physics 121 newtonian mechanics lecture notes are posted on instructor karine chesnel april 2, 2009
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Physics 121
Newtonian Mechanics
Lecture notes are posted on
www.physics.byu.edu/faculty/chesnel/physics121.aspx
InstructorKarine Chesnel
April 2, 2009
Review 3
Mid-term exam 3
• Friday April 3 through Tuesday April 7
• At the testing center : 8 am – 9 pm
• Closed Book and closed Notes
• Only bring: - Math reference sheet - Pen / pencil- Calculator- your CID
• No time limit (typically 2 – 3 hours)
Midterm exam 3
Review: ch 9 – ch 13
Ch. 10 Rotation of solid• Moment of inertia• Rotational kinematics• Rolling motion• Torque
Ch. 12 Static equilibrium and elasticity• Rigid object in equilibrium• Elastic properties of solid
Ch. 11 Angular momentum• Angular momentum• Newton’s law for rotation• Isolated system• Precession motion
Ch. 9 Linear Momentum & collision• Center of mass• Linear momentum• Impulse• Collisions 1D and 2D
Ch. 13 Universal gravitation• Newton’s law of Universal gravitation• Gravitational Field & potential energy • Kepler’s laws and motion of planets
Linear Momentum & Impulse
Review 3 4/2/09
• Newton’s second law Fdt
pd Vmp
• The linear momentum of a particle is the product of its mass by its velocity
Units: kg.m/s or N.s
cstp
• For an isolated system 0
dt
pd
• The impulse is the integral of the net force, during an abrupt interaction in a short time
f
idtFI
Modeling of an impulse
t1 t2t
F
avgF
t
tFdtFIf
i avg .
Ip
• According to Newton’s 2nd law:
Collisions
iiff pppp ,2,1,2,1
1. Conservation of linear momentum
2,22
2,11
2,22
2,11 2
1
2
1
2
1
2
1iiff VmVmVmVm
2. Conservation of kinetic energy
(2)
Elastic collision
iiff VmVmVmVm ,22,11,22,11
(1)
V1,i
V2,i
V1,
f
V2,f
Review 3 4/2/09
Collisions 1D
if Vmm
mmV ,1
21
21,1
if V
mm
mV ,1
21
1,2
2
• If one of the objects is initially at rest:
iif Vmm
mV
mm
mmV ,2
21
2,1
21
21,1
2
• Combining (1) and (2), we get expression for final speeds:
iif Vmm
mV
mm
mmV ,1
21
1,2
21
12,2
2
V1,i
V1,f V2,f
V2,i
Collisions 2D
• 3 equations
• 4 unknow parametersV1,i
x
y V1,f
V2,f
Inelastic collision: the kinetic energy K is NOT conserved
Review 3 4/2/09
Center of Mass
m1
m2
m3
m4
m5
m6
iirm
M OC
O
C
y
x
z
Ensemble of particles
r1
r6
Ch.9 Momentum and collision 03/05/09
dmr
M OC
Solid objectC
O
dm
rP
Solid characteristics
M OC dmr
• The center of mass is defined as:
C
O
dm
rdVdm
Ctot VMp
FaM C
• The moment of inertia of the solid about one axis:
dmrI 2
2' MRII
Review 3 4/2/09
Rotational kinematics
dt
d
dt
d
• Solid’s rotation
Angular position
Angular speed
Angular acceleration
RVt
• Linear/angular relationship
Velocity
Acceleration
• Tangential
• Centripetal
Rat 2RaC
For any point in the solid
• Rotational kinetic energy2
2
1 IK
Review 3 4/2/09
Motion of rolling solid
P
C
R
Non- sliding situation
• The kinetic energy of the solid is given by the sum of the translational and rotational components:
Ksolid = Kc + Krot
22
2
1
2
1 IVMK Csolid
222
2
1
2
1 IMRK solid
22 )(2
1 IMRK solid
cstKU sol If all the forces are conservative:
Review 3 4/2/09
Torque & angular momentum
Fr
The torque is defined as
F
The angular momentum is defined as
dt
Ld
Deriving Newton’s second law in rotation
angular momentum Linear momentum
prL
sinFr
When a force is inducing the rotation of a solid about a specific axis:
For an object in pure rotation
Inet IL
Review 3 4/2/09
Precession
sin.mgR
Top view
LdL d
IL
is the projection of the angular momentum in the horizontal
plane
hL
The angular momentummoves along a cone
h
pLdt
d
The precession speed is
mg
Side view
L
If an object spinning at very high speed is experiencing a torque
in a direction different than its angular momentum L, then it will precess about a second axis
Review 3 4/2/09
Solving a problem
Static equilibrium
• Define the system
• Locate the center of mass (where gravity is applied)
• Identify and list all the forces
0
F• Apply the equality
• Choose a convenient point to calculate the torque (you may choose the point at which most
of the forces are applied, so their torque is zero)
• List all the torques applied on the same point.
0
• Apply the equality
Review 3 4/2/09
Example 3Beam and cable tension
0
F
0
• We do not know the force R that the hinge applies to the beam.P is a convenient point to calculate the torque
00
RR
gMCPW
2/MgDW
TQPT
sinTDT
sin2
MgT 0 WTR
RP
• Find the tension on the cable
T
Mg C Q
Ch.10 Rotation of Solids 3/24/09
Example 3Beam and cable tension
0
F
• Find the magnitude and direction for the force R
exerted by the wall on the beam
• in the horizontal direction
0cos TRx
tan2cos
MgTRx
• in the vertical direction
0sin TMgRy
2sin
MgTMgRy
tantan x
y
R
R 2tan12
Mg
R
RP
T
Mg C Q
Ch.10 Rotation of Solids 3/24/09
Gravitational laws
rMg ur
mMGrgmF
2
)(
Any object placed in that field experiences a gravitational force
Any material object is producing a gravitational field
rM ur
MGrg
2
)( M
r
ur
m
Fg
The gravitational field created by a spherical object is centripetal
(field line is directed toward the center)
The gravitational potential energy is
r
mMGU g
Ug
0 r
Review 3 4/2/09
Kepler’s Laws
“The orbit of each planet in the solar system is an ellipse with the Sun as one focus ”
First Law
0LcstL
“The line joining a planet to the sun sweeps out equal areas during equal time intervals as the planet travels along its orbit.”
Second Law
cstm
LdtdA
20
“The square or the orbital period of any planet is proportional to the cube of the semimajor axis of the orbit”
Third Law
32
2 4R
GMT
S
Review 3 4/2/09
Kepler’s laws
First law
0
• Physical observations (Brahe and Kepler, early 17th ) showed that orbits are elliptical
• This phenomenon could be demonstrated later (late 17th )using the Newton’s laws of motion
The motion of a body orbiting around another body under the only influence gravitational force must be in a plane
r
V
L0
L0
L0
Solar system
0LcstL
Review 3 4/2/09
0LcstL
cstVmr
0
.L
dt
drrm
cstm
L
dt
dA
20
Kepler’s laws
Second law
cstdt
rdrm
V
r
2/.drrdA
The area swept by the radius during the time interval dt is
“The line joining a planet to the sun sweeps out equal areas during equal time intervals as the planet travels along its orbit.”
Review 3 4/2/09
Case of circular orbit
2
2
r
mMG
r
Vmma S
c
r
GM
T
r S
22
32
2 4R
GMT
S
Kepler’s laws
Third law
Fg
2r
mMGma S
r
Applying Newton’s law of motionWith gravitational force
T
rrV
2Also
r
GMV S2
“The square or the orbital period of any planet is proportional to the cube of the semi-major axis of the orbit”
sS
KGM
24
The proportionality constant is
Solar system Ks =2.97 10-19 s2/m3
Review 3 4/2/09