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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
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
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