4 - kinetics of particles - impulse & momentum
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Principle of Impulse and Momentum
Ken Youssefi MAE 1
The third method of solving engineering problems deals
with force, mass, velocity and time.
The method is particularly effective when dealing with
problems involving impulsive motion, force applied over a
short interval of time, and problems involving impact.
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Concept of Impulse and Momentum
Ken Youssefi MAE 2
Vector equation. Scalar equation.
The concept of workrelates
force to displacement.
The concept of impulse relates
force to time.
Greater force or greater
displacement is associated with
more work done.
Greater force or greater time of
action is associated with more
impulse applied.
More work done changes the
motion of a system to a greater
degree.
More impulse changes the
motion of a system to a greater
degree.
That which is changed is called
kinetic energy.
That which is changed is called
momentum.
Work & Energy Impulse & Momentum
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Principle of Linear Impulse and Momentum
Ken Youssefi MAE 3
Consider Newtons 2nd law.
mv (vector quantity) is called the linear
momentum of the particle
Linear impulse
Principle of linearImpulse & Momentum
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Components of Impulse
Ken Youssefi MAE 4
Rectangular coordinate system
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Principle of Impulse and Momentum
Ken Youssefi MAE 5
The final momentum of a particle is obtained by adding vectoriallyits initial
momentum and the impulse of the force F acting during the interval considered.
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Conservation of Linear Momentum
Ken Youssefi MAE 6
When the sum of the external impulses acting on a system of particle is
zero, the equation for the principle of linear impulse and momentum
reduces to the following:
Consider two boats initially at rest, which are pulled together
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Impulsive forces and Motion
Ken Youssefi MAE 7
Impulsive force is a force that acts on a particle during a very short
time interval and produces a definite change in momentum. The
resulting motion is called an impulsive motion. Baseball hitting a bat.
Non impulsive forces like weight of the body, the force exerted by spring, orany other force which is known to be small compared with the impulsive
force may be neglected.
In case of the impulsive motion of several particles, we can write:
No impulsive external
forces acting on the body
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Example
Ken Youssefi MAE 8
A car weighing 4000 lb is moving down a 5o incline at a
speed of 60 mi/h when the brakes are applied, causing
a constant braking force of 1500 lb (applied by the road
on the tires). Determine the time required for the car to
come to stop.
Apply the principle of impulse and momentum
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Example
Ken Youssefi MAE 9
A 4-oz baseball is pitched with a velocity of 80 ft/s. After
the ball is hit by the bat atB, it has a velocity of 120 ft/s in
the direction shown. If the bat and the ball are in contact
for .015 second, determine the average impulsive forceexerted on the ball during the impact.
Apply the principle of impulse and momentum to the ball, weight of the
ball can be neglected (weight is nonimpulsive force)
xx
t
t x mvmvdtF 12
2
1
yyt
t y mvmvdtF 12
2
1
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Example
Ken Youssefi MAE 10
x components
y components
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Example
Ken Youssefi MAE 12
x components
y components
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example
Ken Youssefi MAE 13
Initial energy
Final energy
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Direct Central Impact
Ken Youssefi MAE 15
Consider the impact of two particles
Before impact
The total momentum of the two particles is conserved
Scalar components
During impact
Same velocity
after impact
Restoration takes place
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Velocities after the Impact & the Coefficient of Restitution (e)
Ken Youssefi MAE 17
The ratio of the magnitude of the impulses corresponding to
the period of restitution and to the period of deformation is
called the coefficient of restitution, e is always between 0 and 1.
Same approach for particle B gives:
Substitute for the impulses
Eliminate u
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Oblique Central Impact
Ken Youssefi MAE 20
Velocities of the two colliding particles are not directed
along the line of impact.
Resolve the velocities along the n direction (along the line of impact)and the tdirection (along the common tangent, perpendicular to the
n direction).
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Oblique Central Impact
Ken Youssefi MAE 21
1. The tcomponents of the velocity of each particle
remains the same.
2. The component along the n axis of the total momentum
of the two particles is conserved.
3. The coefficient of restitution equation can be used inthe n direction
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Example
Ken Youssefi MAE 24
The magnitude and direction of the velocities
of two balls before they collide are shown in
the figure. If the coefficient of restitution is 0.9,
determine the magnitude and direction of thevelocity of each ball after the impact (neglect
friction)
Resolve the velocities into components along the line of impact and
along the common tangent to the surface in contact
BallA
BallB
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Example
Ken Youssefi MAE 25
Motion along the common tangent
Motion along the line of impact
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Using Conservation of energy and Momentum
Ken Youssefi MAE 27
Consider a pendulumA released with no velocity from positionA1.
PendulumA hits pendulumB which is initially at rest. After the impact
pendulumB swings through angle . Determine the angle .
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Using Conservation of energy and Momentum
Ken Youssefi MAE 28
2. PendulumA hits pendulumB. Use conservation of momentum and theequation for coefficient of restitution to solve for velocities of the two
pendulums after the impact, (vA)3 and (vB)3
1. PendulumA swings fromA1 toA2. Use conservation of energy to find the
velocity of pendulumA, (vA)2 , just before it collides with pendulumB.
3. PendulumB swings from B3 toB4. Use conservation of energy for
pendulum B to determine the heighty4. Use trigonometry to find .
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Example
Ken Youssefi MAE 29
A 30 kg block is dropped from a height of 2 m onto the 10 kg pan of a spring
scale. If the impact is perfectly plastic, determine the maximum deflection of
the pan. The spring stiffness is k= 20 kN/m.
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Example
Ken Youssefi MAE 30
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Example
Ken Youssefi MAE 31
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Example
Ken Youssefi MAE 32
Conservation of energy
Initial deflection of the spring due to the scale pan (WB)
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l
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Angular Momentum
Ken Youssefi MAE 35
The linear momentum was defined as mv. The moment of the
vector mvabout point O is called the angular moment or the
moment of momentum of the particle about point O
Ho = (d)(mv)
Consider a 2D motion, particle
moving along a path onxy plane.
The magnitude of the angularmoment is;
l
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Angular Momentum
Ken Youssefi MAE 36
Ho
is a vector perpendicular to the
plane containing vectors r and mv
Plane containing
vector r and
vector mv
A l M
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Angular Momentum
Ken Youssefi MAE 37
r = rx i + ry j + rzk , v = vx i + vy j + vzk
Position and velocity vectors in rectangular coordinate system
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A l M t
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Angular Momentum
Ken Youssefi MAE 39
Relationship between moment of a force and angular momentum.
Equation of motion
Moments of the forces about point O
Same result for
linear momentum
P i i l f A l I l d M t
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Principle of Angular Impulse and Momentum
Ken Youssefi MAE 40
Modt= dHo
At time t= t1 Ho = (Ho)1 , At time t= t2Ho = (Ho)2
Principle of
angular impulse
and momentum
Angular Impulse
Linear Impulse
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P i i l f Li & A l I l d M t
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Principle of Linear & Angular Impulse and Momentum
Ken Youssefi MAE 42
Linear and Angular impulse & momentum
3D vector
equations
2D scalarequations
Linear
Angular
Conser ation of Ang lar Moment m
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Conservation of Angular Momentum
Ken Youssefi MAE 43
If the angular impulse acting on a particle is
zero the principle equation reduces to the
following:
Conservation ofAngular Momentum
If force F is directed toward
point O the moment is zero
so the angular momentum is
conserved. Force F is knownas a Central Force.
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Example
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Example
Ken Youssefi MAE 46
A 2 kg bob is given a horizontal speed of 1.5 m/s.
It rotates on a circular pathA. If the force F on
the cord is increased, the bob rises and rotates
around the horizontal circular pathB. Determine
the speed of the bob around pathB. Also, find
the work done by force F.
There are three forces involved in this problem, W,F,
and the cord force. WandFproduce no moment
about thezaxis. Also, the cord force is along the cord
so it does not produce a moment about thezaxis.
Therefore, the conservation of momentum applies.
r1mv1 = r2mv2
r1 = l1 sin1 r2 = l2 sin2
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Example
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Example
Ken Youssefi MAE 48
Position 2
l= l2 = .3 m , v = v2 and = 2
Eliminate v2
(1)
r1mv1 = r2mv2
r1 = l1 sin1 r2 = l2 sin2
Conservation of momentum
(2)
Example
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Example
Principle of Work and Energy
When changes to 1 to 2 , Wdisplaces vertically upward.
Therefore, Wdoes negative work
h = .6 cos (34.21) - .3 cos (57.866) = .3366 mh