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DYNAMICS
CHAPTER 13
Kinetics of Particles:
Energy and Momentum Method
§13.1 Principle of Work and Energy
§13.2 Conservation of Energy
§13.3 Principle of Impulse and Momentum
2008/10/6 CH-13 1
§13.4 Impact
DYNAMICS
(1) Previously, problems dealing with the motion of particles were
solved through the fundamental equation of motion, SF = ma.
(2) Current chapter introduces two additional methods of analysis.
(3) Method of work and energy: directly relates force, mass, (3) Method of work and energy: directly relates force, mass,
velocity and displacement.
(4) Method of impulse and momentum: directly relates force,
mass, velocity, and time.
(5) The advantage of the two methods lie in the fact that they make
the determination of the acceleration unnecessary.
2008/10/6 CH-13 2
the determination of the acceleration unnecessary.
DYNAMICS
§13.1 Principle of Work and Energy
[I](1) Consider a force F that acts on a particle as it moves from point A to point A’. Then dU ≡ the work that the force, F , performs
when the particle undergoes an when the particle undergoes an infinitesimal displacement, dr
≡ F．dr
= F ds cos awhere F & ds are the magnitude of F & dr, respectively,
and a is the angle between F and dr.
--- (13.1-1)
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and a is the angle between F and dr. (2) Work is a scalar quantity, i.e., it has magnitude and sign but
not direction.(3) Dimensions of work are length × force.
In SI systems, the unit of work is 1 Joule (J) = 1 N． m.
DYNAMICS
(4) Consider the relationship dU = F ds cosa, then
(A) dU is positive if a is an acute angle,
(B) dU is negative if a is an obtuse angle,
(C) dU = 0 if F dr , i.e., a = 90o.
(5) For rectangular Cartesian coordinate system, (5) For rectangular Cartesian coordinate system,
F = Fxi + Fyj + Fzk & dr = dxi + dyj + dzk, then
(13.1-1) => dU = F．dr = Fxdx + Fydy + Fzdz --- (13.1-2)
2
1
A
A
21
d
ntdisplaceme finite a during force a of Work U)6(
rF
2008/10/6 CH-13 4
1A
2
1
s
s)ds(Fcos
2
1
A
A zyx dz)FdyFdx(F
--- (13.1-3)2
1
s
s tdsF
DYNAMICS
[II](1) When a particle moving in a
straight line from A1 to A2 and is
acted upon by a constant force F,
then the work done by F is
U = (F cos a)Dx --- (13.1-4) U1→2 = (F cos a)Dx --- (13.1-4)
where a = the angle between F and the direction of motion;
& Dx = displacement from A1 to A2.
(2)(A) If the force acting on the particle is its weight only , then
dU = –W dy
2yWyWyWdyU
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2
1y 2121 WyWyWdyU
= –W(y2 –y1) = –WΔy=> Work of the weight is equal to product
of weight W and vertical displacement Dy.
(13.1-5)
DYNAMICS
(B) ＊ When the particle moves down=> Dy < 0 => U1→2 > 0＊ When the particle moves up=> Dy > 0 => U1→2 < 01→2 ＊ When the particle moves horizontal=> Dy = 0 => U1→2 = 0
(3)(A) Consider a body A attached to a fixed point B by a spring. Itis assumed that the spring is undeformedwhen the body is at A0. If the deflection of the spring measured from the position A is x, then the magnitude of force F
U1→2 = –WΔy
2008/10/6 CH-13 6
A0 is x, then the magnitude of force Fexerted by the spring on body is
F = kx --- (13.1-6)where k is the spring constant (N/m or kN/m)
DYNAMICS
(B) Then, the work of the force exerted by the
spring during a finite displacement of the
body from A1(x = x1) to A2(x = x2)
=> dU = –F dx = –kx dx 11 22x2 7)-(13.1--- kx2
1kx
2
1kxdxU 2
221
x
x21
2
1
(C) ＊ When the spring is returning to its undeformed position
=> x1 > x2 => U1→2 > 0＊ When the spring is moving away from its undeformed position
=> x1 < x2 => U1→2 < 0(E) Work of the force exerted by the spring is equal
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(E) Work of the force exerted by the spring is equal to negative of area under curve of F plotted against x,
U1→2 = –(F1 + F2) Δx /2 --- (13.1-8)
DYNAMICS
(4)(A) Assume a particle M occupies a fixed position O & the other particle m moves along the path shown. Let F be the gravitational force between M & m, then
F = F = GMm/r2 --- (12.3-7)
(B) U1→2 = the work of F exerted on m during
a finite displacement of the particle
m from A1 (r = r1) to A2 (r = r2)
2
1
A
ArdF r
A
A r2edre
r
GMm2
1
GMm GMmGMm
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2
1
r
r 2dr
r
GMm
12 r
GMm
r
GMm
9)-(13.1--- r
GMm
r
GMm
12
DYNAMICS
(C) ＊If m moves toward to the M
=> r1 > r2 => U1→2 > 0＊If m moves away from the M
=> r1 < r2 => U1→2 < 0(5) Forces which do not do work (5) Forces which do not do work
(ds = 0 or cos a = 0):(A) Reaction at frictionless pin supporting
rotating body;(B) Reaction at frictionless surface when body
in contact moves along surface;12
21 r
GMm
r
GMmU
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(C) Reaction at a roller moving along its track; and(D) Weight of a body when its center of gravity moves
horizontally.
(6) The work done by friction forces is always negative.
DYNAMICS
[III](1) Consider a particle of mass m acted upon
by force F
dt
dvmmaF tt
ds
dvmv
dt
ds
ds
dvm
dsmv
dtdsm
mvdvdsFt 21
22
v
v
s
s t mv2
1mv
2
1vdvmdsF
2
1
2
1
assuming m is a constant
=> U1→2 = T2 –T1 --- (13.1-10) where T ≡kinetic energy of the particle = mv2/2--(13.1-11)
=> The work of the force F is equal to the change in kinetic
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=> The work of the force F is equal to the change in kineticenergy of the particle.
=> T1 + U1→2 = T2 --- (13.1-12) => Principle of work and kinetic energy=> The principle relates the forces & the velocities directly.
DYNAMICS
(2) When several forces acted on the particle, the expression U1→2 in(13.1-10) & (13.1-12) represents the total work of the forcesacting on the particle; it is obtained by adding algebraically the work of the various forces.
(3)(A) Units of work and kinetic energy are the same:(3)(A) Units of work and kinetic energy are the same:T = mv2/2 = kg(m/s)2 = (kg． m/s2)m = N． m = J
(B) The kinetic energy of a particle moving with a speed v represents the work which must be
done to bring the particle from rest to the speed v.
(4) The pendulum OA consisting a bob of
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(4) The pendulum OA consisting a bob of weight W attached to a cord of length l. To determine velocity of pendulum bob at A2 by using work & kinetic energy.
DYNAMICS
=> T1 + U1→2 = T2
=> Tension P acts normal to path and
does no work.
=> 0 + mgl = mv22/2
=> v2 = (2gl)1/2
=> The advantages of the principle of work & kinetic energy:
(A) Velocity can be found without determining expression for
acceleration and integrating.
(B) All quantities in the equation are scalars and can be added
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(B) All quantities in the equation are scalars and can be added
algebraically.
(C) Forces which do no work are eliminated from the problem.
DYNAMICS
=> The disadvantages of the principle of work & kinetic energy:
(A) Cannot be applied to directly determine the acceleration of
the pendulum bob.
(B) Calculating the tension in the cord requires supplementing
the method of work and energy with an application of the method of work and energy with an application of
Newton’s second law.
nmamgP
l
22v
m
l2g
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l
l2gmmgP
3mg
DYNAMICS
[IV](1) Power ≡ time rate at which work is done
t
Ulim
0t dt
dU
dt
drF 13)-(13.1--- vF
dt
drF
dt dt
(2)(A) Dimensions of power are work/time or force × velocity.
(B) Units for power are
1 W (watt) = 1 J/s = 1 (N．m)/s = 1 N． (m/s)
koutput wor
)3(
efficiency mechanical
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input power
output power
input work
koutput wor
--- (13.1-14)
DYNAMICS
[V](1) Example 13.1-1
A 2000 kg automobile is driven down a 50 incline at a speed
of 100 km/h when the brakes are applied, causing a constant
total braking force (applied by the road on the tires) of 7 kN.
Determine the distance traveled by the automobile as it Determine the distance traveled by the automobile as it
comes to a stop.
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DYNAMICS
Solution:
mg = (2000)(9.81) = 19620 N = 19.62 kN
m/s78.27h
km100 v: 1 1Position
)(27.78)2000(1
mv1
T 22
kJ 771.73J771730
)(27.78)2000(2
1mv
2
1T 22
11
0T 0 v: 22 2Position
kN)x(5.29)x 5n (19.62)(si)x (7U 021
:Energyand WorkofPrinciple
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T1 + U1→2 = T2
=> 771.73 –5.29x = 0
=> x = 145.9 m
:Energyand WorkofPrinciple
DYNAMICS
(2) Example 13.1-2
Two blocks are joined by an inextensible cable as shown. If the
system is released from rest, determine the velocity of block A
after it has moved 2 m. Assume that the coefficient of kinetic
friction between block A and the plane is m = 0.25 and that thefriction between block A and the plane is mk = 0.25 and that the
pulley is weightless and frictionless.
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DYNAMICS
Solution:
For block A:
=> mA = 200 kg => WA = (200)(9.81) = 1962 N
=> FA = mk NA = mkWA = 0.25(1962) = 490 N
=> T + U = T => 0 + F (2) –F (2) = m v2/2=> T1 + U1→2 = T2 => 0 + FC(2) –FA(2) = mAv2/2
=> FC(2) –(490)(2) = (200)v2/2 --- (a)
For block B:
=> mB = 300 kg => WB = (300)(9.81) = 2940 N
=> T1 + U1→2 = T2 => 0 + WB(2) –FC(2) = mBv2/2
=> (2940)(2) –F (2) = (300)v2/2 --- (b)
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=> (2940)(2) –FC(2) = (300)v2/2 --- (b)
(a) & (b) => 2 unknowns for 2 equations
=> v = 4.43 m/s.
0 + [–(490)(2) + (2940)(2)] = (200 + 300)v2/2
DYNAMICS
(3) Example 13.1-3
A spring is used to stop a 60-kg package which is sliding on a
horizontal surface. The spring has a constant k = 20 kN/m and
is held by cables so that it is initially compressed 120 mm.
Knowing that the package has a velocity of 2.5 m/s in the Knowing that the package has a velocity of 2.5 m/s in the
position shown and that the maximum additional deflection of
the spring is 40 mm, determine
(a) the coefficient of kinetic
friction between the package and
the surface, (b) the velocity of the
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the surface, (b) the velocity of the
package as it passes again through
position shown.
DYNAMICS
Solution:
a. Motion from Position 1 to Position 2
Kinetic Energy Position 1: v1 == 2.5 m/s
187.5JmN187.5m/s)kg)(2.5(602
1mv
2
1T 22
11
Position 2: (maximum spring deflection): v2 = 0 => T2 = 0
Work
Frictional Force F. One has
187.5JmN187.5m/s)kg)(2.5(602
mv2
T 11
k2
k
kkk
(588.6N))1m/s(60kg)(9.8
mgWNF
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The work of F is negative and equal tokk (588.6N))1m/s(60kg)(9.8
kkf21 (377J)0.040m)(0.600m(588.6N)Fx)(U
DYNAMICS
Spring Force P. The variable force P exerted by the spring does an amount of negative work equals to the area under the force-deflection curve of the spring force. One has
N2400m)0kN/m)(0.12(20kxP omin
N0023m)0.04m0kN/m)(0.12(20)xk(xP
112.0J)3200)(0.04(24002
1x)P(P
2
1)(U maxmine21
OR (U1→2)e = kx1
2/2 –kx22/2
= (1/2)(20x103) (0.1202 –0. 1602)
N0023m)0.04m0kN/m)(0.12(20)xk(xP omax
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112.0J(377J))(U)(U work totalU ke21f2121
= (1/2)(20x103) (0.1202 –0. 1602) = –112.0 J
DYNAMICS
Principle of Work and Energy
T1 + U1→2 = T2 => 187.5 –(377) mk –112.0 = 0 => mk = 0.2
b. Motion from Position 2 to Position 3
Kinetic Energy. Position 2: v2 = 0 => T2 = 0
Position 3: 11Position 3: 23
233 (60)v
2
1mv
2
1T
Work. Since the distances involved are the same, the numerical values of the work of thefriction force F and of the spring force P are the same as above. However, while the work of F is still negative, the work of P is now positive.
2008/10/6 CH-13 22
of F is still negative, the work of P is now positive.
U2→3 = –(377) mk + 112.0 = –75.5 + 112.0 = 36.5 J
Principle of Work and Energy
T2 + U2→3 = T3 => 0 + 36.5 = (60)v23/2 => v3 = 1.103 m/s ←
DYNAMICS
(4) Example 13.1-4
A 1000 kg car starts from rest at point 1 and moves without
friction down the track shown. (a) Determine the force exerted
by the track on the car at point 2, where the radius of curvature of
the track is 6 m (b) Determine the minimum safe value of thethe track is 6 m (b) Determine the minimum safe value of the
radius of curvature at point 3.
2008/10/6 CH-13 23
DYNAMICS
Solution:
a.Force exerted by the track at Point 2.
Kinetic Energy
T1 = 0 & T2 = mv22/2T1 = 0 & T2 = mv2 /2
Work. The only force which does
work is the weight W.
The work of the weight is
U1→2 = + W(12 m) = mg(12 m)
Principle of Work and Energy
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Principle of Work and Energy
T1 + U1→2 = T2 => 0 + mg(12) = mv22/2
=> v22 = 24g = 24(9.81) => v2 = 15.3 m/s
DYNAMICS
Newton's Second Law at Point 2:
nn maF
nmaNmg 2v 24g
22v
m mg46
24gm
=> N = 5 mg = 5 (1000 kg)(9.81 m/s2) = 49.1 kN ↑
2008/10/6 CH-13 25
DYNAMICS
b. Minimum Value of r at Point 3.
Principle of Work and Energy.
T1 + U1→3 = T3 => 0 + mg(12 –4.5) = mv32/2
=> v23 = 15g = 15(9.81) => v3 = 12.1 m/s
Newton's Second Law at Point 3. Newton's Second Law at Point 3.
The minimum safe value of r occurs when N = 0.
W –N = man
If N = 0, an = (an)max = (an)critical
If an > (an)critical, then N < 0
the car flies away
2008/10/6 CH-13 26
the car flies away
nnmaF
23v
mmg
15gm => r = 15 m
DYNAMICS
(5) Example 13.1-5
The dumbwaiter D and its load have a combined mass of 300 kg,
while the counterweight C has a mass of 400 kg.
Determine the power delivered by the
electric motor M when the dumbwaiter electric motor M when the dumbwaiter
(a) is moving up at a constant speed of
2.5 m/s, (b) has an instantaneous velocity
of 2.5 m/s and an acceleration of 0.75 m/s2,
both directed upward.
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DYNAMICS
Solution:
a. Uniform motion
Free Body C => +↓ ΣFy = 0 => (400)(9.81) –2T = 0
=> T = 1962 N
Free Body D => +↑ΣF = 0 => F + T –(300)(9.81) = 0 Free Body D => +↑ΣFy = 0 => F + T –(300)(9.81) = 0
=> F = (300)(9.81) –1962 = 981 N
(13.1-13) => Power = FvD = (981 N)(2.5 m/s) = 2453 J/s
= 2453 W
b. Accelerated motion
2x + x = constant => 2a + a = 0
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2xC + xD = constant => 2aC + aD = 0
=> aD = 0.75 m/s2↑ aC = –1/2aD = 0.375 m/s2↓
Free Body C => +↓ΣFy = mCaC
=> 3924 –2T = 400(0.375) => T = 1887 N
DYNAMICS
Free Body D => +↑ΣFy = mDaD
=> F + T –2943 = 300(0.75)
=> F + 1887 –2943 = 225
=> F = 1281 N
(13.1-13) => Power = Fv(13.1-13) => Power = FvD
= (1281 N)(2.5 m/s) = 3203 J/s
= 3203 W
2008/10/6 CH-13 29
DYNAMICS
§13.2 Conservation of Energy
[I](1) A force, F, is said to be conservative (i.e. a conservative force ) if the work done, U1→2, by the force on a particle A as it moves from A to A is independent of moves from A1 to A2 is independent of the path the particle takes but depends only on the initial and final positions of A . For example : gravitational force , spring force , universal
gravitational force.(2)(A) Define V(x, y, z) ≡ potential energy (or potential
2008/10/6 CH-13 30
function) of conservative force F= the negative value of the work done by the conservative
force F on the particle that moves from a datum position
(i.e. , potential energy = 0) to a required position .
DYNAMICS
(B) The potential energy V is independent of the actual path followed; it depends only upon the initial &final position of theparticle.
(C) The potential energy is expressed in the same unit as work.(3)(A) Let “0” be defined as the datum position &”1”&”2” be (3)(A) Let “0” be defined as the datum position &”1”&”2” be
referred to as the another two arbitrary positions. r is the position vector & F is the acting force. Then
=> U1→2 ≡ the work done by force from position 1 to position 2= F．(r2 –r1)
V1 ≡ the potential energy at position 1 = –F．(r1 –r0) = –U0→1
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= –F．(r1 –r0) = –U0→1
V2 ≡ the potential energy at position 2 = –F．(r2 –r0) = –U0→2
=> V1 –V2 = (–F．r1 + F．r0) –(–F．r2 + F．r0) = F．r2 –F．r1 = U1→2 --- (13.2-1)
DYNAMICS
=> U1→2 = V1 –V2 --- (13.2-1)
=> The work done by F from position 1 to position 2 equals to the
potential energy at position 1 minus the potential energy at
position 2.
(B) The datum position from which potential energy is measured(B) The datum position from which potential energy is measured
merely affects an additive constant in the potential energy.
However , an additive constant in the potential energy is
irrelevant, since , as shown in eq.(13.2-1), only change of
potential energy are significant.
(4)(A) A force is said to be a nonconservative force if the work
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(4)(A) A force is said to be a nonconservative force if the work
done by the force on a particle depends on the path that the
particle follows.
For example : friction force .
DYNAMICS
(B) In general , nonconservative force may be called dissipative
force.
(5) The total work performed by a conservative
force is zero when the system moves around
a closed path and return to its initial position .a closed path and return to its initial position .
=> For any conservative force applied on a closed path,
F．dr = 0 --- (13.2-2)
(6) Let dU = the work corresponding to the displacement dr from
A: (x, y, z) to A’: (x + dx, y + dy, z + dz)
= V(x, y, z) –V(x + dx, y + dy, z + dz)
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= V(x, y, z) –V(x + dx, y + dy, z + dz)
=> dU = V –(V + dV)
=> dU = –dV(x, y, z) --- (13.2-3)
=> dU = F．dr = (Fxi + Fyj + Fzk)．(dxi + dyj + dzk)
DYNAMICS
dzz
Vdy
y
Vdx
x
VdzFdyFdxF zyx
4)-(13.2--- z
VF
y
VF
x
VF zyx
VVV
5)-(13.2--- V)function scalar of(gradient Vgrad
z
V
y
V
x
VFFF zyx
kjikjiF
(7)(A) F is conservative
The corresponding potential energy V exists.
(B) If the conservative force F is given, the corresponding
if and only if
2008/10/6 CH-13 34
(B) If the conservative force F is given, the corresponding
potential energy V can be obtained from (13.2-4), and vice
versa.
DYNAMICS
[II](1)(A) It is recalled from (13.1-5) that the work
of the force of gravity W is
U1→2 = Wy1 –Wy2 --- (13.1-5)
=> Work is independent of path followed.=> Work is independent of path followed.
=> The force of gravity is a conservative force.
=> Corresponding potential energy exists
=> Vg≡ potential energy of the body with respect to force of
gravity.
= Wy --- (13.2-6)
2008/10/6 CH-13 35
= Wy --- (13.2-6)
=> U1→2 = (Vg)1 –(Vg)2 --- (13.2-7)
(B) The datum from which the elevation y is measured
can be chosen arbitrarily.
DYNAMICS
(2)(A) Let F be the universal gravitational forcebetween two particles m & M and it is recalled from (13.1-9) that the work
of the force F is
9)-(13.1--- GMmGMm
U
=> Work is independent of path followed.=> The universal gravitational force is a conservative force.=> Corresponding potential energy exists=> Let the positions at which r = ∞ is the datum position ,
then the potential energy V at position r corresponding to
9)-(13.1--- r
GMm
r
GMmU
1221
2008/10/6 CH-13 36
then the potential energy Vr at position r corresponding tothe universal gravitational force can be expressed as
8)(13.2r
GMm
r
GMm
r
GMmUV rr
DYNAMICS
=> U1→2 = (Vr)1 –(Vr)2 --- (13.2-9)
(B) The expression (13.2-6) for potential energy of a body with
respect to gravity is only valid when the weight of the body can
be assumed constant.be assumed constant.
(C) When the variation in the force of gravity can not be neglected,
the potential energy corresponding to the force of gravity will be
expressed as
10)-(13.2--- r
WR
r
GMmV
2
g
2008/10/6 CH-13 37
rrg
DYNAMICS
(3) Consider a body attached to a spring, it is recalled from (13.1-7) that the work of the spring force is
7)-(13.1--- kx2
1kx
2
1U 2
22121
=> Work is independent of path followed.=> The spring force is a conservative force.=> Corresponding potential energy exists=> Let the undeformed position (x = 0, i.e. A0) be the datum
position , then the potential energy Ve at position x (measured from the undeformed position ) corresponding to the spring force can be expressed as
22
2008/10/6 CH-13 38
force can be expressed as
=> U1→2 = (Ve)1 –(Ve)2 --- (13.2-12)
11)-(13.2--- kx2
1kx0)k(x
2
1UV 222
AAe 0
DYNAMICS
[III](1)(A) Equations (13.1-10) & (13.2-1)
=> U1→2 = T2 –T1 = V1 –V2
=> T1 + V1 = T2 + V2 --- (13.2-13)
=> When a particle moves under the action of conservative
forces, the sum of kinetic energy & of the potential forces, the sum of kinetic energy & of the potential
energy of the particle remains constant.
(B) T+V ≡ E ≡ the total mechanical energy
Eq. (13.2-13) is called the principle of conservation
of mechanical energy.
(2)(A) Consider a pendulum is
2008/10/6 CH-13 39
(2)(A) Consider a pendulum is
released with zero velocity
from A1 and allowed to swing
in a vertical plane.
DYNAMICS
(B) At A1 T1 = 0, V1 = Wl, T1 + V1 = Wl
A2 v22 = 2gl (see p.12 of § 13.1)
T2 = mv22/2 = (W/g)(2gl)/2 = Wl
V2 = 0, T2 + V2 = Wl
A T = 0, V = Wl, T + V = WlA3 T3 = 0, V3 = Wl, T3 + V3 = Wl
=> T1 + V1 = T2 + V2 = T3 + V3 = constant
(C) The (T + V) of the pendulum remains constant & VA = VA’
=> TA = TA’ speed at A = speed at A’
=> The speed of the pendulum will have the same value at any two
points located on the same level.
2008/10/6 CH-13 40
points located on the same level.
=> Similar, the particle which slides in a
vertical plane along a frictionless track,
will have the same speed at A, A’ & A’’.
DYNAMICS
(3)(A) Since the friction forces are nonconservative forces , it
follows that when a mechanical system involves friction , its
total mechanical energy does not remain constant but
decreases . The energy of the system , however, is not lost , it
is transformed into heat , and the sum of the mechanical
energy and of the thermal energy of the system remains
constant .
(B) Other forms of energy can also be involves in a system. For
instance , a generator converts mechanical energy into electric
2008/10/6 CH-13 41
energy, a gasoline engine converts chemical energy into
mechanical energy ; a nuclear reactor converts mass into
thermal energy .
DYNAMICS
(C) If all forms of energy are considered , the energy of any system
can be considered as constant & the principle of conservation of
energy remains valid under all conditions.
(4)(A) In general ,the work U1→2 is the sum of the work Uc1→2
preformed by conservative forces & the work Uncpreformed by conservative forces & the work Unc1→2
performed by nonconservative force ; that is
U1→2 = Uc1→2 + Unc
1→2 --- (13.2-14)
(B) (13.1-10) U1→2 = T2 –T1
(13.2-1) Uc1→2 = V1 –V2
2008/10/6 CH-13 42
1→2 1 2
=> T1 + V1 + Unc1→2 = T2 + V2 --- (13.2-15)
=> If Unc1→2 < 0 (T2 + V2 ) < (T1 + V1 )
If Unc1→2 > 0 (T2 + V2 ) > (T1 + V1 )
DYNAMICS
(5)(A) When a particle moves under a conservative central force, both the principle of conservation of angular momentum
rmv sinf = r0mv0 sinf0 --- (12.3-2)and the principle of conservation of energyand the principle of conservation of energy
16)-(13.2--- r
GMmmv
2
1
r
GMmmv
2
1
13)-(13.2--- VTVT
2
0
20
00
(B) Given r, the equations may be solved
for v and f.
(C) At minimum and maximum r, f = 90o.
2008/10/6 CH-13 43
(C) At minimum and maximum r, f = 90o.
Given the launch conditions, the
equations may be solved for rmin, rmax,
vmin, and vmax.
DYNAMICS
[IV](1) Example 13.2-1
A 9 kg collar slides without friction along a vertical rod as
shown. The spring attached to the collar has an undeformed
length of 100 mm and a constant of 540 N/m. If the collar is
released from rest in position 1, determine its velocity after it released from rest in position 1, determine its velocity after it
has moved 150 mm to
position 2.
2008/10/6 CH-13 44
DYNAMICS
Solution:Position 1 Kinetic Energy: T1 = 0Potential Energy: V1g = 0 & V1e = kx1
2/2 & V1e = kx12/2
= (540 N/m)(0.2 m –0.1 m)2/2 = 2.7 JPosition 2 Kinetic Energy: T2 = mv2
2/2 = 9v22/2 = 4.5v2
2
Potential Energy: V2g = –Wy = –(9 × 9.81 N)(0.15 m) = –13.3 J
& V2e = kx22/2 = (540 N/m)(0.25 m –0.1 m)2/2 = 6.1 J
2008/10/6 CH-13 45
Conservation of EnergyT1 + (V1e + V1e) = T2 + (V2g + V2e) 0 + (0 + 2.7) = 4.5v2
2 + (–13.3 + 6.1) v2 = ±1.48 m/s v2 = 1.48 m/s↓
DYNAMICS
(2) Example 13.2-2
The 200 g pellet is pushed against the
spring at A and released from rest.
Neglecting friction, determine the
smallest deflection of the spring for smallest deflection of the spring for
which the pellet will travel around the
loop ABCDE and remain at all times in contact with the loop.
NOTE
W+N = man
N = 0, a = (a )
2008/10/6 CH-13 46
N = 0, an = (an)min
If an < (an)min
=> N < 0
=> Pullet drop down
DYNAMICS
Solution:ma W maΣF nnn
/r vga mamg 2Dnn
222D /sm5.89)(0.6)(9.81rgv
Position 1 Position 1 Kinetic Energy: T1 = 0Potential Energy: V1g = 0 & V1e = kx1
2/2 = (540 N/m)(x m)2/2 = 270x2
Position 2 Kinetic Energy: T2 = mv2
2/2 = mvD2/2 = (0.2)(5.89)/2 = 0.589 J
2008/10/6 CH-13 47
2 2 D
Potential Energy: V2g = Wy = (0.2 × 9.81)(1.2 m) = 2.35 J& V2e = 0Conservation of Energy0 + (0 + 270x2) = 0.589 + (2.35 +0) => x = 0.104 m = 104 mm
DYNAMICS
(3) Example 13.2-3
A sphere of mass m = 0.6 kg is attached to an elastic cord of
constant k = 100 N/m, which is undeformed when the sphere is
located at the origin O. Knowing that the sphere may slide
without friction on the horizontal surface and that in the position without friction on the horizontal surface and that in the position
shown its velocity vA has a magnitude of 20 m/s, determine (a)
the maximum and minimum distances from the sphere to the
origin O, (b) the corresponding values of its speed.
2008/10/6 CH-13 48
DYNAMICS
Solution:
Central force motion:
=> rAmvA sin 60° = rmmvm
=> (0.5)(0.6)(20) sin 60° = rm(0.6)vm
=> v = 8.66/r --- (a)=> vm = 8.66/rm --- (a)
Conservation of energy:
At point A: TA = mvA2 /2= (1/2)(0.6)(20)2 = 120 J
VA = krA2/2 = (1/2)(100)(0.5 m)2 = 12.5 J
At point B: TB = mvm2/2 = (1/2)(0.6)vm
2 = 0.3vm2
V = kr 2/2 = (1/2)(100)r 2 = 50r 2
2008/10/6 CH-13 49
VB = krm2/2 = (1/2)(100)rm
2 = 50rm2
=> TA + VA = TB + VB => 120 + 12.5 = 0.3vm2 + 50rm
2 --- (b)
=> (a) & (b) => rm2 = 2.468 or 0.1824
=> rm = 1.571 m, rm’ = 0.427 m & vm = 5.51 m/s, v’m = 20.3 m/s
DYNAMICS
(4) Example 13.2-4
A satellite is launched in a direction parallel to the surface of the
earth with a velocity of 36 900 km/h from an altitude of 500 km.
Determine (a) the maximum altitude reached by the satellite, (b)
the maximum allowable error in the direction of launching if thethe maximum allowable error in the direction of launching if the
satellite is to go into orbit and come no closer than 200 km to the
surface of the earth.
2008/10/6 CH-13 50
DYNAMICS
Solution:
(a) Planetary elliptical motion:
21
20
A'A'AA
r
GMmmv
2
1
r
GMmmv
2
1
VTVT
11
o0 r
mv2r
mv2
1
o0111oo r
rv v mvrmvr&
1
0
02
1
202
0 r
r1
r
GM
r
r1v
2
1
r
2008/10/6 CH-13 51
2001
0
vr
2GM
r
r1
1vr
2GMr
r
200
01
DYNAMICS
=> r0 = 6370 km + 500 km = 6870 km = 6.87 × 106 m
v0 = 36 900 km/h = (36.9 × 106 m)/(3.6 × 103 s) = 10.25 × 103 m/s
GM = gR2 = (9.81 m/s2)(6.37 × 106 m)2 = 398 × 1012 m3/s2
=> r1 = 66.8 × 106 m
=> Maximum altitude = 66.8 × 106 m –6.37 × 106 m => Maximum altitude = 66.8 × 106 m –6.37 × 106 m
= 60.4 ×106 = 60 400 km
(b)
maxmin000
min
2max
0
20
mvrsinmvr
r
GMmmv
2
1
r
GMmmv
2
1
=> r = 6370 km + 200 km = 6570 km = 6.57 × 106 m
2008/10/6 CH-13 52
=> r min = 6370 km + 200 km = 6570 km = 6.57 × 106 m
=> sin f0 = 0.9801
=> f0 = 90° ± 11.5°
=> Allowable error = ±11.5°
DYNAMICS
§13.3 Principle of Impulse and Momentum
[I](1) Consider a particle of mass m acted upon by a force F , then
Newton’s second law yields
)(mdt
d
dt
dmm v
vaF
Assume m is constant
)(mdtdt
mm vaF
)d(mdt vF
where Imp1→2 ≡ The impulse of the force F in the time
interval (t2 –t1) = a vector
1)-(13.3--- mmdt2
1
t
t 1221 vvFImp
2008/10/6 CH-13 53
=> The impulse of the force that acts on a particle in the time
interval t2 –t1 is equal to the change of momentum of the
particle in the same time interval .
=> The principle of impulse and momentum.
DYNAMICS
(2) Eq. (13.3-1) is valid only in Newtonian reference frame.
(3) In SI units,
the unit of impulse is N‧s,
& the unit of momentum is kg‧m‧s–1
Eq. (13.3-1) => N‧s = kg‧m‧s–1
1)-(13.3--- mmdt2
1
t
t 1221 vvFImp
Eq. (13.3-1) => N‧s = kg‧m‧s–1
(4) It is noted that impulse is a quantity that measures the change of
momentum of the particle.
When one says that an impulse is given to a particle, he/she
means that it is a result of an external force applied to the
particle and , in effect , that momentum is transferred from an
2008/10/6 CH-13 54
particle and , in effect , that momentum is transferred from an
external agent to the particle.
Eq. (13.3-1) may be rewritten as
mv1 + Imp1→2 = mv2 --- (13.3-2)
DYNAMICS
(5) By resolving F & v into Cartesian Rectangular components, the
eq. (13.3-2) can be written as
t
2x
t
t x1x
)(mvdtF)(mv
)(mvdtF)(mv
2
2
1
--- (13.3-3)
2z
t
t z1z
2yt y1y
)(mvdtF)(mv
)(mvdtF)(mv
2
1
2
1
--- (13.3-3)
(6)Note:(A) If F is a constant , i.e., constant magnitude & direction , then the
impulse is represented by the vector F(t2 –t1) ,which has the
2008/10/6 CH-13 55
impulse is represented by the vector F(t2 –t1) ,which has the same direction as F.
(B) When several forces are on a particle, the impulse of each of the force must be consider:
=> mv1 + ΣImp1→2 = mv2 --- (13.3-4)
DYNAMICS
(C) When a problem involves two particles or more , each particle
can be considered separately and eq. (13.3-4) can be written for
each particle.
One can also add vectorically the momenta of all the particles &
the impulses of all the forces involved. the impulses of all the forces involved.
=> Σmv1 + ΣImp1→2 = Σmv2 --- (13.3-5)
(D) The impulse of the forces of action & reaction cancel out & only
the impulses of the external forces need be considered.
[II](1) If no external force is exerted on the particles or more
generally, if the sum of the external forces is zero,
2008/10/6 CH-13 56
generally, if the sum of the external forces is zero,
=>ΣImp1→2 = 0
=>Σmv1 = Σmv2 --- (13.3-6)
=> The total momentum of the particles is conserved
DYNAMICS
(2) Consider, for example, two boats, of mass mA & mB, initially at
rest, which are being pulled together,
If the resistance of water is neglected, the only external forces on theboats are their weights& buoyant forces exerted on them. Sincethese forces are balanced , => Σmv1 = Σmv2
=> 0 = mAv’A+ mBv’B
2008/10/6 CH-13 57
where v’A & v’B represent the velocities of the boats after a finitetime interval.
=> The boats move in opposite directions (toward each other ) withvelocities inversely proportional to their masses.
DYNAMICS
[III](1) A force acting on a particle during a very short time
interval that is large enough to produce a definite change in
momentum is called an impulsive force and the resulting
motion is called an impulsive motion.
For example , a baseball be struck by a bat.For example , a baseball be struck by a bat.
(2) When impulsive forces act on a particle, the force SF can be
assumed as constant during the very short time interval, Dt =
t2 –t1, then eq.(13.1-1) can be written as
mv1 +ΣFΔt = mv2 --- (13.3-7)
2008/10/6 CH-13 58
DYNAMICS
(3) Note:
(A) In eq. (13.3-7), Any force which is not an impulsive force may be neglected, since the corresponding impulse ΣFΔt is very small.
(B) Non impulsive force: (B) Non impulsive force:
• the weight of the body;
• the force exerted by a spring;
• any other force which is known to be smaller compared with
an impulsive force.
2008/10/6 CH-13 59
(C) Unknown reactions may or may not be impulsive, their impulses
should therefore be included in eq.(13.3-7) as long as they have
not been proved negligible.
DYNAMICS
[IV] The differences between the principle of impulse & momentum
(13.3-2) and the principle of kinetic energy (13.1-12)
Principle of impulse & momentum Principle of kinetic energy
2)-(13.3--- mdtm2t
21 vFv T1 + U1→2 = T2 --- (13.1-12)
(1) Vector quantities Scalar quantities
(2) Relates force& velocity at Relates force& velocity at
two different times two different positions
2)-(13.3--- mdtm1t
21 vFv T1 + U1→2 = T2 --- (13.1-12)
2008/10/6 CH-13 60
DYNAMICS
[V](1) Example 13.3-1
A 120 g baseball is pitched with a velocity of 24 m/s toward
a batter. After the ball is hit by the bat B, it has a velocity of
36 m/s in the direction shown. If the bat and ball are in
contact 0.015 s, determine the average impulsive force contact 0.015 s, determine the average impulsive force
exerted on the ball during the impact.
2008/10/6 CH-13 61
DYNAMICS
Solution:
Impulsive motion (m = 120 g = 0.12 kg)
=> The weight of the ball can be neglected
=> mv1 + ΣImp1→2 = mv2
At x-direction:At x-direction:
=> –mv1 + FxΔt = mv2 cos 40°
=> –(0.12)(24) + Fx(0.015) = (0.12)(36)cos 40°
=> Fx = + 412.6 N
At y-direction:
=> 0 + F Δt = mv sin 40°
2008/10/6 CH-13 62
=> 0 + FyΔt = mv2 sin 40°
=> Fy(0.015) = (0.12)(36) sin 40°
=> Fy = + 185.1 N
F = 412.6 i + 185.1 j N
DYNAMICS
(2) Example 13.3-2
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 ofimpulse exerted by the cart on the package, (c) the fraction of
the initial energy lost in the impact.
2008/10/6 CH-13 63
DYNAMICS
Solution:
(a) For package and cart
mPv1 + ΣImp1→2 = (mP + mC)v2
=> x components
2008/10/6 CH-13 64
=> mPv1 cos 30° + 0 = (mP + mC)v2
=> (10 kg)(3 m/s) cos 30° = (10 kg + 25 kg)v2
=> v2 = 0.742 m/s →
DYNAMICS
(b) For package
x components mPv1 cos30° + FxΔt = mPv2
=> (10)(3) cos 30° + FxΔt = (10)(0.742)
=> FxΔt = –18.56 N．s
y components m v sin 30° + F Δt = 0
2008/10/6 CH-13 65
y components mPv1 sin 30° + FyΔt = 0
=> –(10)(3) sin 30° + FyΔt = 0
=> FyΔt = +15 N．s
FΔt = –18.56 i + 15 j N．s
DYNAMICS
(c) Fraction of energy lost
45Js)(10kg)(3m/2
1vm
2
1T 22
1P1
9.63J2m/s)25kg)(0.74(10kg2
1)vm(m
2
1T 22
2CP2 22
78.6%45J
9.63J45J
T
TT
1
21
2008/10/6 CH-13 66
DYNAMICS
§13.4 Impact[I](1) Impact: Collision between two bodies which occurs during a
small time interval and during which the bodies exert large forces on each other.
(2) Line of Impact: Common normal to the surfaces in contact during impact.during impact.
(3) Central Impact: Impact for which the mass centers of thetwo bodies lie on the line of impact; otherwise, it is aneccentric impact.
(4) Direct Impact: Impact for which the velocities of the twobodies are directed along the line of impact.
(5) Oblique Impact: Impact for
2008/10/6 CH-13 67
(5) Oblique Impact: Impact for which one or both of the bodies move along a line other than the line of impact.
DYNAMICS
[II](1) Direct central impact
=> Bodies moving in the same straight
line, vA > vB.
=> Upon impact the bodies undergo a=> Upon impact the bodies undergo aperiod of deformation, at the end of
which, they are in contact and moving
at a common velocity.
=> A period of restitution follows during
which the bodies either regain their
2008/10/6 CH-13 68
which the bodies either regain their
original shape or remain permanently
deformed.
DYNAMICS
(2) If one wishes to determine the final velocities of the two bodies,
the total momentum of the two body system is preserved.
=> mAvA + mBvB = mAv’A + mBv’B
=> mAvA + mBvB = mAv’A + mBv’B --- (13.4-1)
since the directions of v’ & v’ are known for direct central since the directions of v’A & v’B are known for direct central
impact.
=> There are 2 unknowns, v’A & v’B, a second relation between
the final velocities is required.
(3)(A) Consider the ball A
=> Period of deformation:
2008/10/6 CH-13 69
=> Period of deformation:2)-(13.4--- umPdtvm AAA
DYNAMICS
& Period of restitution:3)-(13.4--- vmRdtum '
AAA
4)-(13.4--- Pdt
Rdte
nrestitutio oft coefficien
5)-(13.4--- uv
vu
A
'A
Note: 0 e 1.
2008/10/6 CH-13 70
Note: 0 e 1.
The value of e depends to a large extent on the two materials
involved, but it also varies considerably with the impact
velocity and the shape and size of the two colliding bodies.
DYNAMICS
(B) A similar analysis of particle B yields
6)-(13.4--- vu
uve
B
'B
(C) Equations (13.4-5) & (13.4-6) yield'''' vvu)(v)v(u
BA
'A
'B
BA
'B
'A
vv
vv
)v(uu)(v
u)(v)v(ue
=> v’B –v’A = e (vA –vB) --- (13.4-7)
(4) The values of v’A & v’B can be solved from (13.4-1) & (13.4-7)
provides the value of e is given.
(5) Perfectly plastic impact, e = 0
2008/10/6 CH-13 71
(5) Perfectly plastic impact, e = 0
=> There is no period of restitution & both particles stay together
after impact & v’A = v’B = v’
=> mAvA + mBvB = (mA + mB)v’ --- (13.4-8)
DYNAMICS
(6) Perfectly elastic impact, e = 1=> v’B –v’A = vA –vB --- (13.4-9)=> The particles move away from each other after impact with the
same relative velocity with which they approached each otherbefore impact. before impact.
=> The impulses received by each particle during the period of deformation and during the period of restitution are equal.
=> (13.4-1) mA(vA –v’A) = mB(v’B –vB)& (13.4-9) vA + v’A = vB + v’B
=> mA(vA –v’A) (vA + v’A) = mB(v’B –vB) (v’B + vB)=> m v 2 –m (v’ )2 = m (v’ )2 –m v 2
2008/10/6 CH-13 72
=> mAvA2 –mA(v’A )2 = mB(v’B )2 –mBvB
2
=> mAvA2/2 + mBvB
2/2 = mA(v’A)2/2 + mB(v’B)2/2 --- (13.4-10)=> In the case of a perfectly elastic impact, the total energy of the
two particles, as well as their total momentum, is conserved.
DYNAMICS
[III](1) Oblique Central Impact
( Both particles moved freely )
=> The velocities of the two colliding particles are
not directed along the line of impact.
=> Final velocities are unknown in magnitude and=> Final velocities are unknown in magnitude and
direction. Four equations are required
=> Assuming that the particles are perfectly
smooth and frictionless
No tangential impulse component, SFtDt = 0
Each particle’s momentum in tangential
2008/10/6 CH-13 73
Each particle’s momentum in tangential
direction is conserved
(vA)t = (v’A)t --- (13.4-11)
(vB)t = (v’B)t --- (13.4-12)
DYNAMICS
=> Normal component of total momentum of the two
particles is conserved.
mA(vA)n + mB(vB)n = mA(v’A)n + mB(v’B)n
(13.4-13)
=> With a derivation similar to that given in the => With a derivation similar to that given in the
derivation of (13.4-7) for direct central impact
yields
Normal components of relative velocities
before and after impact are related by the
coefficient of restitution
2008/10/6 CH-13 74
coefficient of restitution
(v’B)n –(v’A)n = e[(vA)n –(vB)n] --- (13.4-14)
=> Four unknowns, (v’A)t, (v’A)n, (v’B)t, and (v’B)n,
for four equations, (13.4-11) –(13.4-14).
DYNAMICS
(2) Oblique Central Impact( One or both of the colliding bodies is constrained in its motion )
=> Let the block is constrained to move along horizontal surface and assuming no frictionhorizontal surface and assuming no frictionbetween any two faces.
=> Only Impulses from internal forces along the n axis and from external force exerted by horizontal surface and directed along the vertical to the surface are not zero.
2008/10/6 CH-13 75
=> Final velocity of ball unknown in direction and magnitude and final velocity of block unknown in magnitude.
=> Three equations are required.
DYNAMICS
=> Tangential momentum of ball is conserved.
(vB)t = (v’B)t --- (13.4-15)
=> Total horizontal momentum of block and ball
is conserved.
m v + m (v ) = m v’ + m (v’) mAvA + mB(vB)x = mAv’A + mB(v’B)x
= mAv’A + mB[(v’B)ncosq –[(v’B)tsinq](13.4-16)
=> Consider the block
Period of deformation:
umPdt)cosθ(vm
2008/10/6 CH-13 76
umPdt)cosθ(vm AAA
DYNAMICS
& Period of restitution:
AAA v'mRdt)cosθ(um
Pdt
Rdte
nrestitutio oft coefficien
uv
vu 'A
17)-(13.4--- u)(v
)(vu n'An
For the sphere, similar to the derivation of
(13.4-6), it can be shown that
(v’) –(v’) = e[(v ) –(v ) ] --- (13.4-19)
uvA 17)-(13.4---
u)(v nnA
18)-(13.4--- )v(u
u)(ve
nBn
nn'B
2008/10/6 CH-13 77
(v’B)n –(v’A)n = e[(vA)n –(vB)n] --- (13.4-19)
(v’B)n –(v’A)cosq = e[(vA)cosq –(vB)n] --- (13.4-20)
=> Three unknowns, v’A, (v’B)t, and (v’B)n, for three equations,
(13.4-15), (13.4-16) & (13.4-20).
DYNAMICS
Note: Equation (13.4-19) still validates for the case when both particles are constrained in their motions.
[IV](1) Three methods for the analysis of kinetics problems:(A) Direct application of Newton’s second law(B) Method of work and energy(B) Method of work and energy(C) Method of impulse and momentum
(2) Select the method best suited for the problem or part of a problem under consideration.
2008/10/6 CH-13 78
DYNAMICS
[V](1) Example 13.4-1A 20-Mg railroad car moving at a speed of 0.5 m/s to the right collides with a 35-Mg car which is at rest. If after the collision the 35-Mg car is observed to move to the right at a speed of 0.3 m/s, determine the coefficient of restitution between the two cars. (It is assumed that the friction forcesbetween car & railroad can be neglected)
Solution:
mAvA + mBvB = mAv’A + mBv’B
2008/10/6 CH-13 79
mAvA + mBvB = mAv’A + mBv’B
=> (20)(0.5) + (35)(0) = (20)v’A + (35)(0.3)=> v’A = –0.025 m/s v’A = 0.025 m/s ←
65.00.5
0.325
00.5
0.025)(0.3
vv
v'v'e
BA
AB
DYNAMICS
(2) Example 13.4-2A ball is thrown against a frictionless, vertical wall. Immediately before the ball strikes the wall, its velocity has amagnitude v and forms an angle of 30° with the horizontal. Knowing that e = 0.90, determine the magnitude and direction ofthe velocity of the ball as it rebounds from the wall.
Solution:vn = v cos 30° = 0.866v & vt = v sin 30° = 0.500v
=> Conservation of ball’s momentum in t-direction v’t = vt = 0.500v↑
=> Relation corresponding to e (v’) –(v’) = e[(v ) –(v ) ]
2008/10/6 CH-13 80
(v’B)n –(v’A)n = e[(vA)n –(vB)n] 0 –v’n = e(vn –0) v’n = –0.90(0.866v) = –0.779v = 0.779v ←
=> v’ = 0.926v 32.7o
DYNAMICS
(3) Example 13.4-3
The magnitude and direction of the velocities of two identical
frictionless balls before they strike each other are as shown.
Assuming e = 0.90, determine the magnitude and direction of the
velocity of each ball after the impact. velocity of each ball after the impact.
2008/10/6 CH-13 81
DYNAMICS
Solution:
(vA)n = vA cos 30° = +7.8 m/s
(vA)t = vA sin 30° = +4.5 m/s
(vB)n = –vB cos 60° = –6.0 m/s
(v ) = v sin 60° = +10.4 m/s
note
(vB)t = vB sin 60° = +10.4 m/s
=> Conservation of each ball’s momentum in
tangential direction
(v’A)t = (vA)t = 4.5 m/s↑
& (v’B)t = (vB)t = 10.4 m/s↑=> Conservation of total momentum in normal direction
2008/10/6 CH-13 82
=> Conservation of total momentum in normal direction
mA(vA)n + mB(vB)n = mA(v’A)n + mB(v’B)n
m(7.8) + m(–6.0) = m(v’A)n + m(v’B)n
(v’A)n + (v’B)n = 1.8 --- (a)
DYNAMICS
=> Relation corresponding to e
(v’B)n –(v’A)n = e[(vA)n –(vB)n]
(v’B)n –(v’A)n = (0.90)[7.8 –(–6.0)]
(v’B)n –(v’A)n = 12.42 --- (b)
=> (a) & (b)
(v’A)n = –5.3 = 5.3 m/s ←
& (v’B)n = +7.1 = 7.1 m/s →
=> v’A = 6.95 m/s 40.3o
& v’B = 12.6 m/s 55.6o
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& v’B = 12.6 m/s 55.6
DYNAMICS
(4) Example 13.4-4
Ball B is hanging from an inextensible cord BC. An identical
ball A is released from rest when it
is just touching the cord and
acquires a velocity v before acquires a velocity v0 before
striking ball B. Assuming perfectly
elastic impact (e = 1) and no
friction, determine the velocity of
each ball immediately after impact.
2008/10/6 CH-13 84
DYNAMICS
Solution:sinq = r/2r = 0.5 q = 30o
Ball B is constrained to move in a circle of center C=> Velocity v’B after impact must be horizontal=> 3 unknowns: magnitude of v’B, magnitude & direction of v’A.=> 3 unknowns: magnitude of v’B, magnitude & direction of v’A.=> Impulse-momentum principle for ball A mvA + FΔt = mv’A
t-direction:mv0 sin 30° + 0 = m(v’A)t
(v’A)t = 0.5v0 --- (a)=> Impulse-momentum principle for both ball A & ball B
2008/10/6 CH-13 85
=> Impulse-momentum principle for both ball A & ball B mvA + TΔt = mv’A + mv’B
x-direction:0 = m(v’A)t cos 30° –m(v’A)n sin 30° –mv’B
0.5(v’A)n + v’B = 0.433v0 --- (b)
DYNAMICS
=> Relation corresponding to e
(v’B)n –(v’A)n = e[(vA)n –(vB)n ]
v’B sin 30° –(v’A)n = v0 cos 30° –0
0.5v’B –(v’A)n = 0.866v0 --- (c)
=> (b) & (c) => (b) & (c)
v’B = 0.693v0 = 0.693v0 →
& (v’A)n = –0.520v0 = 0.520v0
=> v’A = 0.721v0
b = 46.1o
a = 46.1o –30o = 16.1o
2008/10/6 CH-13 86
a = 46.1o –30o = 16.1o
v’A = 0.721v0 16.1o
DYNAMICS
(5) Example 13.4-5
A 30-kg block is dropped from a height of 2 m onto the 10-kg
pan of a spring scale. Assuming the impact to be perfectly
plastic, determine the maximum deflection of the pan. The
constant of the spring is k = 20 kN/m. constant of the spring is k = 20 kN/m.
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DYNAMICS
Solution: WA = (30)(9.81) = 294 N & WB = (10)(9.81) = 98.1 N
Position 1 to position 2=> T1 = mA(vA)1
2 /2 = 0 & V1 = WAy = (294)(2) = 588 J
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1 A A 1 1 A
=> T2 = mA(vA)22 /2= (30)(vA)2
2/2 & V2 = 0 => T1 + V1 = T2 + V2
=> 0 + 588 J = (30)(vA)22/2 + 0
=> (vA)2 = + 6.26 m/s = 6.26 m/s↓
DYNAMICS
Position 2 to position 3
=> mA(vA)2 + mB(vB)2 = (mA + mB)v3
=> (30)(6.26) + 0 = (30 + 10)v3
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=> v3 = 4.70 = 4.70 m/s↓
Position 3 to position 4
m104.91N/m1020
98.1N
N/m1020
)1m/s(10kg)(9.8
k
Wx 3
33
2B
3
DYNAMICS
=> T3 = (mA + mB)v32/2 = (30 + 10)(4.70)2 /2 = 442 J
& V3 = Vg + Ve = 0 + kx32 /2 = (20 × 103)(4.91 × 10-3)2 /2 = 0.241 J
=> T4 = 0 & V4 = Vg + Ve = (WA + WB)(–h) + kx42/2
= –(392)h + (20 × 103)x42/2
=> T + V = T + V
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=> T3 + V3 = T4 + V4
=> 442 + 0.241 = 0 –392(x4 –4.91 × 10-3) + (20 × 103)x42/2
=> x4 = 0.230 m => h = x4 –x3 = 0.230 m –4.91 × 10-3 m
=> h = 0.2251 m
