P3 Dynamics
Mark Cannon
Hilary Term 2012
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Lecture 1
Introduction to Dynamics
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Introduction
Dynamics
(dῑ-năm0 ιks)
from Greek:
δυναµικóς – powerful
δ ύναµις – power, strength
Dynamics concerns the calculation of forces and motion for analysis & design
? For stationary objects, use statics & elasticity,
e.g. bridges, buildings . . .
? For problems involving motion, use the laws of dynamics,
e.g. machines, vehicles, robots . . .
Dynamics is a component of Mechanics, which involves:
Kinematics
Dynamics
−→
−→
motion
forces and moments
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Introduction
Dynamics is essentially about Newton’s 2nd law
even gyroscopic forces can be explained using Newton’s 2nd law . . .
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Introduction
. . . . . . despite how counter-intuitive they may appear to be
Henry W. Wallace’s so-called anti-gravity “kinemassic field” generator
from U.S. Patent 3,626,605 (1971)
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Lectures & turorials
8 lectures covering:
Force and momentum as applied to particles
Work, power and energy
Circular motion
Gravity and satellite orbits
Rigid body dynamics
Two tutorial sheets:
1P3H — dynamics of particles
1P3J — dynamics of rigid bodies
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Reading
Favourite books:
Meriam and Kraige Engineering Mechanics Volume 2 Dynamics 5th
edition, SI version, Wiley, 2003.
Meriam Dynamics 2nd edition, SI version, Wiley, 1975.
Other possibilities:
Hibbeler Engineering Mechanics - Dynamics SI edition, Prentice Hall,
1997.
Bedford and Fowler Engineering Mechanics - Dynamics SI edition,
Addison-Wesley, 1996.
..
.
Lecture notes & slides:
For all handouts (lecture notes & these slides), and the tutorial sheets, go
to
http://www.eng.ox.ac.uk/∼conmrc/dcs
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Things you should know by the end of the course
How to:
use the definitions of velocity and acceleration to analyse straight-line and
curvilinear motion of particles
use Newton’s second law to analyse the motion of particles under the
action of a steady or impulsive force
apply the principles of conservation of momentum and conservation of
energy to the motion of a particle
describe planar motion of a particle in rectangular, normal-tangential, and
polar coordinates
calculate the moment of inertia of a planar rigid body from first principles
or from standard cases
apply the principles of conservation of angular momentum and
conservation of energy to the motion of a planar rigid body
analyse the translation and rotation of a planar rigid body under the
action of a steady or impulsive force or moment.
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Force and Motion
A particle is a discrete mass concentrated at a point
treat an object as a particle when considering its translation
e.g. when analysing
? the trajectory of a golf ball
? or the motion of a spacecraft orbiting the earth
? or straight-line motion with variable acceleration
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Force and Motion
A rigid body is a system of particles rigidly connected to each other
treat any object with distributed mass as a rigid body when considering
its rotation
e.g. when analysing
? the motion of gears
? or forces and accelerations of pistons and crankshaft in a car engine
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Force and Motion
A particle moving in a straight line:
0
P
P
x
Dx
? instantaneous speed and acceleration:
∆x
dx
=
= ẋ
∆t→0 ∆t
dt
V = lim
? integrate w.r.t. t: Z
t
a(t) dt
V (t) = V0 +
dV
= ẍ
dt
a=
Z
x(t) = x0 +
0
t
V (t) dt
0
? If a = constant:
V (t) = V0 + at
1
x(t) = x0 + V0 t + at 2
2
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Force and Motion
A particle moving in a straight line:
0
? Or use a =
P
x
Dx
d ẋ
dx d ẋ
d ẋ
=
= ẋ
to get
dt
dt dx
dx
a=V
? integrate w.r.t. x using
dV
d 1 2
V
=
V =a
dx
dx 2
? If a = constant:
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P
1 2
2V
dV
dx
=⇒
1 2
2V
−
− 21 V02 = a(x − x0 )
1 2
2 V0
Z
x
=
a(x) dx
x0
Force and Motion
Newton’s second law
for a particle:
force = rate of change of momentum
where
momentum = mass × velocity
Equivalent vector equation:
F=
dV
dm
d
mV = m
+V
dt
dt
dt
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Force and Motion
If mass is constant (dm/dt = 0), then
F=m
dV
dt
or
F = ma
i.e.
force = mass × acceleration
If force and motion are in only one direction, then use the scalar form:
F =
d
dV
dm
mV = m
+V
dt
dt
dt
and
F =m
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dV
dt
or
F = ma
Force and Motion
Some comments on weight . . .
A dropped object accelerates downwards at g = 9.81 m s−2
(at earth’s surface)
The force of gravity causing this acceleration is the weight of the object:
weight = mass × gravitational acceleration
The SI unit of force is the Newton: 1 N acting on 1 kg produces 1 m s−2
acceleration, so
weight (Newtons) = mass (kg) × 9.81 (m s−2 )
. . . and on speed
Velocity refers to the vector V
Speed is a scalar quantity V , equal to the magnitude of V
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Force and Motion
Example – Motion in a straight line
A boat of mass 1500 kg is launched from a trolley on a sloping ramp.
The trolley is allowed to run down the ramp at 1 m s−1 until the
boat is just afloat. The trolley then stops and the boat continues to
move at 1 m s−1 .
Once afloat, a crew member of mass 70 kg stops the boat by pulling
steadily on a rope with a force equal to 30 % of his own weight.
How long will the boat take to stop, and what length of rope must
be allowed to slip?
1 ms
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force
Force and Motion
Solution: use the scalar form of Newton’s 2nd law for constant mass:
F = ma
−70 × 9.81 × 0.3 = 1500 × a
gives
so the acceleration is
a=
dv
= −0.137 m s−2
dt
? To find the distance: use
? To find the time to stop the boat:
separate variables
dt =
a=
dv
a
separate variables
and integrate
t=
dv ds
dv
dv
=
=v
dt
ds dt
ds
v dv = a ds
v2 − v1
0−1
=
= 7.28 s
a
−0.137
and integrate
1 2
(v2 − v12 ) = a(s2 − s1 )
2
1 02 − 12
= 3.64 m
s = s2 − s1 =
2 −0.137
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Force and Motion
Example – terminal velocity in free fall
A free-fall parachutist has mass m = 75kg and frontal area A = 0.8m2 .
The air density at 2000 m is ρ = 1.007 kg m−3 (HLT p.68) and the
aerodynamic drag is
D=
1
CD ρV 2 A
2
drag coefficient CD = 1.2 .
Find the terminal speed Vt . How far will the parachutist fall before
reaching 90 % of this terminal speed?
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Force and Motion
Example – terminal velocity in free fall
D
Solution:
mg
? Forces: weight mg downwards and aerodynamic drag D upwards.
? Hence Newton’s second law: F = ma gives
mg −
1
CD ρV 2 A = ma
2
? Terminal velocity Vt is reached when a = 0, so
Vt2 =
2mg
2 × 75 × 9.81
=
ρACD
1.007 × 0.8 × 1.2
=⇒
Vt = 39.01 m s−1
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Force and Motion
Example – terminal velocity in free fall
D
Solution contd:
mg
? To find the variation of V with s, use a = V dV /ds:
dV
mV
= mg − 12 CD ρV 2 A
ds
? Simplify by rearranging and writing this in terms ofVt2 :
dV
g
V
= 2 Vt2 − V 2
ds
Vt
Separate variables and integrate between limits V = 0 and V = 0.9Vt :
2
Z 0.9Vt
Vt − (0.9Vt )2
Vt2
V
Vt2
dV = −
ln
s=
g 0
2g
Vt2 − V 2
Vt2 − 0
1 − 0.81
= −77.58 ln
= 128.8 m
1
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Force and Motion
Highest parachute jump
Joseph Kittinger, Aug 16, 1960
Height:
Terminal velocity:
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Impulse and Momentum
What if F is time-varying and the magnitude F (t) is unknown?
This is often true in problems involving collisions
Use the concept of impulse – the integral of F(t) over time
The impulse acting on a body is related to its momentum.
Given that
dV
F=m
dt
for a body of mass m, then
2
Z
F dt = m (V2 − V1 )
1
In words:
impulse = change in momentum
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31.3 km
274 m s−1
(988 km h−1 )
Impulse and Momentum
The impulse-momentum equation requires no details of the time-variation
of force
An impulse can describe an impact involving large forces over a short time
force
impulse = area under force-time graph
time
0
The area under graph of F(t) gives the magnitude of the impulse
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Impulse and Momentum
Example – Impulse with unknown force variation
A batsman is struck by a cricket ball of mass 0.15 kg travelling at
40 m s−1 . The ball is stopped by the impact. Can you estimate the
force exerted by the ball?
Solution:
? the force variation and duration of the impact is unknown
? but the impulse is m∆V = 0.15 × 40 = 6.0 kg m s−1
so if the impact lasts t seconds, then the average force is
Fav =
m∆V
6.0
=
N
t
t
e.g. t = 0.05 s gives an average force of 120 N
? shorter duration =⇒ greater force =⇒ more painful impact!
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Impulse and Momentum
Example – Impulse with constant force
Use impulse & momentum to find the time needed to stop the boat
in the first example.
Solution:
The force F is constant, so
Z 2
Z
F dt = F
1
2
dt = F (t2 − t1 ) = m (V2 − V1 )
1
giving
t = t2 − t1 = m
V2 − V1
0−1
= 1500 ×
= 7.28 s
F
−70 × 9.81 × 0.3
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Impulse and Momentum
Example – Impulse and momentum as vectors
A cannon of mass M is free to roll without friction on horizontal
ground. An explosive charge projects a ball of mass m at speed v
relative to the barrel, which is inclined upward at angle θ.
v
q
u
At the instant after the ball leaves the muzzle find:
(a) the backward recoil speed u of the gun
(b) the absolute velocity components of the ball
(c) the magnitude and direction of any external impulse acting on the system.
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Impulse and Momentum
Solution:
The components of the ball’s absolute velocity
are:
vx = v cos θ − u
horizontally
vy = v sin θ
vertically
u
v
v sinq
q
v cos q - u
(a) Consider cannon plus ball together as a single system
there’s no external horizontal impulse on the system, so the horizontal
momentum before and afterwards is zero
therefore
0 = m (v cos θ − u) − Mu
giving
u=
mv cos θ
M +m
[Note: the impulse between the cannon and ball is internal to the system]
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Impulse and Momentum
Solution contd:
(b) Substituting for u in vx = v cos θ − u gives
vx = v cos θ −
mv cos θ
M
= v cos θ
M +m
M +m
and the other component of absolute velocity is just vy = v sin θ
(c) The only external impulse is vertical: Qy (from the ground)
Qy is equal to the change of upward momentum of the entire system:
Qy = mv sin θ
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Impulse and Momentum
Example – Rowing on a sliding seat
Between strokes, the crew of a boat slide a distance d towards the
stern of the boat. If the crew has mass m and the boat has mass M,
what happens to the boat?
Solution: Assume the crew slides distance d at constant speed for time ∆t
there’s no external impulse so the momentum of boat plus crew is unchanged
hence the velocity v of the centre of mass G is unchanged
x
d
G
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Impulse and Momentum
Solution contd:
x
Relative to G:
? the boat moves forward ∆x at speed
∆x/∆t
Dx
? the crew move backward d − ∆x at speed
(d − ∆x)/∆t
t
Considering momentum before and after gives
(M + m)v + M
∆x
d − ∆x
−m
= (M + m)v
∆t
∆t
so the boat surges forward between strokes by an amount ∆x =
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md
M +m
Summary
Analyse translational motion of an object
by representing it as a particle concentrated at the centre of mass
Newton’s second law
? general case:
force = rate of change of momentum
dV
dm
d
mV = m
+V
dt
dt
dt
⇐⇒
F=
⇐⇒
F=m
? for objects with constant mass:
force = mass × acceleration
dV
dt
Impulse and momentum
2
Z
impulse = change in momentum
⇐⇒
F dt = m (V2 − V1 )
1
Accelerations are measured in an inertial frame of reference,
i.e. non-accelerating, non-rotating
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