Classic Mechanics : Dynamics

DynamicsDynamics

Newton’s lawsNewton’s laws Work and energyWork and energy Momentum & angular momentum

Momentum & angular momentum

Newton’s laws

Newton’s laws applicationapplication

Work and

energy

Work and

energy

Momentum and

angular momentum

Momentum and

angular momentum

Conservation of

energy

Conservation of

energy

Conservation of Momentum

and angular

momentum

Conservation of Momentum

and angular

momentum

"Nature and Nature's laws lay hid in night; God said, Let Newton be! and all was light."

Newton, Sir Isaac (1642-1727), mathematician and physicist, one of the foremost scientific intellects of all time.

a profound impact: a profound impact:

astronomy, physics, and mathematics

achievements: achievements:

• reflecting telescope• three laws of motion; • the law of universal gravitation• the invention of calculus

Key terms: dynamicsforce masssuperpositionnet forceNewton’s law of motionInertiaequilibriumaction-reaction pairelasticitytensionfriction forcegravitational interactionfree-body diagram

Chapter4-5 Newton’s Laws of Motion

1. Newton’s law of motion1.1 Newton’s first law

A body acted on by no net force moves with constant velocity and zero acceleration

inertial force

1.2 Newton’s second lawIf a net external force acts on a body, the body accelerates. The direction of acceleration is the same as the direction of the net force. The net force vector is equal to the mass of the body times the acceleration of the body.

amF

Chapter4-5 Newton’s Laws of Motion

dt

)m(d

dt

pdF

dt

dmv

dt

vdm

dt

mdF

)(

If m is a constant, then:

dt

dmF

am

Caution: This is a vector equation used for a instant. We will use it in component form.

(net force)

In the natural coordinate axis:

2

nn mmaF

dt

dmmaF tt

In the rectangular coordinate axis:

xx maF

yy maF

zz maF

te

m

ne

1.3 Newton’s third law

Whenever two bodies interact, the two forces that they exert on each other are always equal in magnitude and opposite indirection.

1.4 Application area of Newton’s law

low speed

macroscopic

practicality

inertial frame

2. What is a force?

Forces are the interactions between two or more objects.

unified field theory:gravity and

electromagnetic interaction

Einstain :Einstain :

S.L.Glashow S.L.Glashow

physicistsphysicists Grand Unification Theory

weak and electromagnetic interaction

2.1 Fundamental Forces:

x

2.2 the Common forces in mechanics

rr

mmGF ˆ

221

1) Gravitation

okxF

FF

k: force constant

x: elongation

2) elasticity

3) Frictional force

Static friction

Kinetic friction

Nf kk

Nf ss 0

3. Applications of Newton’s law

* Identify the body

Examine the force, draw a free body diagram

Construct a coordinate

* Write the Newton’s law in component form

* Calculate the equation

Problem-solving strategy

Example: A wedge with mass M rests on a frictionlesshorizontal table top. A block with mass m is placed on thewedge, and a horizontal force F is applied to the wedge. What must be the magnitude of F if the block is to remain

at a constant height above the table top?

Solution:

x

yF

M

mN

mg

For m: x:Nsin=ma y:Ncos=mg a=g.tg

For (M+m):

F=(M+m)a =(M+m)g.tg

a

mm N

mg

Fs

Discussion:What must be the magnitude of a if the block does not slide down the wedge?Draw a free body diagram of m.

horizontal: N=maperpendicular: N=mg So: a=g/ 。

If a,then N , then N>mg, will the m go up?

mR

m N

mg

Fs

Discussion:What must be the if the block does not slide down the cylinder?Draw a free body diagram of m.

horizontal: N=mR2

perpendicular: N=mg R

g:so

Example: A man pulls a box by a rope with constant speed along a straight line, known:k=0.6, h=1.5m. Find how long the rope is when F=Fmin.

L

m

hF

fk

N

mg

Solution: horizontal ： Fcos - fk=0 perpendicular ： Fsin + N - mg=0 fk = µN

sincos:

mgFso

So: tg = µ 。that is: when L=h/sin=2.92m F=Fmin

0,02

2

d

Fd

d

dFF=Fmin ：

Example: A parachute man drop into air, the resisting force is approximately proportional to the man’s speed v, find the velocity in any instantaneous time and the final velocity?

Solution:

o

y

Draw a free body diagram of the man, establish Newton’s law in y direction:

,mafmg kvf

dt

dvmkvmg

Suppose:k

mgvt

mg

f

o

y

mg

f

t

0

v

0t

dtm

k

vv

dv

)e1(vv m

kt

t

final velocity: ck

mgvt

Can you describe the man’s motion?

dt

dvmkvmg Suppose:

k

mgvt

Example: Try to find the path of the particle according to the picture.

x

y

m

F=f0 t iv0

O

Solution:

,dtdva,tfma:x xx0x

m2

tfv:so,dt

m

tfdv

0v,0t:conditioninitial2

0x

t

0

0v

0 x

x

x

dtm

tfdv:so 0

x

dt

dxvx dt

m2

tfdx

20

irjvv0000

x

y

m

F=f0 t iv0

O

m6

tfx:so,dt

m2

tfdx

0x,0t:conditionsinitial3

0t

0

20

x

0

330

0 ymv6

fxeqationpathso :

tvyy 0:

Example: A small bead can slide without friction on a Circular hoop that is in a vertical plane and has a radius of R=0.1m. The hoop rotates at a constant rate of =4.0rev/s about a vertical diameter.

a) Find the angle at which the bead is in vertical equilibrium.

b) Is it possible for the bead to “ride” at the same elevation as the center of the loop?

c) What will happen if the hoop rotates at 1.00rev/s

p. 161, 5-103Solution: a) For the bead

normal ： Nsin=m2Rsinperpendicular ： Ncos=mg

cos=g/ (2R) =80.90

=4.0rev/s

R=0.1m

N

mg

b) N can not balance mg, so…

c) When =1.0rev/s=2rad/s,

cos=2.5, so the bead stay at the bottom of the loop

Example: A small bead can slide without friction on a Circular hoop that is in a vertical plane and has a radius of R=0.01m. The hoop rotates at a constant rate of =4.0rev/s about a vertical diameter.

a) Find the angle at which the bead is in vertical equilibrium.

b) Is it possible for the bead to “ride” at the same elevation as the center of the loop?

c) What will happen if the hoop rotates at 1.00rev/s=4.0rev/s

R=0.1m

N

mg

Example: A small bead at rest slide down a frictionless bowl of radius R from point A. Find N, an , at at this position.

R

oA

N

mg

Solution:

dt

dtangential ： mgsin = mat = m

R

2normal ： N-mgcos = man = m

so ： at=gsin

fn=man , ft=mat

dt

dmmg

sindt

d

d

dm

dmgRdm sin0

2

d

d

Rm

d

dm

cos2

1 2 mgRm

4. Noninertial Frame of reference

B

A

a

m kB: mass m is accelerating.

A: mass m is at rest.

Which one is right?

The bus is not a inertial frame of reference.

A frame of reference in which Newton’s first law is valid is called an inertial frame of reference.

Any frame of reference will also be inertial if it moves relative to earth with constant velocity.

Newton’s laws of motion become valid in non-inertial system by applying a inertial force on the object.

B

A

a

m k

amF:assume i

ABmAmB aaa a'a

am'amam mB

F

xkF

amFi

'amamF

Transposition:

F

: real force

iF

: inertial force a

: acceleration of A relative to B

'amFF i

Then

:

Example: A wedge rests on the floor of a elevator. A block with mass m is placed on the wedge. There is no friction between the block and the wedge. The elevator is accelerating upward. The block slides along the wedge. Try to find the acceleration of the block with respect to the elevator.

a

m a

mgma

Nm a

Solution: The elevator is the frame of reference, there are three forces acts on the block.

m(g+a)sin=ma

a=(g+a)sin

Example: A wedge with mass M rests on a frictionlessHorizontal table top. A block with mass m is placed on The wedge. There is on friction between the block and the wedge. The system is released from rest.a) Calculate the acceleration of the wedge and the Horizontal and vertical components of the acceleration of the block.b) Do your answer to part (a) reduce to the correct results When M is very large?c) As seen by a stationary observer, what is the shape of

the trajectory of the block? (see page 182, 5-108)

M

m

mg

N

'Na

MaNxMfor sin:)(

ma

'sincos://

)(

mamgma

mfor

x

y'a

cossin: mgNma

Tracing problem

Plane: x=x0+vt, y=hMissile: dY/dX=(y-Y)/(x-X)

h

v

x

y

O

u

So: dY/dt=k(y-Y) dX/dt=k(x-X)

And: k2[(y-Y)2+(x-X)2]=u2

k=u/[(y-Y)2+(x-X)2]-1/2

So: Y(n+1)=Y(n)+k(y-Y)t X(n+1)=X(n)+k(x-X)t

Y(0)=0, X(0)=0