Newton’s Second Law - Purdue University
Transcript of Newton’s Second Law - Purdue University
Introduction to Dynamics (N. Zabaras)
Newton’s Second Law
Prof. Nicholas Zabaras
Warwick Centre for Predictive Modelling
University of Warwick
Coventry CV4 7AL
United Kingdom
Email: [email protected]
URL: http://www.zabaras.com/
April 10, 2016
Introduction to Dynamics (N. Zabaras)
Equations of Motion
• Newton’s second law provides
amF
• Solution for particle motion is facilitated by
resolving vector equation into scalar component
equations, e.g., for rectangular components,
zmFymFxmF
maFmaFmaF
kajaiamkFjFiF
zyx
zzyyxx
zyxzyx
• For tangential and normal components,
2vmF
dt
dvmF
maFmaF
nt
nntt
2
Introduction to Dynamics (N. Zabaras)
Linear Momentum of a Particle
• Replacing the acceleration by the derivative of
the velocity yields
particle theof momentumlinear
L
dt
Ldvm
dt
d
dt
vdmF
• Linear Momentum Conservation Principle:
If the resultant force on a particle is zero, the
linear momentum of the particle remains
constant in both magnitude and direction.
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Introduction to Dynamics (N. Zabaras)
Sample Problem
The two blocks shown start from
rest. The horizontal plane and the
pulley are frictionless, and the
pulley is assumed to be of
negligible mass.
Determine the acceleration of each
block and the tension in the cord.
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Introduction to Dynamics (N. Zabaras)
Sample Problem
• Write equations of motion for blocks and pulley.
:AAx amF
AaT kg1001
:BBy amF
B
B
BBB
aT
aT
amTgm
kg300-N2940
kg300sm81.9kg300
2
22
2
:0 CCy amF
02 12 TT
SOLUTION:
• Write the kinematic relationships for the dependent
motions and accelerations of the blocks.
ABAB aaxy21
21
x
y
O
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Introduction to Dynamics (N. Zabaras)
Sample Problem
N16802
N840kg100
sm20.4
sm40.8
12
1
221
2
TT
aT
aa
a
A
AB
A
• Combine kinematic relationships with equations of
motion to solve for accelerations and cord tension.
ABAB aaxy21
21
AaT kg1001
A
B
a
aT
21
2
kg300-N2940
kg300-N2940
0kg1002kg150N2940
02 12
AA aa
TT
x
y
O
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Introduction to Dynamics (N. Zabaras)
Angular Momentum of a Particle• moment of momentum or the
angular momentum of the particle about O. VmrHO
• Derivative of angular momentum with respect to time,
O
O
M
Fr
amrVmVVmrVmrH
• It follows from Newton’s second law that the sum
of the moments about O of the forces acting on
the particle is equal to the rate of change of the
angular momentum of the particle about O.
zyx
O
mvmvmv
zyx
kji
H
• is perpendicular to plane containingOH
Vmr
and
2
sin
mr
vrm
rmVHO
7
rr r v e e(in polar coordinates)
Introduction to Dynamics (N. Zabaras)
Eqs of Motion in Radial & Transverse Components
rrmmaF
rrmmaF rr
2
2
• Consider particle at r and , in polar coordinates,
rrmF
rrrm
mrdt
dFr
mrHO
2
22
2
2
• The 2nd Eq. above may also be derived from
conservation of angular momentum,
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Rate of change
of angular momentum
Moments of external
forces around O
Introduction to Dynamics (N. Zabaras)
Conservation of Angular Momentum
• When only force acting on particle is directed
toward or away from a fixed point O, the particle
is said to be moving under a central force.
• Since the line of action of the central force
passes through O, and 0 OO HM
constant OHVmr
• Position vector and motion of particle are in
a plane perpendicular to .OH
• Magnitude of angular momentum,
000 sin
constantsin
Vmr
VrmHO
2
2
constant
angular momentum
unit mass
O
O
H mrv mr
Hr h
m
or
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Central force
rr r
v r
v e e
Introduction to Dynamics (N. Zabaras)
Conservation of Angular Momentum
• Radius vector OP sweeps infinitesimal area
drdA 221
• Define 2
212
21 r
dt
dr
dt
dAareal velocity
• Recall, for a body moving under a central force,
constant2 rh
• When a particle moves under a central force,
its areal velocity is constant.
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Introduction to Dynamics (N. Zabaras)
Newton’s Law of Gravitation• Gravitational force exerted by the sun on a planet
or by the earth on a satellite is an important
example of gravitational force.
• Newton’s law of universal gravitation - two particles
of mass M and m attract each other with equal and
opposite force directed along the line connecting
the particles,
2
3 412 9
2 4
m ft66.73 10 34.4 10
kg s lb s
MmF G
r
G
constant of gravitation
• For particle of mass m on the earth’s surface,
222 s
ft2.32
s
m81.9 gmg
R
MGmW
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Introduction to Dynamics (N. Zabaras)
Sample Problem
A satellite is launched in a
direction parallel to the surface of
the earth with a velocity of 18820
mi/h from an altitude of 240 mi.
Determine the velocity of the
satellite as it reaches it maximum
altitude of 2340 mi. The radius of
the earth is 3960 mi.
SOLUTION:
• Since the satellite is moving under a
central force, its angular momentum is
constant. Equate the angular
momentum at A and B and solve for
the velocity at B.
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Introduction to Dynamics (N. Zabaras)
Sample Problem
SOLUTION:
• Since the satellite is moving under a
central force, its angular momentum is
constant. Equate the angular
momentum at A and B and solve for
the velocity at B.
sin constant
3960 240 mi18820mi h
3960 2340 mi
O
A A B B
AB A
B
rmv H
r mv r mv
rv v
r
hmi12550Bv
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Introduction to Dynamics (N. Zabaras)
Trajectory of a Particle Under a Central Force
• For particle moving under central force directed towards the force center,
022 FrrmFFrrm r
• Second expression is equivalent to from which,,constant 2 hr
rd
d
r
hr
r
h 1and
2
2
2
2
2
• After substituting into the radial equation of motion and simplifying,
ru
umh
Fu
d
ud 1where
222
2
• If F is a known function of r or u, then particle trajectory may
be found by integrating for u = f(), with constants of
integration determined from initial conditions.14
2
2 2 2 2
2 2 2 2 2
1
1 1 1
dr dr d h dr dh
dt d dt r d d r
d r d dr d d d d d h d hh h
dt d dt dt d d r dt d r r d r r
Use:
Introduction to Dynamics (N. Zabaras)
Trajectory of a Satellite
constant
1where
22
2
2
2222
2
h
GMu
d
ud
GMmur
GMmF
ru
umh
Fu
d
ud
• Consider earth satellites subjected to only gravitational
pull of the earth,
• Solution is equation of conic section,
tyeccentricicos11 2
2
GM
hC
h
GM
ru
• Origin, located at earth’s center, is a focus of the conic
section.
• Trajectory may be ellipse, parabola, or hyperbola
depending on value of eccentricity.
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Introduction to Dynamics (N. Zabaras)
Trajectory of a Satellite
tyeccentricicos11 2
2
GM
hC
h
GM
r
• Trajectory of earth satellite is defined by
• hyperbola, > 1 or C > GM/h2. The radius vector
becomes infinite for
2
1111 cos
1cos0cos1
hC
GM
• parabola, = 1 or C = GM/h2. The radius vector
becomes infinite for
1800cos1 22
• ellipse, < 1 or C < GM/h2. The radius vector is finite
for all . For =C=0, we obtain a circle.
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Introduction to Dynamics (N. Zabaras)
Trajectory of a Satellite• Integration constant C is determined by
conditions at beginning of free flight, =0, r =
r0 ,
2
2
0
0 022
0 0 0 0
11 cos 0
1 1,
GM Ch
r h GM
GM GMC where h r v
r h r r v
00
200
2
2
or 1
r
GMvv
vrGMhGMC
esc
• Satellite escapes earth orbit for
• Trajectory is elliptic for v0 < vesc and
becomes circular for = 0 or C = 0,
0r
GMvcirc
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Introduction to Dynamics (N. Zabaras)
Time to Complete an Orbit
• Recall that for a particle moving under a
central force, the areal velocity is constant,
i.e.,
constant212
21 hr
dt
dA
• The Periodic time or time required for a satellite
to complete an orbit is equal to area within the
orbit divided by areal velocity,
h
ab
h
ab
2
2
where
10
1021
rrb
rra
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Introduction to Dynamics (N. Zabaras) 19