Download Announcement I Physics 1408-001 Principles of Physics Chapter 9

Physics 1408-001
Principles of Physics
Lecture notes are on the web
http://highenergy.phys.ttu.edu/~slee/lectures/
Lecture 17
(Serway & Jewett Vol.1) – Ch. 9
February 21, 2007
HW Assignment #5 has been placed on
WebAssign, and is due by 11:30pm on
Tuesday, 2/27
Sung-Won Lee
[email protected]
The solutions for Exam 1 will be posted on the wall;
Next to my office (Sci 117)
The grades for Exam 1 is in progress!!
Chapter 9
Linear Momentum
and
Collisions Energy
1.
2.
3.
4.
5.
6.
7.
Announcement I
Linear Momentum and its Conservation
Impulse and Momentum
Collisions in 1-dimension
2-dimensional Collisions
The Center of Mass
Motion of a System of Particles
Rocket Propulsion
9.5 The Center of Mass
• There is a special point in a system or object, called the
center of mass, that moves as if all mass of the system is
concentrated at that point
• The coordinates of the center of mass are
– where M is the total mass of the system
• The center of mass can be located by its position vector, rCM
• ri is the position of the i th particle, defined by
Center of Mass, Example
• Both masses are on the x-axis
• Center of mass is on the x-axis
• Center of mass is closer to
the particle with the larger mass
Center of Mass, Extended Object,
Position
• The position of the center of mass can also be found by:
• The center of mass of any symmetrical object lies on an
axis of symmetry and on any plane of symmetry
Center of Mass, Example
• An extended object can be considered
a distribution of small mass elements,
∆m
• The center of mass is located
at position rCM
1
1
xcm =
∑ xi ∆m = M ∫ x dm
M i
1
1
ycm =
∑ yi ∆m = M ∫ y dm
M i
As Fig shows, we can divide an extended
object into many small cells or boxes,
each with the very small mass ∆m
The Center of Mass of a Rod
Find the center of mass of a
thin uniform rod of length L
and mass M. Find the
tangential acceleration of one
tip of a 1.6 m rod that rotates
about its center of mass with
an angular speed of 6.0
rad/s2.
dm dx
M
=
so dm =
dx
M
L
L
1
1
M
xcm =
x dm =
x dx
∫
∫
M
M
L
L
L
1
1
= ∫ xdx = ⎡⎢⎣ 12 x 2 = 12 L
0
L0
L
at = rα = 12 Lα
= 12 (1.6 m)(6.0 rad/s 2 )
= 4.8 m/s 2
9.6 Motion of a System of Particles
• Assume the total mass, M, of the system remains constant.
• We can describe the motion of the system in terms of the v
and a of the center of mass of the system.
• We can also describe the momentum of the system and
Newton’s 2nd law for the system.
• The velocity of the center of mass of a system of particles is
• The momentum can be expressed as
Acceleration of the Center of Mass
• The acceleration of the center of mass can be found by
differentiating the velocity with respect to time
Forces in a System of Particles
• The acceleration can be related to a force
• If we sum over all the internal forces, they cancel in pairs
and the net force on the system is caused only by
the external forces
Momentum of System of Particles
• The total linear momentum of the system equals the total
mass multiplied by the velocity of the center of mass
Motion of the Center of Mass,
Example
• A projectile is fired into the air
and suddenly explodes
• The total linear momentum of a system of particles is
conserved if no net external force is acting on the system
• MvCM = ptot = constant when ΣFext = 0
9.7 Rocket Propulsion
• The operation of a rocket depends upon the law of conserva
tion of linear momentum as applied to a system of particles,
where the system is the rocket plus its ejected fuel
• With no explosion,
the projectile would follow
the dotted line
• The initial mass of the rocket plus
all its fuel is M + ∆m at time ti and
velocity v
• The initial momentum of the system
is pi = (M + ∆m) v
• After the explosion,
the center of mass of the
fragments still follows
the dotted line,
the same parabolic path the
projectile would have followed
with no explosion
• At some time t + ∆t,
the rocket’s mass has been
reduced to M and an amount
of fuel, m has been ejected
• The rocket’s speed has
increased by ∆v
Rocket Propulsion, 2
• Because the gases are given some momentum when they
are ejected out of the engine, the rocket receives a compensating momentum in the opposite direction
• Therefore, the rocket is accelerated as a result of the “push”
from the exhaust gases
•
The basic equation for rocket propulsion is
•
The increase in rocket speed is proportional to the speed of the escape
gases (ve)
– So, the exhaust speed should be very high
The increase in rocket speed is also proportional to the natural log of th
e ratio Mi/Mf
– So, the ratio should be as high as possible, meaning the mass of t
he rocket should be as small as possible and it should carry as mu
ch fuel as possible
•
Chapter 10
Rotation of Rigid
Object about
a Fixed Axis
1.
2.
3.
4.
5.
6.
7.
8.
9.
Angular Position, Velocity, and Acceleration
Rotational Kinematics
Angular and Linear Quantities
Rotational Kinetic Energy
Calculation of Moments of Inertia
Torque
Relationship between Toque and Angular Acceleration
Work, Power, and Energy in Rotational Motion
Rolling Motion of a Rigid Object
Thrust
• The thrust on the rocket is the force exerted on it by the
ejected exhaust gases
Thrust =
• The thrust increases as the exhaust speed increases
• The thrust increases as the rate of change of mass increa
ses
– The rate of change of the mass is called the burn rate
Rigid Body Rotation
A rigid body is an extended
object whose size, shape, and
distribution of mass do not
change as the object moves and
rotates.
10.1 Angular Velocity &
Angular Acceleration
Rotation + Translation
Fig illustrates the three basic types of motion of a rigid body
Fig. shows a wheel rotating on an axle. Notice that 2 points on
the wheel turn through the same angle as the wheel rotates.
That’s ∆θ1 = ∆θ2 during same ∆t. So, the 2 points have equal
angular velocities.
Tangential acceleration :
at =
dvt d (rω )
dω
=
=r
dt
dt
dt
Let’s define
the angular
acceleration:
Table shows the resulting kinematic equations for constant
angular acceleration. These equations apply to a particle in
circular motion or to any rigid-body rotation !!
d ω at
=
dt
r
The derivative in
above eq. is the rate
of change of velocity
We see that the tangential
and angular accelerations
are related by
10.2 Rotational Kinematic Equations
α≡
at = rα
10.3 Angular/Linear Quantities
10.4 Rotational Kinetic Energy
• An object rotating about some axis with an angular speed
, ω, has rotational kinetic energy even though it may not
have any translational kinetic energy
• Each particle has a kinetic energy of Ki = ½ mivi2
• Since the tangential velocity depends on the distance, r,
from the axis of rotation, we can substitute vi = ωI r
• The total rotational kinetic energy of the rigid object is
the sum of the energies of all its particles
10.5 Moment of Inertia
• The definition of moment of inertia is
• The dimensions of moment of inertia: [kg.m2]
• We can calculate the I of an object more easily by assuming
it is divided into many small volume elements, each of mass
∆mi
• We can rewrite the expression for I in terms of ∆m
• With the small volume segment assumption,
• If ρ is constant, the integral can be evaluated with known
geometry, otherwise its variation with position must be
known
where I is called the moment of inertia
Notes on Various Densities
• Volumetric Mass Density: mass per unit volume:
ρ = m/V
• Face Mass Density: mass per unit thickness of a sheet
of uniform thickness, t:
σ = ρt
• Linear Mass Density: mass per unit length of a rod of
uniform cross-sectional area:
λ = M/L = ρΑ
Moment of Inertia of
a Uniform Thin Hoop
• Since this is a thin hoop,
all mass elements are the same
distance from the center
Moment of Inertia of
a Uniform Rigid Rod
• The shaded area has a mass
– dm = λdx = M/L dx
• Then the moment of inertia is
Moments of Inertia of
Various Rigid Objects
Moment of Inertia of
a Uniform Solid Cylinder
• Divide the cylinder into concentric
shells with radius r, thickness dr
and length L
• Then for I
The Parallel-Axis Theorem
The I depends on the rotation axis. Suppose you need to know
the I for rotation about the off-center axis in Fig.
I = I cm + Md 2
The parallel-axis theorem
If axis of interest is distance d from
a parallel axis through the center
of mass, the I is
Moment of Inertia for a Rod
Rotating Around One End
• The moment of inertia of the ro
d about its center is
• D is 1/2 L
• Therefore,
End of Lecture 17
‰HW Assignment #4 will be due on 2/20.
‰Before the next lecture, read the text book,
chapter Chapter 10.
‰Next lecture: Fridayday, 2/23