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Prof. Dr. Ersel Canyurt 4 - 1
DYNAMICS
CHAPTER
Kinetics of Particles:
Impuls & Momentum
Dynamics
R.C. Hibbeler
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
Prof. Dr.
O. Ersel Canyurt
Gazi University
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 2
Introduction
- In section 2, the problems associated with the movement of particles,
İmpulse and momentum :
- Method of impulse and momentum will be shown in this section.
Equations of motion F = m a^ was used.
- In section 3, energy method was used.
T
1 + U^1 2 =^ T 2
Work and Energy (Friction exist)
T
1
= T
2
+ V
1
+ V
2
Conservation of Energy (frictionless)
establishes a relationship among
Mass (m)
m
Velocity (v)
F
v Force (F)
t^ and^ time^ (t) 1
t 2
Prof. Dr. Ersel Canyurt 4 - 3
Principles of Impulse and momentum
- Newton's Second Law : F=m.a
Imp (^1) 2
L 1
=m v 1
L 2 =m v 2
m: constant
m v 1
= m v 2
Particle, v 1
If particle is influenced by
external force F ext
for a specific
period of time
Particle has v 2
in the
direction of q
m.v =L, Linear Momentum
a
dv
dt
F = m
F. dt = d (m v )
ʃ F dt
t 1
t 2
= m v 2
Imp (^1) 2 : F impulse force
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 4
Principles of Impulse and momentum
2
m kg s
s
= F. Dt
[N]. [s]
[kg]. [m/s]
kg m
s
kg m
s
Prof. Dr. Ersel Canyurt 4 - 7
Problem 4.
An automobile weighing 4000 N is driven down a 5° incline at
a speed of 30 m/s when the brakes are applied causing a
constant total breaking force of 1500 N.
Determine the time traveled by the automobile as it comes to
a stop?
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 8
Problem 4.1 • Apply the principle of impulse momentum.
m v 1
= m v 2
m v 1
Initial momentum
W=4000 [N]
N
F
s
m v 2
External Forces Final momentum
x
Taking this as
the direction of the x-axis
y
W
x
= W. cos 85
W
y
5 o
W
x
= 4000. cos 85
W
x
= =348.6 [N]
separate W into components
Prof. Dr. Ersel Canyurt 4 - 9
Problem 4.
If we write the principle momentum-impulse of in x-direction
(DIRECTION is important)
m v 1
N
F
s
m v 2
W
x
= 348.6 [N]
W
y
W
x
= 348.6 [N]
W
y
5
o
m v 1x
+ W
x
. Dt – F s . Dt = m v 2
x
Dt = 10.62 [s]
m =
=407.74 [kg]
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 10
Problem 4.
The 250 N crate (sandık) is acted upon by a force having a variable
magnitude force P = 100 t, where t is in seconds. Determine the crate's
velocity 2 s after P has been applied. The initial velocity is v 1
= 1 m/s
down the plane, and the coefficient of kinetic friction between the crate
and the plane is μ k
=0.
v 1 =1 m/s
Prof. Dr. Ersel Canyurt 4 - 13
Problem 4.3a
A 100 gr baseball is pitched with a velocity of 30 m/s. After the ball is hit by
the bat, it has a velocity of 40 m/s in the direction shown.
If the bat and ball are in contact for 0.015 s, determine the average
impulsive force exerted on the ball during the impact?
40 m/s
30 m/s
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 14
Problem 4.
Apply the principle of impulse and momentum in terms of
horizontal and vertical component equations.
x
y
x - direction :
40 m/s
30 m/s
m v 1
= m v 2
m v 1
=
F
y
Dt
F
x
Dt 40 o
m v 2
m v 1x
+ F
x
F
x
= 404.3 [N]
m v 2x
m v 2y
v2x= v 2 cos 40
= 40. cos 40
v 2x
= 30.64 [m/s]
v 2y
= v 2
cos 50
= 40. cos 50
v2y= 25.71 [m/s]
W Dt 0
(DIR )
W Weight Neglegted
W << F
y
Prof. Dr. Ersel Canyurt 4 - 15
Problem 4.
x
y
y - direction:
40 m/s
30 m/s
m v 1
= m v 2
m v 1
F
y
Dt
F
x
Dt 40 o
m v 2
m v 1y
+ F
y
DF
y
= 171.4 [N]
m v 2x
m v 2y
v 2x
= v 2
cos 40
= 40. cos 40
v2x= 30.64 [m/s]
v2y= v 2 sin 40
= 40. sin 40
v 2y
= 25.71 [m/s]
F= 404.3 i + 171.4 j
|F| = 439.1 [N]
W Dt 0
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 16
Problem 4.3b
A 10-kg package drops from a chute into a 25-kg cart with a velocity
of 3 m/s. Knowing that the cart is initially at rest and can roll freely,
determine
(a) The final velocity of the cart,
(b) The impulse exerted by the cart on the package,
(c) The fraction of the initial energy lost in the impact.
Prof. Dr. Ersel Canyurt 4 - 19
Problem 4.3^ • Impulse-Momentum Principle: Package
1 1 2 2
m v m v
p p
(^)
Imp
y - direction
+ F
y
- mP • (v 1 • sin 30) (^) • Dt = 0
In y - direction – No movement
│F • Dt│ = 23.9 [N•s]
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 20
T
1
m P
2
T
2
(m P
) • v 2
2
- Kinetik energy at final position
The fraction of energy lost is =
T 1 – T 2
T
1
Problem 4.3^ The fraction of energy lost
Prof. Dr. Ersel Canyurt 4 - 21
Problem 4.
Block B start its motion from the rest. Determine the velocity of block B
at time 6 sec.
Mass of pulley and rope are neglegted.
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 22
Problem 4.
xA
x B
2 x A
2 v A
= 0
x
y
O
- Pulley D (mass neglegted)
T (^2) T 2
T 1
F
y
= mD •^ a = 0
T
1
= 2 T
2
vA = – 0.5 vB
W
A
= m A
W
A
=29.4 [N]
W
B
= m B
. g = 5 • (9.81)
W
B
=49 [N]
T 1
T 2
Prof. Dr. Ersel Canyurt 4 - 25
Impact
- Impact: Collision between two bodies which occurs during a small time
interval and during which the bodies exert large forces on each other.
Centers of mass of colliding objects :
On the direction of line of impact called central collision;
otherwise, it is an eccentric impact
Touch Surface (TS)
Line of Impact (LI)
Perpendicular to the touch surface
LI
TS
Eccentric Impact
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 26
Central Impact
- Direct Impact (Doğrusal) : Impact for which the velocities of the two
bodies are directed along the line of impact.
Touch Surface
Line of Impact (LI)
v A
v B
VA and VB are on the Line of Impact
Line of Impact
v A vB
Prof. Dr. Ersel Canyurt 4 - 27
Central Impact
Touch
Surface
Line of Impact
v A
v B
One of VA and VB are other than the Line of Impact
Line of Impact
v A
v B
- Oblique Impact (Eğik): Impact for which one or both of the bodies move
along a line other than the line of impact.
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 28
Direct Central Impact
and m B
moving in the same straight line, v A
v B
.
- Upon impact the bodies undergo a period of deformation, at the end of
which, they are in contact and moving at a common u velocity.
- A period of restitution follows during which the bodies either regain their
original shape or remain permanently deformed.
Final velocity
v^ LI A
v B
u
v’A LI v’ B
Initial velocity
Prof. Dr. Ersel Canyurt 4 - 31
Direct Central Impact
- Ratio of impulsif force for period of deformation and for period of
restitution is called ( e) Coefficient of restitution
e=
ʃ P dt
ʃ R dt
(v A
(u – v’ A
e=
- Similar analysis is performed for object B
(v B
(u – v’ B
e=
(u– v B
(v’ B
e=
The ratio of e is same in each equation,
Therefore numerator (Pay) denominator (Payda) can be added
v A
u – v’ A
e=
=
v’ B
v A
v’B – v’A = e • ( vA – vB )
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 32
Direct Central Impact
- Perfectly plastic impact
- Perfectly elastic impact
Two bodies moves togetger.
e = 0
v’B – v’A = e. ( vA – vB )
v’B = v’A = v’ One unknown
m A
vA + m B
vB = (m A
) • v’ 1 One equation
e = 1
mA vA + mB vB = mA v’A + mB v’B 1
v’B – v’A = vA – vB^2
m A
vA + m B
vB = m A
v’A+ m B
v’B
Prof. Dr. Ersel Canyurt 4 - 33
Oblique Central Impact
TS
v A
v B
LI
n
t
can be divided into components
v At
v An
v Bt
v Bn
- In the tangential direction, t No impulse force.
v’ At
n
t
v’ Bt
v’ At
= v At
v’ Bt
= v Bt
- Initial velocities (^) • Final velocities
v’ An
v’ Bn
- In the Line of Impact, n - direction
m A
vAn + m B
vBn = m A
v’An + m B
v’Bn 1
v’Bn – v’An = e • (vAn– vBn )^2
- Velocities does’t change in the tangential direction
- Momentum is conserved.
- Relationship of Relative Velocity
(Bağıl Hız)
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 34
Problems Involving Energy and Momentum
Three methods for the analysis of kinetics problems were shown
- Select the method best suited for the problem or part of a problem under
consideration.
Conservation of Energy Impact Conservation of Energy
- Direct application of Newton’s second law
- Method of work and energy
- Method of impulse and momentum
v A
A 1
A 2
vA
l
B 1
vB=
l
A 2
v A
B 2
v B
=
l
A 3
vA
B 3
vB
l
Before Impact (^) After Impact
v B
=
B 4
A 4
q
y 4
Prof. Dr. Ersel Canyurt 4 - 37
Problem 4.
The magnitude and direction of the velocities of two
identical frictionless balls before they strike each other are
as shown.
Assuming e = 0.9, determine the magnitude and direction
of the velocity of each ball after the impact?
v A
=9 m/s
v B
=12 m/s
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 38
Line of Impact (LI)
v A
v B
30
o
60
o
- Separete initial velocity components of the ball
n
t
v At
v An
v Bt
v Bn
vAn = vAn= 7.79 [m/s]
= v At
vAt = 4.5 [m/s]
vBn = – vBn= – 6.0 [m/s]
= v Bt
vBt = 10.39 [m/s]
Initial Velocities
Final Velocities
n
t
v’ At
v’ An
v’ Bt
v’ Bn
- velocities in the tangential direction are conserved
v’ At
= v At
v’ At
= 4.5 [m/s]
v’ Bt
= v Bt
v’ Bt
= 10.39 [m/s]
- In the Line of Impact
- Final direction is unknown. (+ DIR. )
Problem 4.
Prof. Dr. Ersel Canyurt 4 - 39
Problem 4.
v A
v B
30
o
60
o
n
t
v At
v An
v Bt
v Bn
v An
= 7.79 [m/s]
v At
= 4.5 [m/s]
v Bn
= – 6.0 [m/s]
v Bt
= 10.39 [m/s]
Final Velocities
n
t
v’ At
v’ An
v’ Bt
v’ Bn
- In the line of Impact direction
- Momentum preserved
m A
vAn + m B
vBn = m A
v’An + m B
v’Bn
v’Bn– v’An = e • (vAn– vBn )
- Relationship of relative velocity
v’An + v’Bn =1.
v’Bn– v’An = 12.
Initial Velocities
Prof. Dr. Ersel Canyurt
Mechanical Engineering
4 - 40
Problem 4.
v A
v B
30
o
60
o
n
t
v At
v An
v Bt
v Bn
vAn= 7.79 [m/s]
v At
= 4.5 [m/s]
vBn= – 6.0 [m/s]
v Bt
= 10.39 [m/s]
Final Velocities
n
t
v’ At
v’ An
v’ Bt
v’ Bn
Initial Velocities
v’ Bn
= 7.1 [m/s]
v’ An
= – 5.31 [m/s]
= – 5.31^ = 7.
