Download Lecture 3

P3 Dynamics
Mark Cannon
Hilary Term 2012
0-1
Lecture 3
Energy and Power
3-1
Review
Newton’s second law:
⇐⇒
force = rate of change of momentum
F=
dV
dm
d
mV = m
+V
dt
dt
dt
Impulse and momentum
2
Z
⇐⇒
impulse = change in momentum
F dt = m (V2 − V1 )
1
During a frictionless collision:
?
?
?
?
no impulse acts parallel to the impact surface
coefficient of restitution e determines the relative normal velocities
if no external impulse acts, then momentum is conserved
If e = 1 (perfectly elastic collision), then energy is conserved.
For continuous flow problems:
force = rate of flux of momentum
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Review
Incremental work done by a force F moving distance ds:
dW = F · ds
total work done in moving from position 1 to position 2:
Z 2
W =
F · ds .
1
Work done in moving an object of mass m between heights z1 and z2 :
Z 2
W =
mg dz = mg (z2 − z1 ) .
1
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Energy
Energy can be:
stored in and recovered from a
mechanical system (e.g. flywheel,
spring)
dissipated through mechanical
losses (e.g. friction, collisions)
conserved if there are no losses
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Kinetic Energy
The kinetic energy of a particle is defined as
T =
1
1
× mass × speed2 = mV 2
2
2
where V 2 = V · V
An alternative definition of kinetic energy is
the work required to accelerate from rest to speed V
hence the work required to change the kinetic energy from T1 to T2 is:
W = T2 − T1
work and kinetic energy are both measured in joules (J)
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Kinetic Energy
Proof that Work = Change in Kinetic Energy
m
Consider a variable force F acting on a particle
of mass m at a variable position r
r
F
0
? Newton’s 2nd law gives F = m r̈
? The work W done by F on m moving
from position
1 to position 2 is
Z 2
Z 2
r̈ · dr
F · dr = m
W =
1
but
and
1
2
Z
dr
dr =
dt = ṙ dt
dt
d|ṙ|2
ṙ · r̈ = 12
dt
=⇒
=⇒
r̈ · ṙ dt
W =m
W =
1
2
1Z
2
m
1
d|ṙ|2
dt
dt
so
W =
1
m (|ṙ2 |2 − |ṙ1 |2 ) = T2 − T1
2
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Kinetic Energy
Example – Launching a boat
1 ms
force
-1
Find the distance travelled before stopping from the work done on the boat
? Work done is equal to change in kinetic energy:
Fs =
1
m (V12 − V22 )
2
? but m = 1500 kg, V1 = 1 ms2 , V2 = 0, and F = 70 × 9.81 × 0.3 so
70 × 9.81 × 0.3 × s =
1
× 1500 × (12 − 02 )
2
giving
s = 3.64 m
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Potential Energy
The potential energy U is:
the energy stored in a system due to its position
e.g. in gravitational, electric or magnetic fields
or because of elasticity
Any increase in potential energy is equal to the work done on the system
In a conservative system this energy may be recovered as work
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Potential Energy
Example – hydroelectric energy storage
The pumped-storage plant at Raccoon Mountain, TN
? 80% efficiency (power recovered/power input)
? rapid start-up (e.g. 16 seconds to peak power output)
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Potential Energy
Examples
Potential energy stored in a mass m raised from height z1 to height z2 .
Work done:
2
Z
F dz = mg (z2 − z1 )
W =
1
hence
U2 − U1 = mg (z2 − z1 )
Potential (or strain) energy stored in a stretched spring with stiffness k
and unstretched length L1
Work done in stretching to length L2 :
W =
1
k (L2 − L1 )2
2
but U1 = 0 when unstretched, giving
U2 =
1
k (L2 − L1 )2
2
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Potential Energy
Example – Gravitational Potential Energy
At the surface of the earth (radius R) the acceleration due to gravity
is g . Elsewhere it is inversely proportional to the square of the radius
r from the centre of the earth, so that the weight of a body of mass
m at radius r is mg (R/r )2 .
Find the potential energy U at radius r if U = 0 when r = ∞.
Solution
2
R
U(r ) = mg
dr
r
Z
1
2
= mgR
dr
r2
mgR 2
+C
=−
r
Z
But U → 0 as r → ∞ so C = 0.
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Connection with Thermodynamics
1st law of Thermodynamics for a closed system (i.e. constant mass) gives
heat input − work output = increase in (internal energy + mechanical energy)
Increase in mechanical energy = ∆T + ∆U
Increase in internal energy = energy change due to temperature increase
In mechanics, heat output or increases in internal energy are typically
caused by frictional losses
Hence use
? work input = − work output
? heat output = − heat input
to rearrange:
work input
=
increase in mechanical energy + losses
losses
=
heat output + increase in internal energy
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Work and Energy
Examples
Bending a longbow or crossbow, compressing the spring on an air gun,
stretching the elastic on a catapult
– work done is later released as kinetic energy.
Model of a self-propelling car designed by Leonardo da Vinci
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Work and Energy
Examples
A flywheel on an internal combustion engine is used to compress the next
charge of gas
– kinetic energy is released as work.
BMW M5 engine
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Work and Energy
Examples
Driving a nail with a hammer
– kinetic energy in the hammer provides momentum for a large
impulse.
Mechnical vibrations
– kinetic energy and potential energy are repeatedly exchanged.
A moving car stopped by its brakes
– kinetic energy is lost as the brakes become hot.
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Work and Energy
Example – Mass on a slope
z
m
s
V
N
mg
q
A mass m slides down a frictionless surface of slope θ. Find the speed
V after the mass has descended through a vertical distance h.
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Work and Energy
Example – Mass on a slope
z
m
s
V
N
mg
q
Solution:
First, try solving by using equations of motion. Newton’s second law gives
mg sin θ = ma = mV
where
dV
ds
dz
sin θ
Eliminate ds, and separate variables V and z to give
ds = −
−g dz = V dV
Integrate from z = 0 to z = −h to give
V 2 = 2gh
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Work and Energy
Example – Mass on a slope
z
m
s
V
N
mg
q
Solution contd:
We can get a much simpler solution using energy.
Since there are no losses, the sum of the kinetic and the potential energy remains
constant. Hence
1
mV 2 = mgh
2
and so as before
V 2 = 2gh
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Work and Energy
Example – Riding a bicycle up a hill
D
d
q
A cyclist of mass M rides a bicycle of mass m up a hill.
The gearing is such that the bike moves a distance D along the road
while the pedals rotate 180o . The diameter of the pedal crank is d and
the cyclist rests his full weight on the down-going pedal throughout
the rotation from the highest to the lowest position.
Find the steepest slope that he can climb at a steady speed.
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Work and Energy
Example – Riding a bicycle up a hill
D
d
q
Solution:
During a downward pedal stroke:
? the total energy remains constant,
? kinetic energy is unchanged (since constant speed),
hence potential energy does not change.
? Rider, mass M falls d − D sin θ as the bicycle, mass m rises D sin θ, hence
Mg (d − D sin θ) = mgD sin θ
giving
sin θ =
Md
.
(M + m)D
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Work and Energy
Example – Riding a bicycle up a hill
D
d
q
Comment
The solution approaches sin θ = d/D if m M
then the rider is at the same height at the end of the pedal stroke as at the
beginning
work is done only as the rider steps up a distance d on to the next pedal.
Question:
why ride a bicycle up a hill when you could walk up the steps?
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Power
Power is the rate of doing work, with units of J s−1 = Watts (W)
For linear motion:
the work dW done in moving a force F a distance ds is
dW = F · ds
hence the rate of doing work is
dW
ds
=F·
dt
dt
or
P =F·V
in words
power = force · velocity
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Power
Power is the rate of doing work, with units of J s−1 = Watts (W)
For a rotating shaft:
the work dW done turning a torque M through angle dθ is
dW = M · dθ
so the rate of doing work is
dW
dθ
=M·
=M·ω
dt
dt
or
P =M·ω
in words
power = torque · angular velocity
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Power
Example – Acceleration of a car
The engine of a car develops 100 bhp (brake horsepower) at 5000
rpm (revolutions per minute) and the power varies in direct
proportion with the speed.
The engine drives the road wheels through a gearbox which reduces
the angular velocity by a factor of 7. The road wheels have radius
0.3 m.
If there are no losses due to friction or air resistance, find the time
required for the car to accelerate from 30 mph to 50 mph. Check
that the engine can do this without exceeding 5000 rpm.
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Power
Solution:
Conversion factors: 1 bhp = 746 W, 1 mile = 5280 ft, 1 ft = 0.3048 m
First find the road thrust F
Maximum engine speed:
ωmax =
5000 × 2π
= 523.60 rad s−1
60
Power at engine speed ω
P = 100 × 746 ×
ω
= 142.48 ω W
523.60
Engine torque
Me =
P
= 142.48 N m
ω
Wheel torque
Mw = 142.48 × 7 = 997.33 N m
Road thrust
F =
3 - 20
997.33
Mw
=
= 3324.43 N
rw
0.3
Power
Solution contd:
Next check the engine speed:
Road speed
V30 = 30 mph =
30 × 5280 × 0.3048
=
3600
13.41 m s−1
22.35 m s−1
V50 = 50 mph =
Engine speed at 50 mph
ω50 =
60
22.35
×7×
= 4980.4 rpm < 5000
0.3
2π
=⇒ ok
Finally find the time taken:
Acceleration
F
3324.43
=
= 2.216 m s−2
m
1500
hence the required time is
a=
t=
V50 − V30
22.35 − 13.41
=
= 4.03 s
a
2.216
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Forces and moments
A force F at right angles to a moment arm R produces a moment M:
M
R
F
M = RF
If F is inclined at an angle α then M is reduced to
M
R
a F
M = RF cos α
M is right-handed about an axis facing out of the paper
The moment can be represented as a vector M with magnitude M and
direction along the right-handed axis direction
A moment about the axis of spin of a shaft is called a torque
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Right-Hand Axes Sign Convention
Stretch the thumb and first two fingers of your right hand in perpendicular
directions.
Remember the axis orientation via
z
+ve
axis 1 (or x) = first finger
thumb
+ve
y
axis 2 (or y ) = second finger
second finger
axis 3 (or z) = thumb
first finger
Positive rotations are clockwise to an observer facing
away from the origin.
+ve
x
Hence if you point your right-hand thumb in the direction of any axis,
then positive rotation about that axis is in the direction that your fingers curl.
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Moment vector definition
The components of moment acting about O are
  

Mx
−Rz Fy + Ry Fz
My  =  Rz Fx − Rx Fz 
Mz
−Ry Fx + Rx Fy
The RHS can be replaced by an equivalent matrix
multiplication:
0
  
 
Mx
0
−Rz
Ry
Fx
My  =  Rz


0
−Rx
Fy 
Mz
−Ry
Rx
0
Fz
Fz
z
Fy
Mz
Fx
y
Rx
Rz
Ry
Mx
x
The RHS is the matrix representation of a vector cross product (HLT p.19)
so we can define the moment of F about O by
M=R×F
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My
Summary
Kinetic energy of a particle: T = 12 mV 2
Potential energy in a system: U = energy stored due to
its position in a field
work input: W = ∆T + ∆U + losses
Power input required to move a force F at velocity V:
P =F·V
Representation of moments as vectors
M=R×F
Power input required to rotate a moment M at angular velocity ω:
P =M·ω
3 - 25