Newton s Laws - Tongji...
Transcript of Newton s Laws - Tongji...
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Newton’s Laws
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Agenda today
1. Newton’s Laws
2. Force on a body
3. Centripetal force
4. Non-inertial reference system
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Newton’s First Law of motion:
Any object will keep its state of
movement when it is left along.
)( 0 Fconstv
Inertia(惯性):the property to keep one’s state
of movement
Mass(质量): the measure of the inertia of a
body
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Newton’s second Law of Motion
The acceleration of an object is directly
proportional to the net force and is inversely
proportional to its mass,its direction is the
direction of the net force.
amF
或dt
vdmF
Force(力): The first derivative of linear
momentum
Linear momentum(动量): the product of object’s
mass and velocity
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Newton’s third law of motion
Whenever one object exert a force on another
one ,the another exert a force on the one same
in magnitude and opposite in direction.
FF
These forces( action and reaction) appears in pairs.
This law implies the conservation law of momentum .
the action force and reaction force are in the same
nature
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According to Newton’s third law, the force I exert
on the cart will cancel the force that the cart exert
on me,so there is no need to pull the carriage
a wise horse’s reason
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Forces in NatureBasic interactions
The gravitation force(引力)
The electromagnetic force(电磁力)
The strong nuclear force(强相互作用力)
The weak nuclear force(弱相互作用力)
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Contact Forces
Elastic force(弹力)
xkF
0 x
F
Hook’s Law
K : force constant
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Friction
Kinetic Friction(滑动摩擦力)
Nf
: the coefficient of kinetic Friction
v
fN
gm
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A wonder caused by friction Leonid meteor shower
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Static friction(静摩擦力)
The Maximum static friction
Nfs
max
F
f
N
gm
Ff
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Air drag force
2
2
1AvCD
Where C is drag coefficient
Terminal speed
AC
Fv
g
t
2
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The way to solve problem
amF
steps:
(1)draw a sketch of the situation and free-body diagram.
(3)Resolve the force into component。
(2)choose the coordinate system
(4)apply the Newton’s law to x axis.y axis.
(5)solve the equation, and find the unknown
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As the figure shows,two blocks have masses:
mA=1.0kg and mB=2.0kg。The coefficient of
friction between A and B:1=0.20,The coefficient of
B and ground is 0.30. The magnitude of each block
is 0.15m·s-2。What is the Force exert on the objects?
mAg
Tf 1
N1
mBg N1f 1
f 2
BF
T
N2
A
Ay
xF
B
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amTf A1
01 gmN A
111 Nf
amTffF B 21
012 gmNN B
222 Nf
A: B:
By equation a, we have amTgm AA 1By equation b, we have:
amTgmmgmF BBAA )(21 By solving them : NF 2.13
y
x
mAg
Tf 1
N1
mBgN1f 1
f 2
BF
T
N2
A
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Centripetal force in circular movement
The force gives rise to centripetal acceleration in
circular movement
r
vmF
2
F
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x
y
A ball falls from point A along the half sphere with radius r, find the ball’s velocity and
force exert on the sphere at any position given.
A
mg
N
Solution:
dt
dvmmg cos
R
vmmgN
2
sin
Rd
dvv
dsdt
dvds
dt
dv
dRgvdv cos
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x
yA
mg
N
00
cos dRgvdvv
sin2
1 2 Rgv
sin2Rgv
R
RgmmgN
sin2sin
R
vmmgN
2
sin
sin3mg
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Force in a non-inertial reference system
mg
F
T
Inertial reference
system
A system that Newton’s Law holds true
Non-inertial reference system
A system that Newton’s law does not hold
ay´
x´
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mg
T
Q=ma
A force invented to make
Newton’s law true in a non-inertial
system
pseudoforce
amQ
a
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Newton’s second law in a non-inertial system
amQF
Example:
Suppose a disk rotating with a constant angular velocity ,
a man is walking along the radical way with constant speed
v from the center to the edge, what is friction force exerted
on the man? If we choose the disk as the reference system,
what is the pseudoforce?
Solution:Choose a rectangular coordinate system,we can
describe the man ‘s position function as:
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By derivative action, we have
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Thus, the friction force exert on the man is ma
The pseudoforce is
The first term is called centrifugal force(离心力), the
second one Coriolis force(科里奥利力).
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Why the shape of roller roaster not a circle?
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Using computer to solve physical problem
Modeling progress
Algorithm (算法)choosing
Use physical formula to write
differential equations
Write flowing chart or pseudocode
programming
The way to make your calculation faster
or more accurate
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An example revisit:
Example A ball bump into some kind of liquid with
initial velocity: its acceleration
is given by: . Find how far can
it go before it is at rest?
)(100 1
0
smiv
)(10 1 smiva
Euler’s method:
A Solution not very good:
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#include <stdlib.h>
#include <stdio.h>
#include <string.h>
void main()
{
double v0=100,a,r=0;
double rlast, err;
do{
a=-10.0*v0;
v0+=a*1E-6;
rlast=r;
r+=v0*1E-6;
err=r-rlast;
}while(err>=1E-7);
printf( "The displacement is %3.4f m before stopped\n",
r);
}
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Solve problem with matlab
Example A ball bump into some kind of liquid with
initial velocity: its acceleration
is given by: . Find how far can
it go before it is at rest?
Solution:By the definition of velocity and
acceleration, we have:
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Build m-file test1.m as following:
function dydt = test1(t,y)
dydt=[y(2)
-10*y(2)];
Save it in the working directory, then type these sentences
in the working space
tspan=[0,2];
y0=[0;10];
[t,y]=ode45(@test1,tspan,y0);
axis([0 2 0 2]);
plot(t,y(:,1))
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Or even simpler:
y2=dsolve('D2y=-10*Dy','y(0)=0','Dy(0)=10');
ezplot(y2,[0,2])
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Another Example by Matlab:
Project one : Draw the trajectory curve of the ball in
projection experiment at different height; with different
angle; considering the air resistance. (the ball’s initial speed
exceeds 200m/s)