Transcript Chapter 4 Kinetics of a particle

Chapter 4 Kinetics of a Particle
slope  lim
f df

 tangent
x dx
df
differenti al  df  slope  x 
dx
dx
df
Local maximum occurs, when
 0 and f "  0
dx
df
Local minimum occurs, when
 0 and f "  0
dx
max
f(x)
f
x
min
x
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Integration: the reverse of differentiation
To calculate the area under the curve from xoto x :
x
 f ( xi )xi   f ( x)dx
i 1
xo
N
lim
x 0
i
x
d  f ( x)dx
xo
dx
 lim
x  0
f ( x)x

 f ( x)
x
f(x)
x  x
x
xo
xo
 f ( x)dx   f ( x)dx
x
xo
x
x
x x+x
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d cos x
 - sin x,
dx
d tan x
 sec 2 x
dx
d cot x
 - csc 2 x
dx
dx n
 n x n -1
dx
d ln x 1

dx
x
de x
 ex
dx
  sin x dx   cos x
  sec 2 x dx  tan x  C
  csc 2 x dx   cot x  C
n 1
x
  x n dx 
 C, n  - 1
n 1
1
  dx  ln x  C
x
  e x dx  e x  C
C is a constant
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Newton’s 2nd law


 Fi  ma

F2
i
Fx  ma x , Fy  ma y and Fz  ma y
where Fx  F1x  F2 x
Fy  F1 y  F2 y
Fz  F1z  F2 z

F1

FR
st law
Newton’s
1




 Fi  0, ma  0, a  0, v  constant
i
4
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Newton’s 3rd law
action = reaction
5
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Work done
 
dU  F  dr  Fds cos ,  Fx dx  Fy dy  Fz dz

where ds  dr
Total work done U  
 
F  dr
Example 1 What is the work done by a

force on a article:
v
(a) in circular motion?

(b) horizontal motion?
A
F
(c) from A to B?

dr


F
B

g
h
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Kinetic energy K.E.
Work done by an external force
 
 
2
2
W1  2   1 F  dr
  1 ma  dr
 
 
dv
2
2
  1 m  dr   1 mv  dv
dt 
d(v  v )
2
2 m
 1 m
  1 dv 2
2
2
1
2 1
 mv2  mv12
2
2

F

dr
 K .E. (change in K.E.)
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Power P
Energy dissipated per unit time

dU
dt

 
F  dr
 
F v
dt
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
force f (e.g.
Dissipative
friction): work done from one
point to another point depends on the path.
 
WA  B( path1)   path 1 f  dr


   path 1 mg dr ( f is opposite to dr )
 - mg S1
path 1
WA  B( path2)    path 2 mg dr
B
 - mg S2
path 2
 WA  B( path 2)
A
"" sign means the force is dissipativ e.
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
Non-dissipative force Fc (conservative force): work done
from one point to another point is independent on the path.
 
 
path 1
 path1 Fc  d r   path 2 Fc  d r
B
 
 
 path1 Fc  d r   path 2 Fc  d r  0
 
 
 path1 Fc  d r   reversedpath 2 Fc  d r  0
path 2
A
 
close path Fc  d r  0
After completion of a closed path under a conservati ve force, K .E.  0
 
 any path Fc  d r is the same.
 
Therefore,   any path Fc  dr is a function of initial and end points only,
It is defined as the change of potential energy, P.E.
P.E. between two points is equal to the work done by an external
force against the field of a conservative force for bringing
 the particle
from the starting point
to the end point,  any path (- Fc )  dr , with the

external force = - Fc .
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Example 2 (gravitational potential)

mM
F   G 2 rˆ ;
r
P.E.  V (r )
M
mM

ˆ
 V    [ G 2 r ]  dr

r
r
mM
 V    [ G 2 rˆ]  drrˆ

r
The reference point is set at , such that V  0 .
r
r

mM
mM
mM
V (r )   G 2 dr   G
G

r
r 
r
r
r̂
r
m
R
Example 3 Find V of a spring. Example 4 Potential energy of
Ans. kx2/2
a mass m, positioned at h from
the ground. Ans. mgh
X
X=0
X
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 
 F  dr  ΔK.E.
In general, the two types of forces coexist:




 
F dr  ΔK.E.

(Fc  f )  dr  ΔK.E.


f  dr  ΔK.E.   (-Fc )  dr

f  dr  ΔK.E.  ΔP.E.
If there is no dissipative force, K.E. + P.E. = 0,
i.e. conservation of mechanical energy.
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Example
The rod is released at rest from  = 0, find :
(a) velocity of m when the rod arrives at the horizontal position.
(b) the max velocity of m.
(c) the max. value of .
P.E.=-2mgr sin+mg(r-r cos )
K.E. = (2m+m)v2/2
(P.E.+ K.E.) = 0
3mv2/2 – mgr(2 sin + cos  -1) = 0
v = [2gr(2 sin + cos  -1)/3]1/2
r

r
m
(a) At  = 45o, v = 0.865 (gr)1/2
d
2

(b)
v 2 / gr   2 cos θ  sin θ   0  tan θ  2 or θ  63.4o
dθ
3
 vmax  0.908( gr )1 / 2
(c) θ has maximum value when v  0
2 sin θ max  cos θ max  1  0
2 1  cos 2 θ max  1  cos θ max
5 cos 2 θ max  2 cos θ max  3  0
cos θ max  0.2  0.8  1 or  0.6,
θ max  0 or θ max  126.9o
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2m
From definition of potential energy: dV(x) = -Fdx
From the concept of differential dV =
Fx  
dV
dx
dx
dV
dx
A

dr
B
Examples
mM
r
dV ( r )
d
mM
d 1
F 

( G
)  G mM
( )  G mM / r 2
dr
dr
r
dr r
Gravitatio nal force with V (r )   G
Restoring force of a spring with V ( x)  kx2 / 2
dkx2 / 2
F 
  kx
dx
Gravitatio nal force with reference on the ground with V ( h)  mgh
dmgh
F 
  mg
dh
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Linear momentum

dG


 d

F  mv  (mv ) 
G
dt
dt


With G  m v defined as the linear momentum
1. In a motion, linear momentum can be conserved,
t
2 

G  0, or  Fdt  0
t
1
When (i) the total (external) force is zero, or (ii)
the collision time t1  t2 is extremely short.
2. Define impulse = change in linear momentum:
2
1
 


F dt   dG  G2  G1
F
time
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Collision between systems A and B.
External force  0

t2
change in linear moment of A  G A  
t1

t2
change in linear moment of B  GB  
t1





t2
G  G A  GB   ( FA  FB )  dt
 t1


Since FA  - FB , G  0 (conservat ion of

FA  dt

FB  dt
linear momentum)
B
A
FR Fa
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Angular Momentum
Take moment about O
Angular momentum about O is :

   
H o  mr  v  r  G  mrv sin  Hˆ o

v

r
O

v


vr
m

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Torque = Moment of force about O is defined as :

 
M o  r  F 
 r  ma 

dv
 r m
 dt
d(r  mv )


 dt
 dG
dH o

(analogous to F 
)
dt
dt
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Example: Prove that the angular momentum of a
particle under a central force is conserved.

mm0
F  G 2 rˆ
r

v m
In polar coordination system :



m a  m r  rθ 2 rˆ  m(rθ  2rθ) θˆ
 Gmmo /r 2  m(r  rθ 2 )
(1)
0  m(rθ  2rθ)
( 2)
( 2) implies : 0 
1 d
r dt
(mr 2θ)
r
mo

F

mr 2θ  mrvθ  cons tan t
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