Impulse & Momentum - WordPress.com...2015/03/05 · propulsion with variable mass. 25/03/58 ME212...
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Impulse & Momentum
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Objectives
• To develop the principle of linear impulse and
momentum for a particle.
• To study the conservation of linear momentum for
particles.particles.
• To analyze the mechanics of impact.
• To introduce the concept of angular impulse and
momentum.
• To solve problems involving steady fluid streams and
propulsion with variable mass.
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Principle of linear impulse and momentum
∫ ∫=Σ
==Σ
2 2t
t
v
vvdmdtF
dt
vdmamF
vv
vvv
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∫ ∫=Σ1 1t v
vdmdtF
∫ −=Σ2
112
t
tvmvmdtFvvv
Linear Impulse Linear Momentum
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Principle of linear impulse and momentum
)()(2t
vmdtFvm =+∑∫
21
2
1
vFvvvvmdtm
t
t=+∑∫
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21
21
21
)()(
)()(
)()(
2
1
2
1
2
1
z
t
tzz
y
t
tyy
x
t
txx
vmdtFvm
vmdtFvm
vmdtFvm
=+
=+
=+
∑∫
∑∫
∑∫
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Procedure for Analysis
• Select the inertial coordinate system
• Draw FBD to account for all forces
• Establish direction and sense of acceleration
• Solve for unknowns• Solve for unknowns
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Problem 15-10
A man kicks the 200-g ball such that it leaves the
ground at an angle of 30⁰ with the horizontal and
strikes the ground at the same elevation a distance
of 15 m away. Determine the impulse of his foot F on the ball. the ball.
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Problem 15-11
The particle P is acted upon by its weight of 30 N and
forces F1 and F2, where t is in seconds. If the particle
originally has a velocity of v1 = (3i + 1j + 6k) m/s, determine its speed after 2 s.
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Problem 15-27
Block A weighs 100 N and block B weighs 30 N. If B is
moving downward with a velocity (vB)1 = 1 m/s at
t=0, determine the velocity of A when t = 1 s. Assume that the horizontal plane is smooth.
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Principle of Linear Impulse and Momentum for a System of Particles
∑∑ =d
m iii
vF
vv
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∑∑ =dt
miiF
( ) ( )21
2
1∑∑∫∑ =+ ii
t
tiii mdtm vFv
vvv
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Conservation of Linear Momentum for a System of Particles
• When the sum of the external impulses acting
on a system of particles is zero, the equation is
( ) ( )∑∑ = mm vvvv
• This equation is referred to as the conservation of linear momentum.
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( ) ( )∑∑ =21 iiii mm vv
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Example 1
The 15-Mg boxcar A is coasting at 1.5 m/s on the
horizontal track when it encounters a 12-Mg tank B
coasting at 0.75 m/s toward it. If the cars meet and
couple together, determine (a) the speed of both
cars just after the coupling, and (b) the average force cars just after the coupling, and (b) the average force
between them if the coupling takes place in 0.8 s.
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Problem 15-52
The boy B jumps off the canoe at A with a velocity of
5 m/s relative to the canoe as shown. If he lands in
the second canoe C, determine the final speed of
both canoes after the motion. Each canoe has a mass
of 40 kg. The boy’s mass is 30 kg, and the girl has a of 40 kg. The boy’s mass is 30 kg, and the girl has a mass of 25 kg. Both canoes are originally at rest.
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Impact
• Central Impact
– Velocities are colinear (a)
– Deforms during collision (b)
– Restore after deformation (b)– Restore after deformation (b)
– Coefficient of restitution
= Irestitution/Ideform
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Coefficient of Restitution
• Consider m1
01
10
101
011
)]([
)]([0
vv
vv
vvm
vvm
dtF
dtF
et
d
t
t
r
−
′−=
−−−
−−′−==
∫
∫
• Consider m2
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0t
d∫
20
02
202
022
][
][
0
0
vv
vv
vvm
vvm
dtF
dtF
et
t
d
t
t
r
−
−′=
−
−′==
∫
∫
21
12
vv
vve
−
′−′=
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Impact
• Oblique Impact
vv )()( ′−′
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nn
nn
vv
vve
)()(
)()(
21
12
−
′−′=
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Problem 15-64
• If the girl throws the ball with a horizontal
velocity of 3 m/s, determine the distance d so
that the ball bounces once on the smooth
surface and then lands in the cup at C. Take surface and then lands in the cup at C. Take
e=0.8.
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Angular Momentum
• If the particle is moving along a space curve,
the vector cross product can be used to determine the angular momentum about O.
vrHvvvm×=
kjir zyx rrr ++=
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vrHvvvmO ×=
kjir zyx rrr ++=
kjiv zyx vvv ++=
θsin0 rvmH =
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Angular Momentum
• RecallLvm
dt
dF &vv
==Σ )(
Hvmrd
Fr &vvvv=×=×Σ )()(
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OHvmrdt
Fr &=×=×Σ )()(
OO HM &v
=Σ
The moment about the fixed point O of all forces acting on m equals to the rate of change of angular momentum.
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Principle of Angular Momentum
dt
HdHM OOO
v
&vv
==Σ
OO HddtMvv
=Σ
2t vvv
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12
2
1
OO
t
t
O HHdtMvvv
−=Σ∫
2
2
1
1 O
t
t
OO HdtMHvvv
=Σ+ ∫2
2
1
1 O
t
t
OO HdtMHvvv
Σ=Σ+Σ ∫
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Conservation of Angular Momentum
• Finally, if all angular impulse is zero.
21 OO HHvv
=
• For a system of particles
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21 OO HHvv
Σ=Σ
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Example 2
The 2 kg disk rests on a smooth horizontal surface
and is attached to an elastic cord that has a stiffness
kc = 20 N/m and is initially unstretched. If the disk is
given a velocity (vD)1 = 1.5 m/s, perpendicular to the
cord, determine the rate at which the cord is being D
cord, determine the rate at which the cord is being
stretched and the speed of the disk at the instant the
cord is stretched 0.2 m.
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Meriam 3-226
The small spheres, which have the masses and initial
velocities shown in the figure, strike and become
attached to the spiked ends of the rod, which is
freely pivoted at O and is initially at rest. Determine
the angular velocity ω of the assembly after impact. the angular velocity ω of the assembly after impact. Neglect the mass of the rod.
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Meriam 3-230
The 6-kg sphere and 4-kg block are secured to the arm of negligible mass which rotates in the vertical plane about a horizontal axis at O. The 2-kg plug is released from rest at A and falls into the recess in the block when the into the recess in the block when the arm has reached the horizontal position. An instant before engagement, the arm has an angular velocity ω0 = 2 rad/s. Determine the angular velocity ω of the arm immediately after the plug has wedged itself in the block.
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