Download 32. (5.1, 5.4) Newton`s second law In an inertial reference frame, the

32. (5.1, 5.4) Newton's second law
2
F41
1
4
F43
3
F42
a
F net
In an inertial reference
frame, the acceleration of a
particle is proportional to the
net force (the sum of all
forces) exerted on the
particle
and
inversely
proportional to the mass of
the particle:
r
r
r
Fnet = ∑ Fi = ma
i =all
(the SI unit of force is 1N=1kg⋅1m/1s2)
33. (5.6) Newton's third law
If one body exerts a force on another body, the second
body exerts an opposite force on the first one:
r
r
F12 = − F21
F 12
F 21
1
2
41
34. (6.6, 5.1) Mechanisms of interaction (forces)
On the microscopic scale there are only four kinds of
interaction between particles: gravitational (attractive),
electromagnetic (attractive or repulsive), strong nuclear
force, weak nuclear force.
To simplify the analysis of interactions between bodies
(systems of particles) it is convenient to introduce complex
interactions.
35. (14.1) The gravitational force
a) A particle with mass m1,
F 21
separated from a particle with 1
r12
mass m2 exerts
r an attractive
2
force (vector) F21 on the other
particle
r
F21 = − G m12m2 ⋅ r$12
r12
The symbol
r G denotes the universal gravitational constant,
and r$12 = r12 a unit vector in the relative position of the
r12
particles.
b) Objects with spherically distributed mass, also obey
the above equations with r representing the distance
between the centers of the objects. The force is exerted
toward the center of the object exerting the force.
42
36. (5.5, 14.3) Weight
The product of the mass of an object and the free fall
acceleration at the location of the object is called the
weight of the object:
r
r
W = mg
If an objects moves near the surface of a planet, then
its distance to the center of the planet, for all practical
purposes, does not change. On a planet with radius R and
mass M the weight of a body is approximately equal to the
gravitational force exerted on the object by the planet:
W = GM
m
R2
Therefore
g = GM
R2
On earth g = 9.80 m/s2.
37. (5.6, 12.4) The normal force
W
r
The
normal
force
F
N is the
N
vector component of a force that a
rigid surface exerts (due to
deformation; strain) on an object
with which it is in contact, namely,
in the direction perpendicular to
the surface.
The normal force prevents objects from crossing the
surface, therefore it is dependent on the other forces
applied to the body.
43
38. (5.8) Static frictional force
F
Static frictional force is a
component of a force that a rigid
surface exerts (due to strain) on
fs
the surface of an object with
which it is in contact, namely, in
the direction parallel to the
surface. Within certain limits the
static friction (interaction) prevents the object's surface
from moving along the rigid surface
f s ≤ µs N
N
The coefficient µs is called the coefficient of static friction.
Static frictional force depends on the other forces
applied to the body.
39. (5.8) Kinetic frictional force
Kinetic frictional force is a component of a force that
a rigid surface exerts (due to deformation) on the surface of
an object with which it is in contact, namely, in the
direction parallel to the surface.
fk = µk N
The coefficient µk is called the coefficient of kinetic
friction. The kinetic frictional force is always in the
direction opposite to the relative velocity of the surfaces.
44
40. (5.7, 12.4) Tension force
T
-T
r
Tension force T is a force that a surface exerts (due to
deformation; strain) on an object with which it is in contact,
namely, in the direction perpendicular to the surface.
Tension force prevents the object from leaving the
surface, therefore it is dependent on the other forces
applied to the body.
41. (5.7) Free-Body Diagram
In order to successfully predict the motion of a body,
one has to recognize all the forces acting on the body. A
figure with all the forces marked is called a free body
diagram. In identifying the forces affecting the motion of
the body make sure that the forces are exerted on the
considered body. Also remember that each force has its
source in another body.
When using the scalar components of vector quantities
related to a certain body be consistent with the choice of
the reference frame. (If two or more bodies are discussed
in a single problem, they can be considered in separate
reference frames.)
45
42. (6.3) Motion in noninertial reference frames
y'
Let's relate the scalar
components of velocity and
Ω
y
acceleration in case of an
arbitrary motion of the primed
r'
x'
reference frame with respect to
O'
the unprimed reference frame.
r
R
z'
The primed reference frame can
x
move, rotate and accelerate, but
O
z
let's maintain that time passes by
identically in both reference frames.
In the unprimed reference frame we change constantly
the base ˆi ', ˆj', kˆ ' of the unprimed reference frame. Notice
v
that all three base vectors rotate with angular velocity Ω ,
therefore the time derivatives of those base vectors
represented in the unprimed reference frame are
dˆi' r ˆ dˆj' r ˆ dkˆ ' r ˆ
i)
= Ω × i',
= Ω × j' ,
= Ω × k'
dt
dt
dt
If in the primed reference frame the components of the
position vector are r ' (t ) = [x ' (t ), y ' (t ), z' (t )] then the position
vector in the unprimed reference frame is
r
ii)
r '( t ) = x ' (t )ˆi ' (t ) + y' (t )ˆj' (t ) + z' (t )kˆ ' (t )
j'
k'
i'
j
i
k
{
}
Therefore, we can relate the position vector (oriented
segment) in the unprimed reference frame in terms of the
scalar components in the primed reference frame and the
primed base:
r
r
iii)
r( t ) = R (t ) + x ' (t )ˆi ' (t ) + y' (t )ˆj' (t ) + z' (t )kˆ ' (t )
46
In order to relate velocity in both frames we have to
find calculate the time derivative of both sides:
r
r
r
dr( t ) dR  dx ' ˆ dy' ˆ dz' ˆ   dˆi '
dˆj'
dkˆ ' 
=
v(t ) =
=
+
i'+
j'+ k '  +  x' + y' + z'
dt
dt  dt
dt
dt   dt
dt
dt 
r
r
r
r
= V ( t ) + v' x ˆi '+ v' y ˆj'+ v' z kˆ ' + x ' Ω × ˆi ' + y' Ω × ˆj' + z' (Ω × kˆ ') =
r
r
r
v
= V ( t ) + v' x ˆi '+ v' y ˆj'+ v' z kˆ ' + Ω × x ' ˆi ' + Ω × y' ˆj' + (Ω × z' kˆ ') =
r
r
= V ( t ) + v' x iˆ'+ v' y ˆj'+ v' z kˆ ' + Ω × x' ˆi'+ y' ˆj'+ z' kˆ '
(
(
(
iv)
) ( ) (
) (
) (
) ( (
)
)
))
r
r
r
r
r
v (t ) = V( t ) + v ' (t ) + (Ω(t ) × r ' (t ))
In order to find the relation between acceleration we have
to calculate the time derivative of velocity.
( (
)
(
))
r
r
d r
a(t ) =
V + v' x ˆi '+ v' y ˆj'+ v'z kˆ ' + Ω × x' ˆi '+ y' ˆj'+ z' kˆ ' =
dt
r
dV  dv' x ˆ dv ' y ˆ dv' z ˆ   dx ' r ˆ dy' r ˆ dz ' r ˆ 
(Ω × k') +
=
+
i'+
j'+
k ' + 
Ω × i' +
Ω × j' +
dt  dt
dt
dt   dt
dt
dt

r
r   dx ' dy' dz' 
r
r
r
dΩ
ˆj'+ kˆ '  + x' Ω × iˆ' + y' Ω × ˆj' + z' (Ω × kˆ ')  =
+
× x' ˆi'+ y' ˆj'+ z' kˆ ' + Ω ×   iˆ'+
dt
dt
dt 
  dt

r
r
dV
=
+ a' x ˆi '+ a' y ˆj'+ a' z kˆ ' + 2 ⋅ Ω × v' x ˆi '+ v' y ˆj'+ v' z kˆ ' +
dt
r
r r
dΩ
+
× x' ˆi'+ y' ˆj'+ z' kˆ ' + Ω × Ω × x' ˆi'+ y' ˆj'+ z' kˆ '
dt
(
(
v)
(
)
(
(
)
( (
)
)
)
(
) (
)
( (
))
r r r r r r r r r r
a = A + a'+ 2Ω × v '+ α × r '+ Ω × (Ω × r ')
47
)
)
Let's assume that the unprimed reference frame is an
inertial reference frame. The net force of all the force
exerted on a particle by other bodies is equal to the product
of the particle's mass and acceleration (in unprimed
reference frame):
r
r
vi)
Fnet = ma
However, in the primed reference frame product of mass
and acceleration is equal to the sum of the net force exerted
on the particle and a few other terms!
r
r r
r r r
r r
r r
vii) ma' = Fnet − mA − 2 mΩ × v '− mα × r − mΩ × (Ω × r ')
Each of the terms represent a certain kind of a fictitious
"force" (also called "inertial force"). Notice that other
bodies do not exert these “forces”.
48
WORK AND ENERGY
43. (7.1, 7.3) Work (scalar) performed on an object by
another object (by the body exerting the force) is a measure
of cumulative interaction. It depends on the force and the
path of a particular point of the body. Consequently it
depends on the choice of the reference frame.
a) The work dW done on a
particle
displaced
along
r
differential pathr dr, by an object
exerting force F is defined as
F
A
dr
B
r r
dW ≡ F ⋅ dr
r r
W
F
We can define work in an integral form
= ∫ ⋅ dr .
path
The SI unit of work is 1J=1N⋅1m
b) For the work done on an object, one must specify on
which point associated with the object the work is done (the
center of mass, the point of the force application ... ).
49
Example 1: Work done by the gravitational force (near the
earth's surface) on the center of mass of an object with
mass m is r
r
Wg = ∫ W ⋅ dr = ∫ − mg ⋅ dy = − mg ( h f − h i ) = − mg ∆h
path
path
where ∆h is the change in the object's altitude.
Example 2: Force independent of position
r r r
r r r
W = ∫ F ⋅ dr = F ⋅ ∫ dr = F ⋅ ∆r
path
path
Example 3. Work done by a spring being stretched.
Fx = -kx
x
0
x
The
force
created in a spring is
related
to
the
deformation of the
spring by Hooke's law:
r r
r
F(x ) = − kx
r
where the position vector x represents the displacement of
the free end from the relaxed position in the direction of the
spring. The proportionality coefficient k is called the spring
constant.
Work done by the spring is therefore
r r x
kx 2
W = ∫ F ⋅ d r = ∫ (− kx )dx = −
2
path
0
50
44. (7.4) Definition of the kinetic energy (classical)
A particle with mass m, moving with speed v has
kinetic energy K of
mv 2
K≡
2
(The SI unit of kinetic energy is 1J)
Note. Kinetic energy is used to describe the motion of an
object. When an object is approximated by a particle, the
kinetic energy defined above is called the translational
kinetic energy of the object.
45. (7.5) The Work-Energy Theorem I (for the net force
of all forces exerted on a particle).
In an inertial reference frame, the work dW done by
all the forces exerted on the particle (the net force) is equal
to the change in the kinetic energy dK of the particle
dW = dK
(or ∆ Wnet = ∆ K)
The work-energy theorem is a direct consequence of
the definition of kinetic energy, the definition of work and
Newton's second law.
proof.
r
r
 m( rv )2 
r
r r
dv r
r r
 = dK
dWnet = Fnet ⋅ dr = ma ⋅ dr = m ⋅ vdt = mv ⋅ dv = d

dt
 2 
51
Example: What is the speed of an object after falling 5m
from a resting position?
a)
from Newton's 2nd law
Assume the initial location of the object is the reference
point and the y-axis is in the vertical direction;
t0 = 0s
t1 = ?
- the time of the release;
- the time corresponding to the instant when the
object is at a position 5m below the release point;
v0=0 m/s - the initial velocity of the object
v1=? m/s - the velocity of the object at instant t1
a=[0,-g] - the acceleration (constant) of the object
Only the y-direction provides nontrivial equations.
v y ( t1 ) = 0 − 10 t1
2
−
10
t
1
5 = ( y( t1 )) = 0 + 0 ⋅ t1 +
2
From the second equation, we find that the object is 5m
below the release point 1s after it was released. Using this
value in the first equation we find that the vertical
component of velocity is
v y ( t1 ) = − 10 m / s2
Since the other components are zero, the speed of the
object is 10m/s.
52
b) from the work-energy theorem
K0 = 0J
K1= ?
m =?
- the initial (translational) kinetic energy of the
object
- the final (translational) kinetic energy of the
object
- mass of the object
Since the object is in the free fall motion, the only work
performed on the object is the gravitational work (done by
earth). The gravitational force is constant throughout the
motion therefore the net work is
Wnet = Wg = − mg ∆ h = − m ⋅ 10 m / s2 ⋅ ( − 5 m )
That work causes a change in the kinetic energy and
according to the work-energy theorem I
mv 22 mv12
−
= − mg ∆h
2
2
Therefore
v = − 2g∆ h = − 2 ⋅ 10 m / s2 ⋅ ( − 5 m ) = 10 m / s2
53
46. (7.5) Power
a) Definition
The power of a force is defined as the rate at which
work is done by that force.
dW
P( t ) ≡
dt
(The SI unit is 1W=1J/1s)
b) From the definition it follows that the work ∆W done
in a certain time interval ∆ t = t 2 − t1 is
t2
∆W = ∫ P( t )dt
t1
c)
The average power during this interval is
t2
Pav =
∫ P(t )dt
t1
t 2 − t1
=
∆W
∆t
r
d) When a force F is exerted by a body on a particle
r
moving with velocity v, the power delivered by this body is
r r r
r
P = dW = F ⋅ dr = F ⋅ v
dt
dt
54
47. (8.intro., 8.2) Conservative and nonconservative
forces
B
A
a) (Definition) If the work done
by a force on an object moving
between
two
positions
is
independent of the path of the
motion, the force is called a
conservative force.
All other
forces are called nonconservative
forces.
b) (Theorem) The work done by a conservative force
around a loop (the object returns to its initial position) is
zero.
48. (8.2) Definition of potential energy
If a force exerted on a particle is conservative, the
change in potential energy dU from one position to another
is defined by the work dW performed by that force as
follows:
dU ≡ - dW
(or ∆U = -∆U)
This definition assigns potential energy only with
accuracy to a constant. We have to choose an arbitrary
position of the object (the reference point for potential
energy) to which we assign zero potential energy.
55
49. (8.1) Gravitational potential energy (near the surface)
Ug
The gravitational potential
energy Ug is the potential energy
that an object has, by virtue of
position, above the surface of the
earth.
h
The gravitational potential
energy Ug of a particle with
mass m, placed at a position with a vertical component
different by h from the reference location is
Ug = mgh
proof.
r
yo +h
r r r
r
r
r
Ug (r ) = Ug , ref − Wro →r = − r∫ mgdr = − ∫ − mgdy = mgh
ro
y0
50. (14.7) Gravitational potential energy (in general)
The
gravitational
potential Ug
energy of a particle with mass m,
placed at distance r from another
particle with mass M is
r
Mm
UG (r ) = −G
r
(The reference point for the potential
energy is far away from the object)
proof.
r
Mm
r
Mm
Mm
1 1 
U G = − ∫ G 2 rˆ ⋅ dr = − ∫ − G 2 dr = −GMm  −  = −G
r
r
r
r ∞
path
∞
56
r
51. (8.2) The ideal spring
Fx = -kx
x
x
0
r
a) The restoring force F of an ideal spring depends on the
r
displacement x, by which the spring is stretched or
compressed from the unstrained position
r
r
F = −kx
(the coefficient k is called the spring constant.)
Us
b) The elastic potential energy
U s that an ideal spring has by
virtue of being stretched or
compressed is
2
x
U s = kx
2
where x is the change in length of the spring from its
unstrained length.
proof.
kx 2
Us = − Ws = − ∫ − kx ' dx ' =
2
0
x
57
52. (8.4) Definition of mechanical energy;
The sum of the kinetic and the potential energy of a
particle is called the total mechanical energy of the particle.
E≡K+U
53. (8.5) Work-energy theorem II
If some forces exerted on a particle are conservative,
the work Wnc, done by all forces not included in the
potential energy, is equal to the change in the mechanical
energy ∆E of the particle.
Wnc = ∆ E
proof.
∆ E = ∆K + ∆ U = Wnet − Wc = Wnc
58
Example: Is it safe for an m = 100 kg person to bungee-jump from
a height of h = 200 m, if the bungee-cord has a length of l = 50 m
and its spring constant is k = 20 N/m ?
solution 1: work-energy theorem I
h
From the definition of potential
energy we can express the work done
by the gravitational force and the
elastic force in terms of the change in
position of the person
U
Wg = − ∆U g = mg (l + ∆l )
k ( ∆l ) 2
We = −
2
Ue
Ug
U
(The second expression is valid only
when the cord is stretched)
If we ignore the resistance of air, no other forces are
performing work on the person. The change in the kinetic energy of
the person between the top of the tower and the lowest point of the
jump is zero. Therefore from the work energy theorem we obtain a
quadratic equation with one unknown ∆l.
k ( ∆l ) 2
mg (l + ∆l ) −
=0
2
k (∆l ) 2 − 2 mg∆l − 2 mg l = 0
(
This equation has two mathematical solutions
59
)
mg ± ( mg) 2 + 2kmgl
∆l =
=
k
100kg ⋅10m / s 2 ± (100kg ⋅10m / s 2 ) 2 + 2 ⋅ 20 N / m ⋅100kg ⋅10m / s 2 ⋅ 50m
=
=
20N / m
 136 m
=
 − 37m
Only the positive solution is physically acceptable, because the
formula for elastic potential energy Ue is valid only for stretching. At
the lowest position the person will be at height
y = 200m-(50m+136m) = 14 m above the ground.
solution 2: work-energy theorem II
From the definition of potential energy, we can express the work
done by the elastic force in terms of the change in position of the
person
k ( ∆l ) 2
We = −
2
We can include the gravitational potential energy Ug into the
mechanical energy E. During motion, the mechanical energy changes
from its initial value
Ei = 0 J + 0 J
to its value at the lowest point
Ef = 0J + mg [ − ( l + ∆ l )]
Using the work-energy theorem II, we obtain the same quadratic
equation
k ( ∆ l )2
−
= − mg ( l + ∆ l )
2
60