MEG 206 Dynamics (New Material).pdf is a

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1. Lecture Note
Engineering Mechanics II
(Dynamics)
Dr. Collins N. NWAOKOCHA
Olabisi Onabanjo University, Ago-Iwoye

2. Description of Engineering Mechanics - Dynamics
Statics is the branch of mechanics that deals with the
analysis bodies at rest, while dynamics deals with the
analysis od bodies in motion.
Dynamics includes:
●Kinematics – study of geometry of motion without
reference to the cause of the motion.
•Kinetics – study of the relation existing between the
forces acting on a body, the mass of the body, and the
motion of the body.
Rectilinear motion: position, velocity, and acceleration of
a particle as it moves along a straight line.
Curvilinear motion: position, velocity, and acceleration of
a particle as it moves along a curved line in two or three
dimensions.

3. I. Dynamics of Particles – Kinematics of Particles
The word particles does not restrict our study to
small corpuscles; rather, the motion of bodies –
possibly as large as rockets, cars, buses or airplanes
– will be considered without regard to their size.
For bodies analyzed as particles, only their motion
as an entire unit will be considered; rotations about
their own mass center will be neglected.
In cases, when such a rotation is not negligible;
such bodies cannot be considered as particles, it is
analyzed as dynamics of rigid bodies.

4. Rectilinear Motion: Position, Velocity & Acceleration
•Rectilinear motion: particle moving
along a straight line
•Position coordinate: defined by
positive or negative distance from a
fixed origin on the line.
•The motion of a particle is known if
the position coordinate for particle is
known for every value of time t.
•May be expressed in the form of a
function, e.g.,
326ttx−=
or in the form of a graph x vs. t.

5. Rectilinear Motion: Position, Velocity & Acceleration
11 - 0
•Instantaneous velocity may be positive or
negative. Magnitude of velocity is referred
to as particle speed.
•Consider particle which occupies position P
at time t and P’ at t+Dt,
txvtxtΔΔ==ΔΔ=→Δ0lim
Average velocity
Instantaneous velocity
•From the definition of a derivative,
dtdxtxvt=ΔΔ=→Δ0lim
e.g.,
2323126ttdtdxvttx−==−=

6. Rectilinear Motion: Position, Velocity & Acceleration
•Consider particle with velocity v at time t and
v’ at t+Dt,
Instantaneous acceleration
tvatΔΔ==→Δ0lim
tdtdvattvdtxddtdvtvat612312e.g.lim2220−==−===ΔΔ=→Δ
•From the definition of a derivative,
•Instantaneous acceleration may be:
- positive: increasing positive velocity
or decreasing negative velocity
-negative: decreasing positive velocity
or increasing negative velocity.

7. Concept Quiz
What is true about the kinematics of a particle?
a) The velocity of a particle is always positive
b) The velocity of a particle is equal to the slope
of the position-time graph
c) If the position of a particle is zero, then the
velocity must zero
d) If the velocity of a particle is zero, then its
acceleration must be zero

8. Exercise 1.1
•Consider particle with motion given by
326ttx−=2312ttdtdxv−==tdtxddtdva61222−===
•at t = 0, x = 0, v = 0, a = 12 m/s2
•at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0
•at t = 4 s, x = xmax = 32 m, v = 0, a = -12
m/s2
•at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2

9. Determination of the Motion of a Particle
11 - 0
•We often determine accelerations from the forces applied
(kinetics will be covered later)
•Generally have three classes of motion
- acceleration given as a function of time, a = f(t)
- acceleration given as a function of position, a = f(x)
- acceleration given as a function of velocity, a = f(v)

• Can you think of a physical example of when force is a
function of position?When force is a function of velocity?
a spring
drag

10. 11 - 0
Determine:
•velocity and elevation above ground at
time t,
•highest elevation reached by ball and
corresponding time, and
•time when ball will hit the ground and
corresponding velocity.
Ball tossed with 10 m/s vertical velocity
from window 20 m above ground.
SOLUTION:
•Integrate twice to find v(t) and y(t).
•Solve for t when velocity equals zero
(time for maximum elevation) and
evaluate corresponding altitude.
•Solve for t when altitude equals zero
(time for ground impact) and evaluate
corresponding velocity.
Exercise 1.2

11. 11 - 0
()()tvtvdtdvadtdvttvv81.981.9sm81.90020−=−−=−==∫∫()ttv⎟⎠⎞⎜⎝⎛−=2sm81.9sm10()()()2210081.91081.91081.9100ttytydttdytvdtdyttyy−=−−=−==∫∫()22sm905.4sm10m20ttty⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛+=
SOLUTION:
•Integrate twice to find v(t) and y(t).

12. •Solve for t when velocity equals zero and evaluate
corresponding altitude.
()0sm81.9sm102=⎟⎠⎞⎜⎝⎛−=ttvs019.1=t
•Solve for t when altitude equals zero and evaluate
corresponding velocity.
()()()2222s019.1sm905.4s019.1sm10m20sm905.4sm10m20⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛+=⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛+=ytttym1.25=y

13. •Solve for t when altitude equals zero and evaluate
corresponding velocity.
()0sm905.4sm10m2022=⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛+=ttty()s28.3smeaningles s243.1=−=tt()()()s28.3sm81.9sm10s28.3sm81.9sm1022⎟⎠⎞⎜⎝⎛−=⎟⎠⎞⎜⎝⎛−=vttvsm2.22−=v

14. Motion of Several Particles
We may be interested in the motion of
several different particles, whose motion
may be independent or linked together.
•For particles moving along the same line, time
should be recorded from the same starting
instant and displacements should be measured
from the same origin in the same direction.
=−=ABABxxx
relative position of B
with respect to A
ABABxxx+==−=ABABvvv
relative velocity of B
with respect to A
ABABvvv+==−=ABABaaa
relative acceleration of B
with respect to A
ABABaaa+=

15. Exercise 1.3
11 - 0
Ball thrown vertically from 12 m level
in elevator shaft with initial velocity of
18 m/s. At same instant, open-platform
elevator passes 5 m level moving
upward at 2 m/s.
Determine (a) when and where ball hits
elevator and (b) relative velocity of ball
and elevator at contact.
SOLUTION:
•Substitute initial position and velocity
and constant acceleration of ball into
general equations for uniformly
accelerated rectilinear motion.
•Substitute initial position and constant
velocity of elevator into equation for
uniform rectilinear motion.
•Write equation for relative position of
ball with respect to elevator and solve
for zero relative position, i.e., impact.
•Substitute impact time into equation
for position of elevator and relative
velocity of ball with respect to
elevator.

16. 11 - 0
SOLUTION:
•Substitute initial position and velocity and constant
acceleration of ball into general equations for
uniformly accelerated rectilinear motion.
222210020sm905.4sm18m12sm81.9sm18ttattvyytatvvBB⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛+=++=⎟⎠⎞⎜⎝⎛−=+=
•Substitute initial position and constant velocity of
elevator into equation for uniform rectilinear motion.
ttvyyvEEE⎟⎠⎞⎜⎝⎛+=+==sm2m5sm20

17. •Write equation for relative position of ball with respect to
elevator and solve for zero relative position, i.e., impact.
()()025905.418122=+−−+=tttyEB()s65.3smeaningles s39.0=−=tt
•Substitute impact time into equations for position of elevator
and relative velocity of ball with respect to elevator.
()65.325+=Eym3.12=Ey()()65.381.916281.918−=−−=tvEBsm81.19−=EBv

18. II. Dynamics of Particles – Kinetics of Particles
•Newton’s first and third laws are sufficient for the study of bodies at
rest (statics) or bodies in motion with no acceleration.
•When a body accelerates (changes in velocity magnitude or direction),
Newton’s second law is required to relate the motion of the body to the
forces acting on it.
•Newton’s second law:
-A particle will have an acceleration proportional to the magnitude of
the resultant force acting on it and in the direction of the resultant
force.
-The resultant of the forces acting on a particle is equal to the rate of
change of linear momentum of the particle.
-The sum of the moments about O of the forces acting on a particle is
equal to the rate of change of angular momentum of the particle
about O.

20. Kinetics of Particles – Newton’s Second Law of Motion
•Newton’s Second Law: If the resultant force acting on a
particle is not zero, the particle will have an acceleration
proportional to the magnitude of resultant and in the
direction of the resultant.
•Consider a particle subjected to constant forces,
maFaFaF mass,constant332211=====!
•When a particle of mass m is acted upon by a force
the acceleration of the particle must satisfy
,F!amF!!=
•Acceleration must be evaluated with respect to a
Newtonian frame of reference, i.e., one that is not
accelerating or rotating.
•If force acting on particle is zero, particle will not
accelerate, i.e., it will remain stationary or continue on a
straight line at constant velocity.

21. Linear Momentum of a Particle
•Replacing the acceleration by the derivative of the
velocity yields
()particle theof momentumlinear ====∑LdtLdvmdtddtvdmF!!!!!
•Linear Momentum Conservation Principle:
If the resultant force on a particle is zero, the linear
momentum of the particle remains constant in both
magnitude and direction.

22. 12 - 0
Systems of Units
•Of the units for the four primary dimensions (force,
mass, length, and time), three may be chosen arbitrarily.
The fourth must be compatible with Newton’s 2nd Law.
•International System of Units (SI Units): base units are
the units of length (m), mass (kg), and time (second).
The unit of force is derived,
()22smkg1sm1kg1N1⋅=⎟⎠⎞⎜⎝⎛=
•U.S. Customary Units: base units are the units of force
(lb), length (m), and time (second). The unit of mass is
derived,
ftslb1sft1lb1slug1sft32.2lb1lbm1222⋅===

23. Equations of Motion
•Newton’s second law provides
amF!!=∑
•Solution for particle motion is facilitated by resolving
vector equation into scalar component equations, e.g.,
for rectangular components,
()()zmFymFxmFmaFmaFmaFkajaiamkFjFiFzyxzzyyxxzyxzyx!!!!!!!!!!!!======++=++∑∑∑∑∑∑∑
•For tangential and normal components,
ρ2vmFdtdvmFmaFmaFntnntt====∑∑∑∑

24. Dynamic Equilibrium
•Alternate expression of Newton’s second law,
ectorinertial vamamF 0≡−=−∑!!!
•With the inclusion of the inertial vector, the system
of forces acting on the particle is equivalent to
zero. The particle is in dynamic equilibrium.
•Methods developed for particles in static
equilibrium may be applied, e.g., coplanar forces
may be represented with a closed vector polygon.
•Inertia vectors are often called inertial forces as
they measure the resistance that particles offer to
changes in motion, i.e., changes in speed or
direction.
•Inertial forces may be conceptually useful but are
not like the contact and gravitational forces found
in statics.

25. Exercise 2.1
A 200-lb block rests on a horizontal
plane. Find the magnitude of the force
P required to give the block an accelera-
tion or 10 ft/s2 to the right. The coef-
ficient of kinetic friction between the
block and plane is mk = 0.25.
SOLUTION:
•Resolve the equation of motion for the
block into two rectangular component
equations.
•Unknowns consist of the applied force
P and the normal reaction N from the
plane. The two equations may be
solved for these unknowns.

26. NNFgWmk25.0ftslb21.6sft2.32lb20022==⋅===µ
x
y
O
SOLUTION:
•Resolve the equation of motion for the block
into two rectangular component equations.
:maFx=∑()()lb1.62sft10ftslb21.625.030cos22=⋅=−°NP:0=∑yF0lb20030sin=−°−PN
•Unknowns consist of the applied force P and
the normal reaction N from the plane. The two
equations may be solved for these unknowns.
()lb1.62lb20030sin25.030coslb20030sin=+°−°+°=PPPNlb151=P
Solution 2.1

27. Exercise 2.2
The two blocks shown start from rest.
The horizontal plane and the pulley
are frictionless, and the pulley is
assumed to be of negligible mass.
Determine the acceleration of each
block and the tension in the cord.
SOLUTION:
•Write the kinematic relationships for the
dependent motions and accelerations of
the blocks.
•Write the equations of motion for the
blocks and pulley.
•Combine the kinematic relationships
with the equations of motion to solve for
the accelerations and cord tension.

28. 12 - 0
Solution 2.2
•Write equations of motion for blocks and pulley.
:AAxamF=∑()AaTkg1001=:BByamF=∑()()()()BBBBBaTaTamTgmkg300-N2940kg300sm81.9kg3002222==−=−:0==∑CCyamF0212=−TT
SOLUTION:
•Write the kinematic relationships for the dependent
motions and accelerations of the blocks.
ABABaaxy2121==
x
y
O

29. 12 - 0
()N16802N840kg100sm20.4sm40.81212212=======TTaTaaaAABA
•Combine kinematic relationships with equations of
motion to solve for accelerations and cord tension.
ABABaaxy2121==()AaTkg1001=()()()ABaaT212kg300-N2940kg300-N2940==()()0kg1002kg150N29400212=−−=−AAaaTT
x
y
O

30. III. Kinetics of Particles – Energy and Momentum Methods
•Previously, problems dealing
with the motion of particles
were solved through the
fundamental equation of
motion,
Current chapter introduces two
additional methods of
analysis:
•Method of work and energy:
directly relates force, mass,
velocity and displacement.
•Method of impulse and
momentum: directly relates
force, mass, velocity, and
time.

32. Work of a Force
•Work of the force is
dzFdyFdxFdsFrdFdUzyx++==•=αcos!!
•Work is a scalar quantity, i.e., it has magnitude and
sign but not direction.
force. length ×
•Dimensions of work are Units are
()()()J 1.356lb1ftm 1N 1 J 1=⋅=joule
Consider a particle which moves from a point A to a neighboring point
A’. If r denotes the position vector corresponding to point A, the small
vector joining A and A’ can be denoted by the differential dr; the vector
dr is called the displacement of the particle. Now, let us assume that a
force F is acting on the particle.

33. Work of a Force
•Work of a force during a finite displacement,
()()∫∫∫∫++===•=→
21212121cos21AAzyxsstssAAdzFdyFdxFdsFdsFrdFUα!!
•Work is represented by the area under the
curve of Ft plotted against s.

34. 13 - 0
Work of a Force
•Magnitude of the force exerted by a spring is
proportional to deflection,
()lb/in.or N/mconstant spring ==kkxF
•Work of the force exerted by spring,
222121212121kxkxdxkxUdxkxdxFdUxx−=−=−=−=∫→
•Work of the force exerted by spring is positive
when x2 < x1, i.e., when the spring is returning to
its undeformed position.
•Work of the force exerted by the spring is equal to
negative of area under curve of F plotted against x,
()xFFUΔ+−=→212121

36. Particle Kinetic Energy: Principle of Work & Energy
dvmvdsFdsdvmvdtdsdsdvmdtdvmmaFttt=====
•Consider a particle of mass m acted upon by force
F!
•Integrating from A1 to A2 ,
energykineticmvTTTUmvmvdvvmdsFvvsst==−=−==→∫∫2211221212122212121
•The work of the force is equal to the change in
kinetic energy of the particle.
F!
•Units of work and kinetic energy are the same:
JmNmsmkgsmkg22221=⋅=⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛==mvT

37. 13 - 0
Power and Efficiency
• rate at which work is done.
vFdtrdFdtdUPower!!!!•=•===
•Dimensions of power are work/time or force*velocity.
Units for power are
W746slbft550 hp 1orsmN 1sJ1 (watt) W 1=⋅=⋅==
•
inputpower outputpower input workkoutput worefficiency===η

38. 13 - 0
Exercise 3.1
An automobile weighing 4000
lb is driven down a 5o incline at
a speed of 60 mi/h when the
brakes are applied causing a
constant total breaking force of
1500 lb.
Determine the distance traveled
by the automobile as it comes to
a stop.
SOLUTION:
•Evaluate the change in kinetic energy.
•Determine the distance required for the
work to equal the kinetic energy change.

39. Solution 3.1
SOLUTION:
•Evaluate the change in kinetic energy.
Position 1:
()()lbft481000882.324000sft88s 3600hmift 5280hmi60221212111⋅====⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=mvTv
.
Position 2:
()0lb1151lbft4810002211=−⋅=+→xTUTft 418=x
•Determine the distance required for the work to
equal the kinetic energy change.
()()()()xxxUlb11515sinlb4000lb150021−=°+−=→0022==Tv

40. 13 - 0
Exercise 3.2
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 friction
between block A and the plane is
mk = 0.25 and that the pulley is
weightless and frictionless.
SOLUTION:
•Apply the principle of work and
energy separately to blocks A and B.
•When the two relations are combined,
the work of the cable forces cancel.
Solve for the velocity.

41. Solution 3.2
SOLUTION:
•Apply the principle of work and energy separately to
blocks A and B.
()()()()()()()()()22122122112kg200m2N490m2m2m20:N490N196225.0N1962sm81.9kg200vFvmFFTUTWNFWCAACAkAkAA=−=−+=+======→µµ()()()()()()()()22122122112kg300m2N2940m2m2m20:N2940sm81.9kg300vFvmWFTUTWcBBcB=+−=+−=+==→

42. Solution 3.2
•When the two relations are combined, the work of the
cable forces cancel. Solve for the velocity.
()()()()221kg200m2N490m2vFC=−()()()()221kg300m2N2940m2vFc=+−()()()()()()221221kg500J 4900kg300kg200m2N490m2N2940vv=+=−sm 43.4=v

43. 13 - 0
Exercise 3.3
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. The
package has a velocity of 2.5 m/s
in the position shown and the
maximum deflection of the spring
is 40 mm.
Determine (a) the coefficient of
kinetic friction between the
package and surface and (b) the
velocity of the package as it
passes again through the position
shown.
SOLUTION:
•Apply the principle of work and energy
between the initial position and the
point at which the spring is fully
compressed and the velocity is zero.
The only unknown in the relation is
the friction coefficient.
•Apply the principle of work and energy
for the rebound of the package. The
only unknown in the relation is the
velocity at the final position.

44. Solution 3.3
SOLUTION:
•Apply principle of work and energy between initial position
and the point at which spring is fully compressed.
()()0J5.187sm5.2kg60222121211====TmvT()()()()()kkkfxWUµµµJ377m640.0sm81.9kg60221−=−=−=→()()()()()()()()()J0.112m040.0N3200N2400N3200m160.0mkN20N2400m120.0mkN2021maxmin21210max0min−=+−=Δ+−===Δ+====→xPPUxxkPkxPe()()()J112J377212121−−=+=→→→kefUUUµ()0J112J 377-J5.187:2211=−=+→kTUTµ20.0=kµ

45. Solution 3.3
•Apply the principle of work and energy for the rebound of
the package.
()2321232132kg600vmvTT===()()()J36.5J112J377323232+=+−=+=→→→kefUUUµ()23213322kg60J5.360:vTUT=+=+→sm103.13=v

46. Classwork

47. 13 - 0
Potential Energy
•Work of the force exerted by a spring depends
only on the initial and final deflections of the
spring,
2221212121kxkxU−=→
•The potential energy of the body with respect to
the elastic force,
()()2121221eeeVVUkxV−==→
•Note that the preceding expression for Ve is
valid only if the deflection of the spring is
measured from its undeformed position.

48. Exercise 3.4
A 20 lb collar slides without friction along a
vertical rod as shown. The spring attached to
the collar has an undeflected length of 4 in.
and a constant of 3 lb/in.
If the collar is released from rest at position 1,
determine its velocity after it has moved 6 in.
to position 2.
SOLUTION:
•Apply the principle of conservation of
energy between positions 1 and 2.
•The elastic and gravitational potential
energies at 1 and 2 are evaluated from
the given information. The initial kinetic
energy is zero.
•Solve for the kinetic energy and velocity
at 2.

49. Solution 3.4
SOLUTION:
•Apply the principle of conservation of energy between
positions 1 and 2.
Position 1:
()()0lbft20lbin.24lbin.24in. 4in. 8in.lb3112212121=⋅=+⋅=+=⋅=−==TVVVkxVgee
Position 2:
()()()()22222221222212221311.02.322021lbft 5.5lbin. 6612054lbin. 120in. 6lb 20lbin.54in. 4in. 01in.lb3vvmvTVVVWyVkxVgege===⋅−=⋅−=−=+=⋅−=−==⋅=−==
Conservation of Energy:
lbft 5.50.311lbft 20222211⋅−=⋅++=+vVTVT↓=sft91.42v

50. IV. Systems of Particles
Here, we will study the motion of systems of particles, i.e., the motion of a
large number of particles considered together.
The first part is devoted to systems consisting of well defined particles; while
the second part considers the motion of variable systems, i.e., systems
which are continually gaining or losing particles, or doing both at the same
time.
•The effective force of a particle is defined as the product of it mass and
acceleration. It will be shown that the system of external forces acting on
a system of particles is equipollent with the system of effective forces of
the system.
•The mass center of a system of particles will be defined and its motion
described.
•Application of the work-energy principle and the impulse-momentum
principle to a system of particles will be described. Result obtained are
also applicable to a system of rigidly connected particles, i.e., a rigid
body.

51. Exercise 4.1
A 20-lb projectile is moving with a velocity
of 100 ft/s when it explodes into 5 and 15-lb
fragments. Immediately after the explosion,
the fragments travel in the directions qA =
45o and qB = 30o.
Determine the velocity of each fragment.
SOLUTION:
•Since there are no external forces, the
linear momentum of the system is
conserved.
•Write separate component equations
for the conservation of linear
momentum.
•Solve the equations simultaneously
for the fragment velocities.

52. Solution 4.1
SOLUTION:
•Since there are no external forces, the
linear momentum of the system is
conserved.
x
y
•Write separate component equations for
the conservation of linear momentum.
()()()0020155vgvgvgvmvmvmBABBAA!!!!!!=+=+
x components:
()1002030cos1545cos5=°+°BAvv
y components:
030sin1545sin5=°−°BAvv
•Solve the equations simultaneously for the
fragment velocities.
sft6.97sft207==BAvv

53. V. Plane Motion of Rigid Bodies: Energy &
Momentum Methods
•Method of work and energy and the method of impulse and momentum will be
used to analyze the plane motion of rigid bodies and systems of rigid bodies.
•Principle of work and energy is well suited to the solution of problems
involving displacements and velocities.
2211TUT=+→
•Principle of impulse and momentum is appropriate for problems involving
velocities and time.
()()21212121OttOOttHdtMHLdtFL!!!!!!=+=+∑∫∑∫
The work done by
friction reduces
the kinetic energy of
the wheel.

54. Exercise 5.1
For the drum and flywheel,
The bearing friction is equivalent to a
couple of At the instant shown, the block
is moving downward at 6 m /s
2kgm5.10=ImN60⋅
Determine the velocity of the block after it
has moved 4 m downward.
SOLUTION:
•Consider the system of the
flywheel and block. The work
done by the internal forces exerted
by the cable cancels.
•Apply the principle of work and
kinetic energy to develop an
expression for the final velocity.
•Note that the velocity of the block
and the angular velocity of the
drum and flywheel are related by
ωrv=

55. 17 - 0
Solution 5.1
SOLUTION:
•Consider the system of the flywheel and block. The work
done by the internal forces exerted by the cable cancels.
•Note that the velocity of the block and the angular velocity of
the drum and flywheel are related by
1.25srad80.4m1.25m/622211vrvsrvrv======ωωω
•Apply the principle of work and kinetic energy to develop an
expression for the final velocity.
()()()mN96.552srad80.4kgm5.1021m/s6m/s10N240212222212121211⋅=+=+=ωImvT
22222222212221236.1525.15.10211024021vvvIvmT=⎟⎠⎞⎜⎝⎛+=+=ω

56. 17 - 0
Solution 5.1 contd.
mN96.552212121211⋅=+=ωImvT2222212221236.15vIvmT=+=ω
•Note that the block displacement and pulley
rotation are related by
rad20.3m25.1m422===rsθ
•Principle of work and energy:
m/s27.936.51mN768mN96.5522222211==⋅+⋅=+→vvTUTm/s27.92=v()()()()()()mN768rad20.3mN60m4N240121221⋅=⋅−=−−−=→θθMssWU
Then,

57. 17 - 0
Exercise 5.2
mm80kg3mm200kg10====BBAAkmkm
The system is at rest when a moment
of is applied to gear B.
Neglecting friction, a) determine the
number of revolutions of gear B before
its angular velocity reaches 600 rpm,
and b) tangential force exerted by gear
B on gear A.
mN6⋅=M
SOLUTION:
•Consider a system consisting of the two
gears. Noting that the gear rotational
speeds are related, evaluate the final
kinetic energy of the system.
•Apply the principle of work and energy.
Calculate the number of revolutions
required for the work of the applied
moment to equal the final kinetic energy
of the system.
•Apply the principle of work and energy to
a system consisting of gear A. With the
final kinetic energy and number of
revolutions known, calculate the moment
and tangential force required for the
indicated work.

58. Solution 5.2
SOLUTION:
•Consider a system consisting of the two gears. Noting
that the gear rotational speeds are related, evaluate the
final kinetic energy of the system.
()()srad1.25250.0100.08.62srad8.62mins60revrad2rpm600=====ABBABrrωωπω()()()()222222mkg0192.0m080.0kg3mkg400.0m200.0kg10⋅===⋅===BBBAAAkmIkmI()()()()J9.1638.620192.01.25400.02212212212212=+=+=BBAAIITωω

59. Solution 5.2 contd.
•Apply the principle of work and energy. Calculate the
number of revolutions required for the work.
()rad32.27163.9JJ602211==+=+→BBTUTθθrev35.4232.27==πθB
•Apply the principle of work and energy to a system
consisting of gear A. Calculate the moment and
tangential force required for the indicated work.
()()J0.1261.25400.02212212===AAITω()mN52.11J0.261rad10.9302211⋅===+=+→FrMMTUTAAArad93.10250.0100.032.27===ABBArrθθN2.46250.052.11==F

60. VI. Mechanical Vibrations
•Mechanical vibration is the motion of a particle or body which
oscillates about a position of equilibrium. Most vibrations in
machines and structures are undesirable due to increased stresses
and energy losses.
•Time interval required for a system to complete a full cycle of the
motion is the period of the vibration.
•Number of cycles per unit time defines the frequency of the vibrations.
•Maximum displacement of the system from the equilibrium position is
the amplitude of the vibration.
•When the motion is maintained by the restoring forces only, the
vibration is described as free vibration. When a periodic force is applied
to the system, the motion is described as forced vibration.
•When the frictional dissipation of energy is neglected, the motion
is said to be undamped. Actually, all vibrations are damped to
some degree.

61. Free Vibrations of Particles. Simple Harmonic Motion
19 - 0
•If a particle is displaced through a distance xm from its
equilibrium position and released with no velocity, the
particle will undergo simple harmonic motion,
()0=+−=+−==kxxmkxxkWFmast!!δ
•General solution is the sum of two particular solutions,
()()tCtCtmkCtmkCxnnωωcossincossin2121+=⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛=
•x is a periodic function and wn is the natural circular
frequency of the motion.
•C1 and C2 are determined by the initial conditions:
()()tCtCxnnωωcossin21+=02xC=nvCω01=()()tCtCxvnnnnωωωωsincos21−==!

62. Exercise 6.1
19 - 0
A 50-kg block moves between vertical
guides as shown. The block is pulled
40mm down from its equilibrium
position and released.
For each spring arrangement, determine
a) the period of the vibration, b) the
maximum velocity of the block, and c)
the maximum acceleration of the block.
SOLUTION:
•For each spring arrangement, determine
the spring constant for a single
equivalent spring.
•Apply the approximate relations for the
harmonic motion of a spring-mass
system.

63. Solution 6.1
mkN6mkN421==kk
SOLUTION:
•Springs in parallel:
- determine the spring constant for equivalent spring
mN10mkN1042121==+==+=kkPkkkPδδδ
- apply the approximate relations for the harmonic motion
of a spring-mass system
nnnmkωπτω2srad14.14kg20N/m104====s 444.0=nτ()()srad 4.141m 040.0==nmmxvωsm566.0=mv2sm00.8=ma()()22srad 4.141m 040.0==nmmaxa

64. mkN6mkN421==kk
•Springs in series:
- determine the spring constant for equivalent spring
- apply the approximate relations for the harmonic motion
of a spring-mass system
nnnmkωπτω2srad93.6kg20400N/m2====s 907.0=nτ()()srad .936m 040.0==nmmxvωsm277.0=mv2sm920.1=ma()()22srad .936m 040.0==nmmaxamN10mkN1042121==+==+=kkPkkkPδδδ
Solution 6.1 contd.