Transcript (linear) momentum

4. Momentum and Impulse
The linear momentum of an object is defined
as the product of its mass and its velocity. The
conservation law of momentum is another most
important physical laws in physics.
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Momentum
（动量）
The (linear) momentum of an object is defined to be the
product of the mass and velocity:
p  mv
Momentum is a vector quantity and its direction is the
same as that for velocity; And it has dimension ML/T. In SI
system, the momentum has the units kg·m/s.
p  mv ,   x, y , z .
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Momentum and Force
As pointed out before, the Newton’s second law can be
rewritten as:
dp
F
dt
From above equation, we see that if the net force on an
object is zero, the time derivative of the momentum is zero,
and thus the momentum of the object must be constant.
Of course, if the particle is isolated, then no forces act on it
and the momentum remains unchanged—this is Newton’s
first law of motion.
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Momentum and Isolated Systems
The momentum of an isolated system is a constant.
pt   pi
The total momentum
i
for an isolated system
Ft  0
Thus, we have dpt  0  p   p =Const
t
i
dt
i
pt  Const,   x, y, z.
The law of conservation of linear momentum!
（动量守恒定律）
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 Example 4.1 A 10000-kg railroad car traveling at a speed
of 24.0 m/s strikes an identical car at rest. If the cars lock
together as a result of the collision, what is their common
speed afterward?
Solution: The initial momentum is simply m1v1. After the
collision, the total momentum will be the same. Since the
twe two cars attached, they will have the same speed v '.
Then:  m1  m2  v '  m1v1 ,
which gives out
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v '  v1  12 m/s .
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Impulse and Momentum
Assuming a net varying force acts on a particle, then
dp
 Ft  dp  Ft dt
dt
the change in the momentum of the particle during the time
interval Δt = tf − ti reads:
tf
p  p f  pi   Ft dt
ti
The Impulse of a force is defined as:
tf
I   Ft dt  p
ti
impulse-momentum theorem
（冲量-动量定理）
Also valid for a system of particles.
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 Example 4.2 
Calculate the final speed of a 110 kg football
player initially running at 8.00 m/s who collides head on with
a padded goalpost and experiences a backward force of 1.76
104 N for 5.5 102 s.
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The Impulse is an interaction between the system and its
environment. As a result of this interaction, the momentum
of the system changes.
The impulse approximation: We assume that one of the
forces exerted on a particle acts for a short time but is much
greater than other force present. this simplification model
allows us to ignore the effects of other forces, because these
effects are be small during the short time during which the
large force acts.
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Collisions
When two objects collide, it is a good approximate in
many cases to assume that the forces due to the collision
are much larger than any external forces present, so we
can use the simplified model: the impulse approximation.
Collisions
Elastic collision
Inelastic collision
Completely inelastic collision
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Momentum is conserved in all cases,
but kinetic energy is conserved only in
elastic collisions!
（动量在所有的碰撞过程中守恒，而动能仅在弹性碰撞中守恒）
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 Example 4.3
A proton collides elastically with another
proton that is initially at rest. The incoming proton has an
initial speed of 3.50  10 m/s and makes a glancing collis-ion with the second proton, as in the figure. After the
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collision, one proton moves off at angle of 37.0 to the
original direction of motion, and the second deflects at
an angle of  to the same axis. Find the final speeds of
the two protons and the angle  .
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x component: m1v1i  0  m1v1 f cos   m2v2 f cos 
y component: 0  0  m1v1 f sin   m2v2 f sin 
For the collision is elastic, we get the third equation for
conservation of kinetic energy:
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1
1
2
2
2
m1v1i  m1v1 f  m2 v2 f
2
2
2
Solve the set of equations composed of above three
equations, we obtain:
v1 f sin 
tan  
v1 f cos   v1i
2
m
1
1
m1v12i  m1v12f  1  v12i  v12f  2v1i v1 f cos  
2
2
2m2
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The Centre of Mass
The center of mass of a system of particles is defined as:
rCM 

m r
i i
i
 xCM i  yCM j  zCM k
M
i  mi xi  j  mi yi  k  mi zi
i
For an extended object reads:
i
i
M
rCM
1

rdm

M
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The centre of mass of a homogeneous, symmetric body
must lie on an axis of symmetry.
The centre of mass is different from its centre of gravity.
Each portion of a system is acted on by the gravitational
force. The net effect of all of these forces is equivalent to
the effect of a single force Mg acting at a special point
called the center of gravity. The centre of gravity is the
average position of the gravitational force on all parts of
the object. If g is uniform over the system, the centre of
gravity coincides with the centre of mass. In most cases,
for objects or systems of reasonable size, the two points
can be considered to be coincident.
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E.g. 0.1 A system consists of three particles located at
the corners of a right triangle as in the figure. Find the
centre of mass of the system.
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E.g. 0.2 A rod of length 30.0 cm has a linear density:
  50.0 g/m  20.0 x g/m
2
where x is the distance from one end, measured in meters.
(a)What is the mass of the rod?
(b)How far from the x = 0 end is its center of mass?
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Motion of a System of Particles
rCM 
MaCM
 mi ri
i
M
dpt
  mi ai  Ft 
dt
i
dri
mi

drCM
dt
i

 vCM
dt
M
MvCM
dri
  mi
 pt
dt
i
Newton’s second law for a system of particles
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The sum of all the forces acting on the system is equal
to the total mass of the system times the acceleration of
its center of mass.
The center of mass of a system of particles (or bodies)
with total mass M moves like a single particle of mass M
acted upon by the same net external force.
The general motion of a system of particles can be
considered as the sum of the translational motion of the
CM, plus rotational, vibrational, or other types of motion
about the CM.
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Motion of a System of Particles
rCM 
MaCM
 mi ri
dri
mi

drCM
dt
i

 vCM
dt
M
i
M
dpt
  mi ai  Ft 
dt
i
MvCM
dri
  mi
 pt
dt
i
质心运动定律
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Outline of Rotational Motion
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Rigid body model: A rigid body is any system of particles in
which the particles remain fixed in position with respect to
one another.
Rotation about a fixed axis: Every particles on a rigid body
has the same angular speed and angular acceleration.
 d 
  lim

t  t
dt
 d 
  lim

t  t
dt
rigid body 刚体
rotation about a fixed axis 定轴转动
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Rotational kinematics
d
d
     0 ,  
,
;
dt
dt
t
t
t0
t0
  0    dt ,    0    dt.
Rotations with constant angular acceleration
  Const
  =0   t
1 2
    0  0 t   t
2
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Rotational and Translational Quantities
ds
d
tangential velocity: v 
r
 v  r .
dt
dt
dv
d
tangential acceleration: a 
r
 a  r .
dt
dt
2
v
2
2
radial acceleration: ac   r  ac  r .
r
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Rotational Kinetic Energy
（转动动能）
I   mi ri
the moment of inertial
2
i
1
rotational kinetic energy K R  I  2
2
For an extended system:
I  lim
mi 0
 r m   r dm    r dV
2
i
2
2
i
i
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The Rigid Body under a Net Torque
The net torque acting on the rigid body is proportional
to its angular acceleration.
 t  I
(转动定律)
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The Rigid Body in Equilibrium
The torque vector
（力矩）
  r F
Two conditions for complete equilibrium of an object:
Ft  0;  t  0
translational equilibrium
rotational equilibrium
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Work and Energy in Rotational Motion
Work done by a torque
dW   d  Id
dW
The power of a torque
 
dt
f
f
i
i
W    d  
1 2 1 2
I  d   I  f  I i .
2
2
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Angular Momentum
The angular momentum of the particle relative to the
origin is defined as:
Lrp
dL
dp
r

dt
dt
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Conservation of Angular Momentum
（角动量守恒）
The total angular momentum of a system remains constant
if the net external torque acting on the system is zero.
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Eg. 01 In the classical model of the hydrogen atom, the electron
(mass m) is held in a circular orbit about the nucleus (a proton) by
the electric force on it. Show the angular momentum of the
electron with respect to the necleus is conserved.
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一个刚体的运动，可以视为二种运动所组成，即质量
中心，受到所有外力作用所引起的运动，加上物体在外力作用
下，绕质心的转动。一个刚体的动能，系质心的平移运动的动
能，加上绕质心运动的转动动能。
引自《古典动力学》，吴大猷
刚体的运动总可以分解为质心的平动刚体的转动的合成。
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A Brief Summary of Part I
① Kinematics of a Particle
② Newton’s Laws of Motion
③ Work and Energy
④ Momentum and Impulse
⑤ Motion of a System of Particles
⑥ Rotations of a Rigid Body about a Fixed Axis
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The End
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Problems
P255
5, 21, 40, 47, 54, 70, 74
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