Emm 222 - Dynamics Mechanism - Chapter 4
-
Upload
pathmaprasad-prasad -
Category
Documents
-
view
216 -
download
0
Transcript of Emm 222 - Dynamics Mechanism - Chapter 4
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
1/39
Kinetics of a Particle
Impulse and Momentum
Chapter Four
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
2/39
In this chapter, you will learn how to: -
derive the principle of impulse and
momentum for a particle from
Newtons second law.
express the conservation oftotal
momentum of the particles whenimpulse of the external forces is zero.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
3/39
Kinetics of a Particle
Principle of Linear Impulse
and Momentum
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
4/39
Principle of Linear Impulse and
Momentum
21
12
MomentumImpulseLinear
mvFdtmv
mvmvFdt
dvmFdt
mdvdtF
dt
dvmF
maF
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
5/39
Resolving into x,y,z components
21
21
21
2
1
2
1
2
1
z
t
t
zz
y
t
t
yy
x
t
t
xx
vmdtFvm
vmdtFvm
vmdtFvm
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
6/39
Example 4.1:
A 20 N block slides down a 30 inclined
plane with an initial velocity of 2 m/s.
Determine the velocity of the block in 3 s
if the coefficient between the block andthe plane is k = 0.25.
30
2 m/s
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
7/39
Solution 4.1:
FBD
N
Ff= 0.25 N
W=20 N
y
x
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
8/39
Solution 4.1: (continued)
___
030cos20
0
___
030cos200
3
0
21
2
1
N
N
maF
or
NN
dtN
vmdtFvm
yy
y
t
t
yy+ y
17.32 N
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
9/39
Solution 4.1: (continued)
smv
vdtN
vmdtFvm x
t
t
xx
/___
)(81.9
2025.030sin20)2(
81.9
20
3
0
21
2
1
+ x
10.3
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
10/39
Example 4.2:
A 20 N ball is thrown in the direction shown
with an initial speed vA=6 m/s. Determine
the time needed for it to reach its highest
point B and the speed which it is travelling
at B. Use the principle of impulse and
momentum for the solution.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
11/39
Solution 4.2:
st
dt
vmdtFvm
t
y
t
t
yy
___
020)30sin6(81.9
20
0
21
2
1
+
0.306
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
12/39
Solution 4.2: (continued)
smv
v
vmdtFvm
B
B
x
t
t
xx
/___
)(81.9
200)30cos6(
81.9
20
21
2
1
+
5.2
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
13/39
Example 4.3:
A hammer head having a
weight of1 N is movingvertically downward at 10 m/s
when it strikes the head of a
nail of negligible mass and
drives into a block of wood.
Find the impulse on the nail if
it is assumed that the grip at A
is loose, the handle has anegligible mass, and the
hammer stays in contact with
the nail while it comes to rest.
Neglect impulse that
caused by hammer
head.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
14/39
Solution 4.3:
NsFdt
Fdt
vmdtFvm
t
y
t
t
yy
___
0)10(81.9
1
0
21
2
1
+
1.02
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
15/39
Kinetics of a Particle
Conservation of
Linear Momentum
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
16/39
When the sum of the external impulses acting
on a system of particles iszero, the equation is
21
2
1
0
iiii
t
t
mm
Fdt
vv
This equation is referred to as the conservation
of linear momentum.
It states that the total momentum for a system of
particles remains constant during the time period
t1 to t2.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
17/39
Example 4.4:
The car A has a weight of22.5 kN and is
travelling to the right at 1 m/s. Meanwhile a
15 kN car B is travelling at 2 m/s to the left.
If the cars crash, determine their commonvelocity just after the collision. Assume
brakes are not applied during collision.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
18/39
Solution 4.4:
smv
v
vmmvmvm BABBAA
/___
81.9
1500022500)2(81.9
15000)1(81.9
22500
2
2
211
- 0.2
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
19/39
Example 4.5:
Given:Two rail cars with masses ofmA = 15
Mg and mB = 12 Mg and velocities as
shown.
Find:The speed of the cars after they meet
and connect. Also find the averageimpulsive force between the cars if the
coupling takes place in 0.8 s.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
20/39
Plan:
1. Use conservation of linear momentum
to find the velocity of the two cars after
connection (all internal impulses
cancel).
2. Then use the principle of impulse and
momentum to find the impulsive forceby looking at only one car.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
21/39
Solution 4.5:Conservation of linear momentum (x-dir):
mA(vA)1 + mB(vB)1 = (mA + mB) v2
15,000(1.5) + 12,000(- 0.75) = (27,000)v2
v2 = 0.5 m/s
Impulse and momentum on car A (x-dir):
mA(vA)1 + F dt = mA(v2)
15,000(1.5) - F dt = 15,000(0.5)
F dt = 15,000 Ns
The average force is
F dt = 15,000 Ns = Favg(0.8 sec); Favg = 18,750 N
Solution:
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
22/39
Kinetics of a Particle
Impact
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
23/39
Impactoccurs when two bodies collide
with each other during a very shortperiod
of time, causing relatively large (impulsive)
forces to be exerted between the bodies.
There are two types of impact.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
24/39
1.Central impactoccurs when the direction of
motion of the mass centers of the two
colliding particles is along theline of impact.
2.Oblique impactoccurs when one or both of
the particles is at an angle with the line of
impact.
The line of impact passes through the masscenters of the particles.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
25/39
In most problems, the initial velocities of the particles,
(vA)1 and (vB)1, are known, and it is necessary to
determine the final velocities, (vA)2 and (vB)2. So the firstequation used is the conservation of linear momentum,
applied along the line of impact.
(mA vA)1 + (mB vB)1 = (mA vA)2 + (mB vB)2
This provides one equation, but there are usually two
unknowns, (vA)2 and (vB)2. So another equation isneeded. The principle of impulse and momentum is
used to develop this equation, which involves the
coefficient of restitution, or e.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
26/39
The coefficient of restitution, e, is the ratio
of the particles relative velocity after impact,(vB)2 (vA)2, to the particles relative velocity
before impact, (vA)1 (vB)1.
The equation defining the coefficient of
restitution, e, is
(vA)1(vB)1(vB)2
(vA)2e =
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
27/39
In general, e has a value between zero and one.
Elastic impact (e = 1): In a perfectly elasticcollision, no energy is lost and the relative
separation velocity equals the relative approach
velocity of the particles. In practical situations,this condition cannot be achieved.
Plastic impact (e = 0): In a plastic impact, the
relative separation velocity is zero. The particlesstick together and move with a common velocity
after the impact.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
28/39
IMPACT: ENERGY LOSSES
During a collision, some of the particles initialkinetic energy will be lost in the form of heat,
sound, or due to localized deformation.
Once the particles velocities before and afterthe collision have been determined, the energy
loss during the collision can be calculated on
the basis of the difference in the particles
kinetic energy. The energy loss is
U1-2 = T2 - T1 where Ti = 0.5mi(vi)2
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
29/39
Example 4.6:
Disk A has a mass of2 kg and is sliding forward
on the smooth surface with a velocity vA = 5m/s when it strikes the 4 kg disk B, which is
sliding towards A at vB = 2 m/s, with direct
central impact. If the coefficient of restitutionbetween the disk is e = 0.4, compute the
velocity of A and B just after collision.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
30/39
Solution 4.6:
)1(42)2(452
MomentumofonConservati
22
2211
BA
BBAABBAA
vv
vmvmvmvm
)2(
)2(5
)()(4.0
)()(
)()(
nRestitutiooftCoefficien
22
11
22
AB
BA
AB
vv
vv
vve
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
31/39
Solution 4.6:(continued)
smv
smsmv
B
A
/27.1
/53.1/53.1
(2)and(1)equationsSolving
2
2
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
32/39
Example 4.7:
Two smooth disks A and B, having mass of 1 kg
and 2 kg respectively, collide with the velocitiesshown. If the coefficient of restitution for the disksis e= 0.75, determine the xand ycomponents ofthe final velocity of each disk just after collision.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
33/39
Solution 4.7:
Resolving each of the initial velocities into x and ycomponents, we have
smv
smv
smv
smv
By
Bx
Ay
Ax
/707.045sin1)(
/707.045cos1)(
/50.130sin3)(
/60.230cos3)(
1
1
1
1
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
34/39
Solution 4.7:(continued)
The four unknown velocity components aftercollision are assumed to act in the positivedirections. Since the impact occurs only in the x
direction (line of impact), the conservation ofmomentum for bothdisks can be applied in thisdirection.
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
35/39
Solution 4.7:(continued)
Conservation of x Momentum.
18.1)(2)(
)(2)(1)707.0(2)60.2(1
)()()()(
22
22
2211
BxAx
BxAx
BxBAxABxBAxA
vv
vv
vmvmvmvm( )
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
36/39
Solution 4.7:(continued)
Conservation of (x) Restitution. Both disks areassumedto have components of velocity in the +xdirection after collision,
48.2)()(
)07.0(60.2
)()(75.0
)()(
)()(
22
22
1122
AxBx
AxBx
BxAxAxBx
vv
vv
vv
vve( )
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
37/39
Solution 4.7:(continued)
Solving the two simultaneous equations,
smv
smsmv
Bx
Ax
/22.1)(
/26.1/26.1)(
2
2
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
38/39
Solution 4.7:(continued)
Conservation of y Momentum.The momentum of each diskis conservedin the ydirection (plane of contact), since the disks aresmooth and therefore noexternal impulse acts in
this direction.
smsmv
vmvmsmv
vmvm
By
ByBByB
Ay
AyAAyA
/707.0/707.0)(
)()(/5.1)(
)()(
2
21
2
21( )
( )
-
8/2/2019 Emm 222 - Dynamics Mechanism - Chapter 4
39/39
Tutorial:Problems (12th Edition)
15-2, 15-38, 15-67