

vx(t) = v0x + axt

cosvx

to

x(t) = v0 cos t x(t) = v0xt + ½axt2

0 since ax=0

0 since ax=0

v0cos x

y

r

vo

Consider Horizontal motion

•velocity

= v0 cos•displacement

-g

const

Use x - xo = v0t + ½ at2

vertical component of displacement

ay= -gUse y - yo = ut + ½ at2

y(t) = v0yt + ½ ayt2

θcosv

xg

2

1

θcosv

xθsinvy(x)

220

2

0

0 Recall

cosvx

to

v0sin

x

y

r

vo

= v0 sin t – ½ gt2

xθtanxθcosv2

g)x(y 2

220

Re-arranging gives

Height (y) as a function of distance (x) for a projectile.

y = -k1 x2 + k2 x

x

y y = kx2

xtanxcosv

gy

2

2202

y = -k1x2 + k2x

xtanxcosv

gy

2

2202

x

y

range

vo

Range

For what x values does y = 0?

i.e. where 0)tanxcosv2

g(x

220

g

vcossin2x

20

0tancos2

2

22

0

xxv

g where

Maximum when 2 = 900.

i.e when = 450cosθ

sinθtanθnow

g

v2sinR

20

x = 0 or 0tanxcosv2

g22

0

tangcosv2

x22

0x=0cosθ

sinθ

R

x

yxx

v

gy

tan

cos22

22

0

x

y

x

yxx

v

gy

tan

cos22

22

0

sine0 angle

http://www.colorado.edu/physics/phet/web-pages/simulations-base.html

R

R’

xxv

gy

tan

cos 2

2202

mxy

y

x

Gradient of slope

Range up a slope

2

2202

xV

gxy

costan mxy mx

)cos

(tan xV

gmx

2220

xV

gm

o

222 costan

2222

0 xV

gxmx

cos.tan

True for x = 0 or

g

Vmx o

222 cos)(tan

y’

If m = 0 (on level ground)

g

Vmx o

22cos2

)(tan' 22 'y'x'R

y’ = mx’

2m1x'R'

R’= x’ = R

R

R’

y

x

x’

y

vo

For collision, x and y for dart and monkey must be the same at instant t

Time to travel distance d is cosov

dt

d

y for dart y - y0 = v0t + ½ at2 2

2

1gttvy o sin 2

2

1)

cos(

cossin

ooo v

dg

v

dvy

2

2

1)

cos(tan

ov

dgdy

2

2

1)

cos(

ov

dgd

d

hy

2

2

1)

cos(

ov

dghy

h2

oo

o )cosθv

dg(

2

1

cosθv

dsinθvy

-g

yo = 0

y

h

vo

d

How far has monkey fallen?

2

2

1)

cos(

ov

dghy y for dart at impact

(ie at time )cosov

dt

Therefore the height of the monkey is

2

o)

cosθv

dg(

2

1 y = h -

dist = v0monkt + ½ at2 2

2

1)

cos(

ov

dg

g

Dynamics

KinematicsKinematicsHOW things moveHOW things move

DynamicsDynamicsWHY things moveWHY things move

40%

QP

. R

If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO

In the absence of a FORCE

a body is at rest

1660 AD

A body only moves if it is driven.

In the absence of a FORCE

A body at rest WILL REMAIN AT REST

350 BC

DynamicsDynamicsAristotle

For an object to MOVE

we need a force.

Newton

For an object to CHANGE its motion

we need a force

Newtons mechanics applies for motion in

an inertial frame of reference! ???????

He believed that there existed an absolute (not accelerating) reference frame, and an absolute time.

The laws of physics are always the same in any inertial reference frame.

His laws applied only when measurements were made in this reference frame…..

Newton clarified the mechanics of motion in the “real world”.

…or in any other reference frame that was at rest or moving at a constant velocity relative to this absolute frame.

Inertial

reference

frame

Inertial

reference

frame

Inertial reference

frame

Newton believed that there existed an absolute (not accelerating) reference frame, and an absolute time.

The laws of physics are always the same in any inertial reference frame.

Einstein recognised that all measurements of position and velocity (and time) are relative.

There is no absolute reference frame.

Frames of Reference

The reference frames may have a constant relative velocity

We need to be able to relate one set of measurements to the other

The connection between inertial reference frames is the “Gallilean transformation”.

In mechanics we need to specify position, velocity etc. of an object or event.

This requires a frame of reference

x

y

z

o

p

The reference frames for the same object may be different

x’

y’

z’

o

Ref. Frame P (my seat in Plane)

T (Lunch Trolley)xTP

Ref. Frame G (ground)

xPG

xTG

xTG = xTP + xPG

Vel. = d/dt(xTG) vTG = vTP + vPG

Accel = d/dt (vTG) aTG = aTP + aPG

In any inertial frame the laws of physics are the same

VPG(const

)

Gallilean transformations

0

N

E

GROUND

N’

E’

AIR

r PG

r AG

r PA

aPG = (vPG) = aPA + aAGdtd

vAG

rPG = rPA + rAG

dtd

vPG = (rPG) = vPA + vAG

PLooking

from above

0

In any inertial frame the laws of physics are the same

AG

PA

PA AG

VPG = VPA + VAG VPA = 215 km/h to East

VAG = 65 km/h to North

hkmv

v

vvv

PG

PG

agpaPG

/225

422546225

22

Tan = 65/215

= 16.8o

Ground Speed of Plane

AG

PA

PA AG

Ground Speed of Plane

In order to travel due East, in what direction relative to the air must the plane travel? (i.e. in what direction is the plane pointing?), and what is the plane’s speed rel. to the ground?

Do this at home and then do sample problem 4-11 (p. 74)

VPA = 215 km/h to East

VAG = 65 km/h to North

A motorboat with its engine running at a constant rate travels across a river from Dock A, constantly pointing East.

FlowN

A

Compare the times taken to reach points X, Y, or Z when the river is flowing and when it is not. Your answer should include an explanation of what might affect the time taken to cross the river, and a convincing justification of your conclusion.

I will quiz you on this next lecture

X

Y

Z

Here endeth

the lessonlecture

No. III

Isaac Newton

1642-1727

Newton’s 1st Law

If a body is at rest, and no force acts on it, IT WILL REMAIN AT REST.

If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO

If a body is moving at constant velocity, we can always find a reference frame where it is AT REST.

At rest moving at constant velocity

An applied force changes the velocity of the body

a F

ForceForceIf things do not need pushing to move at constant velocity, what is the role of FORCE???

Inertial mass

The more massive a body is, the less it is accelerated by a given force.

a = F 1m

F = am

The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line

in which that force is impressed.

a = F/m

Newton’s 2nd Law

a = F/m

aa = = vector SUM of ALL EXTERNAL forces acting on the body

vector SUM of ALL EXTERNAL forces acting on the body

mm______________

F = ma

vector SUM of ALL EXTERNAL forces acting on the body

vector SUM of ALL EXTERNAL forces acting on the body

= ma= ma

Newton’s 2nd Law

F=maF = iFx + jFy + kFz a = iax + jay + kaz

Fx=max

Fy=may

Fz=maz

Applies to each component of the vectors

FA+ FB+ FC= 0 since a= F/m and a = 0

F = 0

Fx = 0 Fy = 0

FCcos - FAcos47= 0

= 220 sin 47 +170 sin28

= 160 +80 240 N

FA sin 47 + FCsin – FB = 0

FB = FA sin 47 +FCsin

220N

? N

170N

Fy = 0

cos= cos47 =280220170

170 cos - 220 cos47= 0

(280)

m2g

N

TM2

m1g

TM1

Apply F = ma to each body

i.e Fx = max and Fy = may

For m2

Vertically

Fy = -N + m2g = m2ay = 0

N = m2g

Horizontally

Fx = T = m2a

21

1

mmgm

a

For mass m1

Vertical (only)

m1g - T = m1a

Analyse the equation!

m1g - m2a = m1a

m1g = m2a + m1a = a(m2 + m1)

41%

26%

33%

2.5 x 103 N 2.5 x 103 N 3.9 x 104 N

2.5 x 103 N 2.5 x 103 NT

3.9 x 104 N

F = ma m is mass of road train

5.0 x 103 N

3.9 x 104 N

F = ma

a = (39 – 5) x 103/4 x 104

a = F/m

a

a = 0.85 m s-2

2.5 x 103 N 2.5 x 103 N

Don’t care!

T

F = ma m is mass of trailer!

F is net force on TRAILER

2.5 x 103 N

Ta = 0 So F= 0

T - 2.5 x 103 = 0

T = 2.5 x 103 N

a = 0

2.5 x 103 N 2.5 x 103 N

3.9 x 104 N

F = ma m is mass of roadtrain

No driving force so F = -5.0 x 103 N a = - 5.0 x 103/4.0 x 104

= - 0.125 m s -2

Use v2 = vo2 + 2a(x – x0)

x – x0 = 400 mv = 0 u = 20 m s-1,