Transcript Lec9

Conservative vs. Non-conservative Forces
The fundamental theorem of calculus can be written as follows:
x2
df ( x)
x dx dx  f ( x2 )  f ( x1 )
1
Why is this important? Because from it you can derive the
definition of conservative forces, which is broadly defined as:
df ( x)
 dx dx  f xi   f x f  xi   0
The circle is used to denote that the limits of integration occur on
a closed path, and thus the positions for evaluating the integral
(i.e. at the initial and final positions) occur at the same point.
9-1
Starting from the force equation, let’s re-write force as the
derivative of a scalar function, U, and evaluate the integral as the
derivative of U is integrated around a closed path.
 dU  x  
x Fdx  x  dx  dx U ( x1 )  U ( x2 )
1
1
x2
x2
What if the force was due to gravity?
x2
x2
x1
x1
 Fdx    mgdx 
x3
 Fdx  mgx
d mgx 
x dx dx  mgx 2  mgx1
1
x2
3
 mgx 2
1
 mgx 3
x1
x3
x2
x2
x1
 Fdx  mgx
x3
9-2
Now let’s assume that the particle’s motion occurred around a
closed path (meaning that it eventually returned to its starting
position). Evaluating the integral
x2
x3
x1
x1
x2
x3
 Fdx   Fdx   Fdx   Fdx
 Fdx  m gx  m gx  m gx  m gx  m gx  m gx  0
2
1
3
2
1
3
shows that the integral of gravitational force around a closed path is
zero. In other words, one can say that the state at which one can
find the particle under the action of gravity is independent the path
it takes between two points. Only the change in the scalar quantity
U is required to compute the change in state of its initial and final
positions. (i.e.)
U  U x f U xi   mgxf  mgxi  mgx f  xi 
Let’s now define the scalar quantity U as the potential energy.
9-3
Now let’s assume that the particle’s motion around a closed path
is due to the force by a spring…
x2
x2
x2
d 1 2
1 2 1 2
x Fdx  x  kxdx  x dx  2 kx dx   2 kx2    2 kx1 
1
1
1
x2
x3
x1
x1
x2
x3
 Fdx   Fdx   Fdx   Fdx
1 2 1 2  1 2 1 2  1 2 1 2 
Fdx


 2 kx2  2 kx1    2 kx3  2 kx2    2 kx1  2 kx3   0
The integral of the force due to a spring around a closed path is
zero, and thus a spring is a conservative force. If we were to
evaluate the potential energy of a spring at two different points:

1 2 1 2 1
U  U x f   U  xi   kx f  kx i  k x 2f  xi2
2
2
2

The scalar quantity U is now called the potential energy of the spring.
9-4
Now let’s generalize this to 3-dimensional vector fields. Assume
the you have a force vector written in Cartesian coordinates.

F x, y, z   Fx xˆ  Fy yˆ  Fz zˆ

U x, y, z 
U x, y, z 
U x, y, z 
ˆ
ˆ
F  x, y , z   
x
y
zˆ
x
y
z

F x, y, z   U x, y, z 
where

rf

rf

ri

ri

is denoted the gradient operator
  
 


 F r   r    U r   r  U ri   U r f

In order to test this concept, let’s look at the 3-D gravitational field...
9-5
The force of gravity can be written as the gradient of a scalar
function:


rˆ
1
F  Gm1m2 2  Gm1m2  U r 
r
r

1
U r   Gm1m2  
r
Which leads to an expression for the gravitational potential
energy defined purely as a scalar of the inverse distance away
from the source.
  
 
1
1
 F r   r   U r   r  Gm1m2 ri  Gm1m2 ri  0
Since the integral of the force over a closed path is zero, gravity
can be defined as a conservative force even in three dimensions.
9-6
What are some examples of non-conservative forces?
Friction and any processes that causes dissipation of heat.
Ff x  Dx  Dv
For example, the energy dissipated in going from position 1 to position 2,
assuming velocity is constant (this assumption does not matter, but just
facilitates the derivation), is given by:
x2
x2
 Ff x dx    Dvdx   Dv x2  x1    D
x1
x2  x1  x
x1
t
2
 x1 
Likewise, the energy dissipated in going from position 2 back to position 1 is:
x1
x1
 Ff x dx    Dvdx   Dv x1  x2    D
x2
x2
x1  x2  x
t
1
 x2 
Thus, the total energy dissipated in a closed path is:
2
2


x2  x1 
x1  x2 
D
 Ff x dx   Ff x dx   Ff x dx   D
x2
x1
0
t
t
which is not zero! Thus friction is a non-conservative force (i.e. is path dependent).
x1
x2
9-7
What about kinetic energy?
Starting
from the previous derivation for potential energy

r2
r

r1

r2

r1

r2
  2



 F r  dr    U r dx U (r1 )  U (r2 )


r2
 
 
 dr
 



r F r  dr  r mr  dr  r mr  dt dt  r mr  r dt
1
1
1
1




r
r2
r
 2    1  2  2 1  2 1  2
dr
  m dt  r   m dr  r   mr   mr2  mr1


dt
2
2
 r1 2
r1
r1

r2


1  1 
U (r1 )  U (r2 )  mr22  mr12  U  T
2
2
Thus the change in potential energy, U, due to movement in a conservative force
from position 1 to position 2 is re-allocated as kinetic energy, T. Another way to
say this is that the sum of potential energy and kinetic energy must always
remain constant. (i.e.) U 2  T2  U1  T1  const. This is a very powerful result
because now we only have to look at the initial and final states instead of
integrating through the entire path of motion.
9-8
9-9
Initial conditions at time t=0
Velocity is zero, so T=0
Potential energy is U=mgy
Constraints: Bar has constant length
xA  xB 2   yA  yB 2  l 2  xB2  yA2


d 2
yA
2
xB  y A  2 x B xB  2 y A y A  0  x B  
y A
dt
xB
Solution by energy method
Ui  Ti  mgyAi  mgyBi  0  U f  Tf
l
1 2 1 2
mg
 m gyA  mx B  my A
2
2
2
We must find the position at which the acceleration of slider B is zero. This may be
an inflection point (i.e. where the velocity is also zero), it may be a point where the
velocity is at a local maxima, or where the velocity is at a global maximum.
9 - 10
Solving for the velocity of slider B:
 l

 l
 2  xB 
2
x  2 g 
 y A   y A  2 g 
 y A   x B  
 2

 2

 yA 
2
2
B
  x 2 
 l

x B2 1   B    2 g 
 yA 
  yA  
 2



2
2
  x 2 
l
l
2
gy



2
2
2
A  l
xB 1   B    x B 2  2 g 

 y A   xB  2 
 yA 
  yA  
yA
l  2
 2




d 2
d 2 gyA2  l
 2g  l

2
xB  2 xB x B 


 yA   2 
2 yA yA  3yA yA 

2
dt
dt l  2
 l  2

gy A
xB  2
2ly A  3 y A2  0
l x B
So, what conditions leads to the acceleration of slider B equal to zero?


9 - 11
The condition that must be satisfied for the acceleration of slider B to equal
zero is shown below.
2ly A  3 y A2  0
There are two solutions. One of them is: y A  0 , however this is an
inflection point, where the velocity is changing its direction. The other
solution is:
2
yA 
l
3
Due to the constraints of the length of the bar, the x-position is therefore:
 2
x B  l 1  
 9
And therefore the maximum velocity is given as:
2
2
gy
8 gl

2
A  l
x B  2 
 yA  
 x B 
54
l  2

8 gl
54
9 - 12
Problem 1
B
r = 600 mm
O
C
q
A
200 g
A small 200-g collar C can slide on a
semicircular rod which is made to rotate
about the vertical AB at the constant rate of
6 rad/s. Determine the minimum required
value of the coefficient of static friction
between the collar and the rod if the collar
is not to slide when (a) q = 90o, (b) q = 75o,
(c) q = 45o. Indicate in each case the
direction of the impending motion.
9 - 13
Problem 1
B
r = 600 mm
O
C
q
A
200 g
A small 200-g collar C can slide on a
semicircular rod which is made to rotate
about the vertical AB at the constant rate of
6 rad/s. Determine the minimum required
value of the coefficient of static friction
between the collar and the rod if the collar
is not to slide when (a) q = 90o, (b) q = 75o,
(c) q = 45o. Indicate in each case the
direction of the impending motion.
1. Kinematics: Determine the acceleration of the particle.
2. Kinetics: Draw a free body diagram showing the applied
forces and an equivalent force diagram showing the vector
ma or its components.
9 - 14
Problem 1
B
r = 600 mm
O
C
q
A
200 g
A small 200-g collar C can slide on a
semicircular rod which is made to rotate
about the vertical AB at the constant rate of
6 rad/s. Determine the minimum required
value of the coefficient of static friction
between the collar and the rod if the collar
is not to slide when (a) q = 90o, (b) q = 75o,
(c) q = 45o. Indicate in each case the
direction of the impending motion.
3. Apply Newton’s second law: The relationship between the
forces acting on the particle, its mass and acceleration is given
by S F = m a . The vectors F and a can be expressed in terms of
either their rectangular components or their tangential and normal
components. Absolute acceleration (measured with respect to
a newtonian frame of reference) should be used.
9 - 15
Problem 1 Solution
Kinematics.
B
w
r = 600 mm
O
C
q
A
B
200 g
r = 600 mm
q
O
C
an
an = (r sinq)
w2
an = (0.6 m) sinq ( 6 rad/s )2
A
r sinq
an = 21.6 sinq m/s2
9 - 16
Problem 1 Solution
B
Kinetics; draw a free body diagram.
r = 600 mm
O
C
q
A
200 g
F
O
q
N
=
man = (0.2) 21.6 sinq
= 4.32 sinq N
(0.2 kg)(9.81 m/s2)
9 - 17
Problem 1 Solution
F
Apply Newton’s second law.
O
q
N
=
man = (0.2) 21.6 sinq
= 4.32 sinq N
(0.2 kg)(9.81 m/s2)
+
+
SFt = 0:
F - 0.2 (9.81) sin q = - 4.32 sinq cos q
F = 0.2 (9.81) sin q - 4.32 sinq cos q
SFn = man: N - 0.2 (9.81) cos q = 4.32 sinq sin q
N = 0.2 (9.81) cos q + 4.32 sin2q
F = mN
For a given q, the values of F , N , and m can be determined
9 - 18
Problem 1 Solution
F
O
q
N
=
man = (0.2) 21.6 sinq
= 4.32 sinq N
(0.2 kg)(9.81 m/s2)
Solution:
(a)
q = 90o,
F = 1.962 N, N = 4.32 N,
m = 0.454 (down)
(b)
q = 75o,
F = 0.815 N, N = 4.54 N,
m = 0.1796 (down)
(c)
q = 45o,
F = -0.773 N, N = 3.55 N,
m = 0.218 (up)
9 - 19
Problem 2
r
O
A
b
Pin B weighs 4 oz and is free to slide
in a horizontal plane along the rotating
arm OC and along the circular slot DE
D B C
friction
.
q of radius b = 20 in. Neglecting
and
assuming that q = 15 rad/s and
.
.
b
q = 250 rad/s2 for the position q = 20o,
determine for that position (a) the
radial and transverse components of
E
the resultant force exerted on pin B,
(b) the forces P and Q exerted on pin
B, respectively, by rod OC and the wall
of slot DE.
9 - 20
r
Problem 2
D
B C
q Pin B weighs 4 oz and is free to slide
O
in a horizontal plane along the rotating
arm OC and along the circular slot DE
b
of radius b = 20 in. Neglecting
friction
.
E
and assuming that q = 15 rad/s and
..
q = 250 rad/s2 for the position q = 20o, determine for that position
(a) the radial and transverse components of the resultant force
exerted on pin B, (b) the forces P and Q exerted on pin B,
respectively, by rod OC and the wall of slot DE.
b
A
1. Kinematics: Examine the velocity and acceleration of the
particle. In polar coordinates:
e
.
q
.
v = r er + r q eq
..
.2
..
..
a = (r - r q ) er + (r q + 2 r q ) eq
r = r er
q
er
9 - 21
r
Problem 2
D
B C
q Pin B weighs 4 oz and is free to slide
O
in a horizontal plane along the rotating
arm OC and along the circular slot DE
b
of radius b = 20 in. Neglecting
friction
.
E
and assuming that q = 15 rad/s and
..
q = 250 rad/s2 for the position q = 20o, determine for that position
(a) the radial and transverse components of the resultant force
exerted on pin B, (b) the forces P and Q exerted on pin B,
respectively, by rod OC and the wall of slot DE.
A
b
2. Kinetics: Draw a free body diagram showing the applied
forces and an equivalent force diagram showing the vector
ma or its components.
9 - 22
r
Problem 2
D
B C
q Pin B weighs 4 oz and is free to slide
O
in a horizontal plane along the rotating
arm OC and along the circular slot DE
b
of radius b = 20 in. Neglecting
friction
.
E
and assuming that q = 15 rad/s and
..
q = 250 rad/s2 for the position q = 20o, determine for that position
(a) the radial and transverse components of the resultant force
exerted on pin B, (b) the forces P and Q exerted on pin B,
respectively, by rod OC and the wall of slot DE.
3. Apply Newton’s second law: The relationship between the
forces acting on the particle, its mass and acceleration is given
by S F = m a . The vectors F and a can be expressed in terms of
either their rectangular components or their radial and transverse
components. With radial and transverse components:
.2
..
..
. .
S Fr = m ar = m ( r - r q )
and
S Fq = m aq = m ( r q + 2 9r- 23
q)
A
b
r
Problem 2 Solution
D
B C
Kinematics.
q
O
b
A
b
B
E
r
b
b
r = 2 b cos q
.
.
..
..
2q
q
O
A
q = 20o
.
r = - 2 b sin q q
.
r = - 2 b sin q q - 2 b cos q q 2
q = 15 rad/s
..
q = 250 rad/s2
9 - 24
B
Problem 2 Solution
..
..
r
b
O
2q
q
A
b
r = 2 b cos q,
For:
r = - 2 b sinq q,
b = 20/12 ft,
.
q = 15 rad/s
.
.
.
r = - 2 b sinq q - 2 b cosq q 2
q = 20o,
..
q = 250 rad/s2
r = 2 (20/12 ft) cos 20o = 3.13 ft
.
r = - 2 (20/12 ft) sin 20o (15 rad/s) = - 17.1 ft/s
..
r = -2(20/12 ft) sin 20o (250 rad/s2 ) - 2(20/12 ft) cos 20o (15 rad/s)2
..
r = - 989.79 ft/s2
9 - 25
Problem 2 Solution
r
D
B C
q
O
b
A
b
(a) Radial and transverse
components of the resultant force
exerted on pin B.
Kinetics; draw a free body diagram.
E
maq
Fq
Fr
r
=
r
B
q
O
mar
B
q
A
O
A
9 - 26
Apply Newton’s second law.
Fq
Problem 2 Solution
maq
Fr
r
=
r
B
q
O
+
Fr =
+
mar
B
q
A
SFr = mar:
A
O
..
.2
Fr = m ( r - r q )
(4/16)
[- 989.79 - ( 3.13 )(152 )] = -13.16 lb Fr = 13.16 lb
32.2
SFq = maq:
..
. .
Fq = m ( r q + 2 r q )
Fq = (4/16) [(3.13)(250) + 2 (-17.1)(15)] = 2.1 lb
32.2
Fq = 2.10 lb
9 - 27
Problem 2 Solution
r
B C
D
q
O
b
A
b
Fq
E
r
q
q
O
Q
Fr
A
=
r
B
q
q
O
B
P
A
Fr = - Q cos q
-13.16 = - Q cos 20o
Fq = - Q sin q + P
2.10 = - 14.0 sin 20o + P
Q = 14.00 lb 40o
P = 6.89 lb
20o
9 - 28
q
Problem 3
A 250-g collar can slide on a
horizontal rod which is free to rotate
about a vertical shaft. The collar is
A
B
initially held at A by a cord attached
to the shaft and compresses a spring
of constant 6 N/m, which is
undeformed when the collar is
.located 500 mm from the shaft. As
the rod rotates at the rate qo = 16 rad/s, the cord is cut and the
collar moves out along the rod. Neglecting friction and the
mass of the rod, determine for the position B of the collar (a) the
transverse component of the velocity of the collar, (b) the radial
and transverse components of its acceleration, (c) the
acceleration of the collar relative to the rod.
400 mm
100 mm
9 - 29
Problem 3
400 mm
100 mm
A
B
The collar is initially held at A by a
cord attached to the shaft and
compresses a spring. As the rod
rotates the cord is cut and the
collar moves out along the rod to B.
1. Kinematics: Examine the velocity and acceleration of the
particle. In polar coordinates:
e
.
q
.
v = r er + r q eq
.2
..
.
.
..
a = (r - r q ) er + (r q + 2 r q ) eq
r = r er
er
q
9 - 30
Problem 3
400 mm
100 mm
The collar is initially held at A by a
cord attached to the shaft and
compresses a spring. As the rod
rotates the cord is cut and the
collar moves out along the rod to B.
2. Angular momentum of a particle: Determine the particle
velocity at B using conservation of angular momentum. In polar
coordinates, the angular momentum HO of a particle about O is
given by
HO = m r vq
The rate of change of the angular momentum is equal to the sum
of the moments about O of the
. forces acting on the particle.
S MO = H O
If the sum of the moments is zero, the angular momentum is
conserved and the velocities at A and B are related by
9 - 31
m ( r v q ) A = m ( r vq ) B
A
B
Problem 3
400 mm
100 mm
A
B
The collar is initially held at A by a
cord attached to the shaft and
compresses a spring. As the rod
rotates the cord is cut and the
collar moves out along the rod to B.
3. Kinetics: Draw a free body diagram showing the applied
forces and an equivalent force diagram showing the vector
ma or its components.
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Problem 3
400 mm
100 mm
A
B
The collar is initially held at A by a
cord attached to the shaft and
compresses a spring. As the rod
rotates the cord is cut and the
collar moves out along the rod to B.
4. Apply Newton’s second law: The relationship between the
forces acting on the particle, its mass and acceleration is given
by S F = m a . The vectors F and a can be expressed in terms of
either their rectangular components or their radial and transverse
components. Absolute acceleration (measured with respect to
a Newtonian frame of reference) should be used.
With radial and transverse components:
S Fr = m ar
.2
.
.
=m(r-rq )
and
S Fq = m aq
..
.
.
=m(rq+2rq
)
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Problem 3 Solution
400 mm
100 mm
A
Kinematics.
B
.
q
.
vq = r q
.2
..
ar = r - r q
.
..
vq
A
r
B
ar
aq
aq = r q + 2 r q
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Problem 3 Solution
400 mm
100 mm
A
(a) The transverse component of
the velocity of the collar.
B
Angular momentum of a particle.
m rA (vq )A = m rB (vq )B
.
since (vq )A = rA q
.
q
(vqA
(vqB
A
B
rA = 0.1 m
rB = 0.4 m
r
(rA)2 .
(vq )B = r
q
B
( 0.1 m )2
(vq )B =
(16 rad/s)
0.4 m
(vq )B = 0.4 m/s
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Problem 3 Solution
400 mm
100 mm
A
(b) The radial and transverse
components of acceleration.
B
Apply Newton’s second law.
Only radial force F (exerted by the
spring) is applied to the collar.
For r = 0.4 m:
Kinetics; draw a free
body diagram.
m aq
F
=
m ar
F = k x = (6 N/m)(0.5 m - 0.4 m)
F = 0.6 N
+
SFr = mar: 0.6 N = (0.25 kg) ar
ar = 2.4 m/s2
+ SFq = maq:
0 = (0.25 kg) aq
aq = 0
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Problem 3 Solution
400 mm
100 mm
A
B
(c) The acceleration of the
collar relative to the rod.
Kinematics.
For r = 0.4 m:
.
.
q
vq
A
r
B ar
.
vq
vq = r q, q =
r
. 0.4 m/s
q=
= 1 rad/s
0.4 m
The acceleration of the collar
..
relative to the rod is r.
.2
..
ar = r - r q
aq
..
(2.4 m/s2) = r - (0.4m)(1 rad/s)2
..r = 2.8 m/s2
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