Momentum, Work and Energy in Mechanical Systems

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Momentum, Work and Energy in Mechanical Systems

• Prof Bill Easson
• http//www.see.ed.ac.uk/bille

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Introduction

• So far, we have explicitly used the force and
  Newtons 2nd Law to analyse dynamic systems.
  However, we do not always know the force, and
  there are other methods available. Here we deal
  with two of these
• Momentum
• Energy

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Momentum
NII resultant force rate of change of momentum
where m is constant
If a constant force, F, acts on a mass m for a
time t, then
where v1 and v2 are initial and final velocities
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Momentum (2)
What happens if F varies with time? We get a
series of products
where the time intervals t are small enough for
the forces to be ? constant over each interval.
Maths gives us an easy way to write this sum
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Momentum (3)
If F is constant then we should get the same
result as before
F
t1
t2
t
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Momentum (4)
If F is a known function of t, then we can solve
Suppose a force acting in the direction, x, gets
larger with time, t. Fkt
F
t
t1
t2
In this case the velocity varies as the square of
time, t
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Momentum (5)
Even if the equation of the force is not known,
the momentum can sometimes be used to solve
dynamical problems where there are no external
forces
Consider two balls colliding
uA
uB
vB
vA
Although we do not know the value of the forces
during the collision, we know that they must be
equal and opposite
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Momentum (6)
The total momentum after the collision is equal
to the total momentum before the collosion.
This turns out to be true however many bodies
there are and is called the Law of Conservation
of Momentum
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Momentum conservation
Consider a truck carrying a large gun which fires
shells of mass 20kg. The truch gun together
weigh 5 Tonnes. If the shell is fired at 300ms-1
at an angle of 60o to the horizontal, what is the
recoil velocity of the truck?
y
u2
x
u1
u1u20
Before
After
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Momentum conservation
As the truck can only move horizontally, we only
consider x values
Before m15000 kg m220 kg u10 ms-1 u20 ms-1
After m15000 kg m220 kg v1? v2300cos60o150
ms-1
Momentum before is zero, so momentum after must
also be zero
m1v1m2v20
v1-m2v2/m1
-20.150/5000
v1-0.6 ms-1
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Energy (and Work)
Here we look at the effect of applying a force,
F, over a distance, x.
The work done is the energy transferred, by the
force, to the body units are joules(J). For a
constant force, as in the figure WF x d
F
x1
x2
x
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Work and Power
If F varies with x, as in the figure, then we add
all the small elements to get the area under the
curve,
F
Power is the rate at which work is done and is
measured in Watts (W)
(since integration is the opposite of
differentiation)
x1
x2
x
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Potential Energy
In dynamics, potential energy is work done
against gravity, often called gravitational
potential energy.
Consider a mass being raised at a constant speed
on a cable
Before
After
T
T
h
mg
?Umgh
mg
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Spring Energy
Empirically (from experiment), we find that Fkx
F(x)
x
F(x)
This is referred to as spring potential energy
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Kinetic Energy
Consider Newtons second law in the integral for
work
This is the work done accelerating a body
It is normally called kinetic energy
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Conservation of Energy
In many of the problems you encounter mechanical
energy will be conserved, and this can be a
useful tool.
TUconst
Example A ball of mass 1kg is dropped from a
balcony of height 10m. Ignoring drag, calculate
its velocity just before it hits the ground.
energy before energy after
v14ms-1
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Elastic Collisions
If the collision is elastic, then kinetic energy
is conserved, and momentum is always conserved,
so we have two equations
CoM
(1)
CoE
(2)
For an in-line (1D) collision, divide equation 2
by equation 1
and
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Elastic collisions - examples
Substituting the equations for v1 and v2 back
into the momentum equation gives
Snooker (m1 m2 u20)
(Second ball moves off with speed of first. First
ball stops. Ignores spin.)
v2 u10 and v1 u2
Ball hitting wall ( m2 gtgt m1 u20)
v2 u20 and v1 -u1
(Wall remains stationary, ball bounces back with
same velocity)
Golf club hitting ball ( m1 gtgt m2 u20)
(Ball goes off at twice clubhead speed and
clubhead velocity remains the same)
v2 2u1 and v1 u1
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Snookered
Consider two smooth, hard balls colliding.
During the collision, which is very short,
friction has a negligible effect. So the force
can only be transferred by the impulse acting
normal to the surface at the collision point.
FA
uA
FB
uB0
In snooker, all the balls have the same diameter
and mass, which makes the problem easy to solve.
The velocity of ball B after impact must be in
the direction of the line of action.
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Snookered
Case 1 In line
y
uA
uB0
x
As the line of action is in the x direction,
there can be no motion in the y direction.
The energy equation gives
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Snookered
Energy
(1)
Momentum
(2)
Substituting for vA from equ 1 into equ 2,
So, vBuA and vA0
The second ball moves off at the speed of the
first, and the first ball becomes stationary
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Snookered
Balls not in line of action
y
vA
?
uA
?
x
uB0
vB
We know that the momentum equation must be
satisifed in x and in y, since it is a vector
equation. The collision is elastic, so energy is
conserved. We also know that B can only move off
on the line of action, ?.
x-momentum is conserved
no y-momentum before, therefore no momentum after
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Snookered
Energy
For any given angle of action, ?, and initial
velocity, uA, we have three unknowns, vA, vB and
the angle of the first ball, ?. However, we also
have three equations so, by substitution and some
algebra, we find that the white ball must move
off at exactly 90o to the red ball.
90o
This enables a simple program to be written to
simulate a snooker/billiards table.