Announcement
Course webpage
http://highenergy.phys.ttu.edu/~slee/1408/
Homework.6 – HW due 3/26 (next Tuesday)
PHYS-1408-001
Ch.9 -- 2, 14, 15, 18, 20, 27, 30, 34, 37, 40, 41
Lab
Lecture 16
Monday 11:00 am – 12: 50 pm (Sci. 105)
Recitation
Monday 3:00 pm – 3:50 pm (Sci. 105)
Mar. 21, 2013
Reminder: Center of Mass, Coordinates
The coordinates of the center of mass
are
Chapter 9
Linear Momentum
& Collisions
∑m x
i
xCM =
i
M
i
i
M
∑m z
i
zCM =
∑m y
i
y CM =
i
i
i
M
!  M is the total mass of the system.
!  Use the active figure to observe
effect of different masses and
positions.
Section 9.6
Center of Mass, Example
•  An extended object can be considered
a distribution of small mass elements,
Δm
•  The center of mass is located
at position rCM
1
xcm =
M
1
ycm =
M
1
∑i xi Δm = M ∫ x dm
1
∑i yi Δm = M ∫ y dm
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.
at = rα = 12 Lα
As Fig shows, we can divide an extended
object into many small cells or boxes, each
with the very small mass Δm
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
= 12 (1.6 m)(6.0 rad/s 2 )
= 4.8 m/s 2
9.7 Motion of Many Particles
Velocity and Momentum of a System of Particles
We can describe the motion of the system (i.e. many particles) in terms of the
velocity and acceleration of the center of mass of the system.
The velocity of the center of mass of a system of particles is
We can also describe the momentum of the system and Newton s Second
Law for the system.



dr
1
vCM = CM = ∑ mi v i
dt
M i
The momentum can be expressed as




M vCM = ∑ mi v i = ∑ pi = p tot
i
i
The total linear momentum of the system equals the total mass multiplied by the
velocity of the center of mass.
Section 9.7
Section 9.7
Acceleration and Force in a System of Particles
Impulse and Momentum of a System of Particles
The acceleration of the center of mass can be found by differentiating the velocity
with respect to time.
The impulse by external forces is



dvCM
1
aCM =
= ∑ mi a i
dt
M i



d rCM
1
vCM =
= ∑ mi v i
dt
M i
The acceleration can be related to a force.


M aCM = ∑ Fi
i

∫ ∑F
ext



dt = M ∫ dvCM → Δptot = I
The total linear momentum of a system of particles is conserved if no net external
force is acting on the system.


M vCM = ptot = constant

when ∑ Fext = 0
Newton s Second Law for a System of Particles
Section 9.7
Section 9.7
9.7 Rocket Propulsion
Motion of the Center of Mass, Example
A projectile is fired into the air and
suddenly explodes.
With no explosion, the projectile would
follow the dotted line.
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.
Section 9.7
•  The operation of a rocket depends upon the law of conservation
of linear momentum as applied to a system of particles,
where the system is the rocket plus its ejected fuel
•  The initial mass of the rocket plus all
its fuel is M + Δm at time ti and
velocity v
•  The initial momentum of the system:
pi = (M + Δm) v "
•  At some time t + Δt, the rocket’s
mass has been reduced to M and
an amount of fuel, Δm has been
ejected
•  The rocket’s speed has increased
by Δv
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!
Thrust (=force)!
•  The thrust on the rocket is the force exerted on it by
the ejected gases!
! !Thrust: !
•  The basic equation for rocket propulsion:
•  The thrust increases as the exhaust speed increases!
•  The increase in rocket speed is proportional to the speed of the
escape gases (ve)
•  The increase in rocket speed is also proportional to the natural
log of the ratio Mi/Mf
–  So, the ratio should be as high as possible,
meaning the mass of the rocket should be as small as
possible and it should carry as much fuel as possible!
•  The thrust increases as the rate of change of mass
increases!
–  The rate of change of the mass is called
the burn rate!
Rigid Body Rotation
Chapter 10
Rotation of a 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
A rigid body (or rigid object) is
an extended object whose
size, shape, and distribution
of mass do not change as the
object moves and rotates.
Rotation + Translation
Fig illustrates the three basic types of motion of a rigid body!
Angular Position
Axis of rotation is the center of the disc
Choose a fixed reference line.
Point P is at a fixed distance r from the
origin.
Polar coordinates are convenient to use
to represent the position of P
P is located at (r, θ) where r is the
distance from the origin to P and θ is
the measured counterclockwise from
the reference line.
Section 10.1
Angular Position, cont.
Angular Displacement
The arc length and r are related:
The angular displacement is defined as
the angle the object rotates through
during some time interval.
!  s = θ r
This can also be expressed as:
θ=
Δθ = θ f − θ i
s
r
Θ = [radian]
Comparing degrees and radians
1 rad =
360°
= 57.3°
2π
Converting from degrees to radians
θ (rad ) =
π
θ (degrees )
180°
Average Angular Speed
Angular Speed
The average angular speed, ωavg, of
a rotating rigid object is the ratio of
the angular displacement to the time
interval.
The instantaneous angular speed is
defined as the limit of the average speed
as the time interval approaches zero.
ωavg
θ − θ i Δθ
= f
=
tf − t i
Δt
ω≡
lim
Δt →0
Δθ dθ
=
Δt
dt
ω will be positive if θ is increasing
(counterclockwise)
ω will be negative if θ is decreasing (clockwise)
Section 10.1
Section 10.1
Angular Acceleration
The average angular acceleration, α avg
The instantaneous angular acceleration
α avg
ω − ωi Δω
= f
=
tf − t i
Δt
α≡
lim
Δt →0
Δω dω
=
Δt
dt
α  will be positive if an object rotating counterclockwise is speeding up.
α will also be positive if an object rotating clockwise is slowing down.
Angular Motion, General Notes
θ, ω, α all characterize the motion of the entire rigid object as well as the
individual particles in the object.
θ=
s
r
ω≡
lim
Δt →0
Δθ dθ
=
Δt
dt
α≡
lim
Δt →0
Δω dω
=
Δt
dt
Directions
The directions are given by the right-hand
rule.
Section 10.1
Section 10.1
Rotational Kinematics
Comparison Between Rotational and Linear Equations
Under constant angular acceleration, we can describe the motion of the rigid
object using a set of kinematic equations.
!  These are similar to the kinematic equations for linear motion.
!  The rotational equations have the same mathematical form as the linear
equations.
ωf = ωi + α t
Rotational Kinematic Equations
Substitutions from translational to
rotational are
1
θ f = θ i + ωi t + α t 2
2
ωf2 = ωi2 + 2α (θf − θ i )
1
(ωi + ωf )t
2
all with consant α
!  x → θ
θf = θ i +
!  v → ω
!  a → α
Section 10.2
Section 10.2
Relationship Between Angular and Linear Quantities
Speed Comparison – Details
Displacements
The linear velocity is always tangent to
the circular path.
!  s = θ r
!  Called the tangential velocity
Speeds
The magnitude is defined by the
tangential speed.
!  v = ω r
Accelerations
v=
!  a = α r
ds
dθ
=r
= rω
dt
dt
Since r is not the same for all points on the
object, the tangential speed of every point is
not the same.
The tangential speed increases as one
moves outward from the center of rotation.
Section 10.3
Acceleration Comparison – Details
The tangential acceleration is the
derivative of the tangential velocity.
at =
dv
dω
=r
= rα
dt
dt
Section 10.3
Speed and Acceleration Note
All points on the rigid object will have the same angular speed, but not the same
tangential speed.
All points on the rigid object will have the same angular acceleration, but not the
same tangential acceleration.
The tangential quantities depend on r, and r is not the same for all points on the
object.
Section 10.3
Section 10.3
Centripetal Acceleration?
Resultant Acceleration
An object traveling in a circle, even though it moves with a constant speed, will
have an acceleration.
The tangential component of the acceleration is due to changing speed.
!  Therefore, each point on a rotating rigid object will experience a centripetal
acceleration.
aC =
v2
= rω2
r
The centripetal component of the acceleration is due to changing direction.
Total acceleration can be found from these components:
a = at2 + ar2 = r 2α 2 + r 2ω 4 = r α 2 + ω 4
at =
dv
dω
=r
= rα
dt
dt
aC =
Section 10.3
v2
= rω2
r
Section 10.3
10.4 Rotational Kinetic Energy
Rotational Motion Example
•  An object rotating about some axis with an angular speed, ω, has rotat
ional KE even though it may not have any translational KE.
For a compact disc player to read a CD,
the angular speed must vary to keep
the tangential speed constant (vt = ωr).
•  Each particle has a KE of Ki = ½ mivi2
At the inner sections, the angular speed
is faster than at the outer sections.
Section 10.3
•  Since the tangential velocity depends on the distance, r, from the axis
of rotation, we can substitute vi = ωI r
•  The total rotational KE of the rigid object is the sum of the energies
of all its particles
where I is called the moment of inertia