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CHAPTER 18 PLANAR KINETICS OF A RIGID BODY

WORK AND ENERGY (Sections 18.1-18.4)

Objectives a) Define the various ways a force

and couple do work. b) Apply the principle of

work and energy to a rigid body.

APPLICATIONS

The work of the torque developed by the driving

gears on the two motors on the mixer is

transformed into the rotational kinetic energy of

the mixing drum.

The work done by the compactor's engine is

transformed into the translational kinetic energy

of the frame and the translational and rotational

kinetic energy of its roller and wheels

KINETIC ENERGY

The kinetic energy of a rigid body in general

plane motion is given by T 1/2 m (vG)2 1/2

IG ?2

- Several simplifications can occur.
- Pure Translation the rotational kinetic energy

is zero (? 0) - T 0.5 m (vG)2

KINETIC ENERGY (continued)

2. Rotation When a rigid body is rotating

about a fixed axis passing through point O, the

body has both translational and rotational

kinetic energy T 0.5m(vG)2 0.5IGw2 Since

vG rGw, T 0.5(IG m(rG)2)w2 0.5IOw2

If the rotation occurs about the mass center

(pure rotation), G, then vG0

T 0.5 IG w2

WORK OF A FORCE

Work of a weight Uw -W?y.

Work of a spring force Us -0.5k(s2)2

(s1)2

FORCES THAT DO NO WORK

1. Reactions at fixed supports because the

displacement at their point of application is

zero.

2. Normal and frictional forces acting on bodies

as they roll without slipping over a rough

surface since there is no instantaneous

displacement of the point in contact with ground.

3. Internal forces because they always act in

equal and opposite pairs.

THE WORK OF A COUPLE

A body subjected to a couple produces work only

when it undergoes rotation.

constant M UM M (q2 q1)

PRINCIPLE OF WORK AND ENERGY

The principle states SU1-2 T12 i.e. the work

done by all external forces and couple moments

equals the change in bodys kinetic energy

(translational and rotational).

This equation is a scalar equation. It can be

applied to a system of rigid bodies by summing

contributions from all bodies.

EXAMPLE

GivenThe disk weighs 40 N, has a radius of

gyration (kG) 0.6 m. A 15 N.m moment is applied.

The spring constant is 10N/m.

Find Angular velocity of the disk when point G

moves 0.5 m. The disk starts from rest and rolls

without slipping. The spring is initially

unstretched.

EXAMPLE (solution)

Free body diagram of the disk

Only the spring force and couple moment M do

work.

Spring will stretch twice the amount of

displacement of G, or 1 m. Why?

EXAMPLE (continued)

Kinematic relation vG r w 0.8w

Kinetic energy T1 0 T2 0.5m (vG)2 0.5 IG

w2 T2 0.5(40/9.8)(0.8w)2 0.5(40/9.8)(0.6)2w2

T2 2.0 w2

Work and energy U1-2 T12 4.4 2.0 w2 w

1.5 rad/s

GROUP PROBLEM SOLVING

Given A sphere weighing 10N rolls along a

semicircular hoop. Its w equals 0 when q 0.

Find The angular velocity of the sphere when q

45 if the sphere rolls without slipping.

GROUP PROBLEM SOLVING (continued)

Solution Draw a FBD and calculate the vertical

distance the mass center moves.

Now calculate the _______________

GROUP PROBLEM SOLVING (continued)

A kinematic equation for finding the velocity of

the mass center is needed. It is

_________ energy

Now apply the principle of work and energy

equation w 13.1

rad/s

End of 18.1-4

Let Learning Continue

PLANAR KINETICS OF A RIGID BODY CONSERVATION OF

ENERGY (Section 18.5)

- Objectives
- Determine the potential energy of conservative

forces. - Apply the principle of conservation of energy.

APPLICATIONS

Torsional spring at the top of the door winds up

as the door is lowered. When the door is raised,

the spring potential energy is transferred into

gravitational potential energy of the doors

weight, thus making it easy to open.

CONSERVATION OF ENERGY

The conservation of energy theorem is a simpler

energy method for solving problems. Once again,

the problem parameter of distance is a key

indicator of when conservation of energy is a

good method for solving the problem.

If it is appropriate, conservation of energy is

easier to use than the principle of work and

energy.

CONSERVATIVE FORCES

A force F is conservative if the work done by the

force is independent of the path.

Typical conservative forces encountered in

dynamics are gravitational forces (i.e., weight)

and elastic forces (i.e., springs).

What is a common force that is not conservative?

CONSERVATION OF ENERGY

When a rigid body is acted upon by a system of

conservative forces, the work done by these

forces is conserved. Thus, the sum of kinetic

energy and potential energy remains constant T1

V1 T2 V2 Constant

i.e. as a rigid body moves from one position to

another when acted upon by only conservative

forces, kinetic energy is converted to potential

energy and vice versa.

GRAVITATIONAL POTENTIAL ENERGY

The gravitational potential energy is a function

of the height of the bodys center above or

below a datum.

The gravitational potential energy is found by

Vg W yG

Gravitational potential energy is ve when yG is

ve, since the weight has the ability to do ve

work when the body is moved back to the datum.

ELASTIC POTENTIAL ENERGY

Spring forces are also conservative forces.

The potential energy of a spring force (F ks)

is found by the equation Ve ½ ks2

Notice that the elastic potential energy is

always ve

EXAMPLE 1

Given The rod AB has a mass of 10 kg. Piston B

is attached to a spring of constant k 800 N/m.

The spring is un-stretched when ? 0. Neglect

the mass of the pistons.

Find The angular velocity of rod AB at ? 0 if

the rod is released from rest when ? 30.

EXAMPLE 1 (solution)

Initial Position

Final Position

Potential Energy put datum in line with rod when

?0. Gravitational and elastic energy0 at 2.

gt V2 0 Gravitational energy at 1 - (10)(

9.81) ½ (0.4 sin 30) Elastic energy at 1 ½

(800) (0.4 sin 30)2 So V1 -

9.81J 16.0 J 6.19 J

EXAMPLE 1 (continued)

Initial Position

Final Position

Kinetic Energy

The rod released from rest (vG10, ?10). Thus,

T1 0. At 2, the angular velocity is ?2 and the

velocity is vG2 .

EXAMPLE 1 (continued)

Thus, T2 ½ (10)(vG2)2 ½ (1/12)(10)(0.42)(?2)

2 At 2, vA0. Hence, vG2 r ? 0.2 ?2

. Then, T2 0.2 ?22 0.067 ?22 0.267 ?22

conservation of energy T1 V1 T2 V2

0 6.19 0.267?22 0 gt ?2 4.82

rad/s

EXAMPLE 2

Given The weight of the disk is 30 N and its kG

equals 0.6 m. The spring has a stiffness of 2 N/m

and an unstretched length of 1m.

Find The angular velocity at the instant G moves

3 m to the left. The disk is released from rest

in the position shown and rolls without slipping.

EXAMPLE 2 (solution)

Potential Energy No changes in the gravitational

potential energy

The elastic potential energy at 1 V1 0.5 k

(s1)2 where s1 4 m. Thus, V1 ½ 2 (4)2

16 N.m.

The elastic potential energy at 2 is V2 ½ 2

(3)2 9 N.m.

EXAMPLE 2 (continued)

Kinetic Energy

Released from rest vG1?10, T10 At 2, ?2 and

vG2

T2 ½ m (vG2)2 ½ IG (?2) 2 ½ (30/9.8)

(vG2)2 ½ (30/9.8) 0.62 (?2) 2

Disk is rolling without slipping vG2 (0.75 ?2)

T2 ½(30/9.8)(0.75 ?2)2 ½(30/9.8) 0.62 (?2)2

1.41 (?2)2

EXAMPLE 2 (continued)

conservation of energy

T1 V1 T2 V2 0 16.0 1.41 ?22 9

Solving , ?2 2.23 rad/s

GROUP PROBLEM SOLVING

Given A 50 N bar is rotating downward at 2

rad/s. The spring has an

unstretched length of 2m and a spring constant of

12 N/m

Find The angle (measured down from the

horizontal) to which the bar rotates before it

stops its initial downward movement.

GROUP PROBLEM SOLVING (solution)

Potential Energy Put the datum in line with the

rod when ? 0.

Gravitational potential energy at 2

Elastic potential energy at 2 So, V2

GROUP PROBLEM SOLVING (continued)

Kinetic Energy

At 1 (when ? 0)

At 2

GROUP PROBLEM SOLVING (continued)

conservation of energy

Thus, ? 49.9 deg.

End of 18.5

Let Learning Continue

CHAPTER 19 PLANAR KINETICS IMPULSE AND MOMENTUM

(Sections 19.1-19.2)

Todays Objectives a) Develop formulations for

the linear and angular momentum of a

body. b) Apply the principle of linear and

angular impulse and momentum.

APPLICATIONS

As the pendulum swings downward, its angular

momentum and linear momentum both increase. By

calculating its momenta in the vertical position,

we can calculate the impulse the pendulum exerts

when it hits the test specimen.

The space shuttle has several engines that exert

thrust on the shuttle when they are fired. By

firing different engines, the pilot can control

the motion and direction of the shuttle.

LINEAR AND ANGULAR MOMENTUM

The linear momentum of a rigid body is L m vG

The angular momentum of a rigid body is HG

IG w

LINEAR AND ANGULAR MOMENTUM (continued)

Translation.

When a rigid body undergoes rectilinear or

curvilinear translation, its angular momentum is

zero because ? 0.

Therefore L m vG HG 0

LINEAR AND ANGULAR MOMENTUM (continued)

Rotation about a fixed axis.

The bodys linear momentum and angular momentum

are L m vG HG IGw

HO ( rG x mvG) IGw IO w

LINEAR AND ANGULAR MOMENTUM (continued)

General plane motion.

The linear and angular momentum computed about G

are required. L m vG HG IG?

The angular momentum about point A is HA

IG? (d)mvG

PRINCIPLE OF IMPULSE AND MOMENTUM

The principle is developed by combining the

equation of motion with kinematics. The

resulting equations allow a direct solution to

problems involving force, velocity, and time.

PRINCIPLE OF IMPULSE AND MOMENTUM (continued)

The previous relations can be represented

graphically by drawing the impulse-momentum

diagram.

EXAMPLE

Given A disk weighing 50 N has a rope wrapped

around it. The rope is pulled with a force P

equaling 2 N.

Find The angular velocity of the disk after 4

seconds if it starts from rest and rolls without

slipping.

EXAMPLE (solution)

Impulse-momentum diagram

Kinematics (vG)2 r w2

GROUP PROBLEM SOLVING

Given A gear set with WA 15 N WB 10 N kA

0.5 m kB 0.35 m M 2(1 e-0.5t) N.m

Find The angular velocity of gear A after 5

seconds if the gears start turning from rest.

Plan Time is a parameter, thus

GROUP PROBLEM SOLVING (continued)

Solution

Impulse-momentum diagrams

Gear A

Gear B

GROUP PROBLEM SOLVING (continued)

Kinematics

Angular impulse momentum relation for gear A

about point A yields For gear B

GROUP PROBLEM SOLVING (continued)

and wA 14.4 rad/s

End of 19.1-2

Let Learning Continue

CONSERVATION OF MOMENTUM (Section 19.3)

Objectives a) Understand the conditions for

conservation of linear and angular

momentum b) Use the condition of conservation of

linear/ angular momentum

APPLICATIONS

A skater spends a lot of time either spinning on

the ice or rotating through the air. To spin

fast, or for a long time, the skater must develop

a large amount of angular momentum.

If the skaters angular momentum is constant, can

the skater vary his rotational speed? How?

The skater spins faster when the arms are drawn

in and slower when the arms are extended. Why?

CONSERVATION OF LINEAR MOMENTUM

This equation is referred to as the conservation

of linear momentum. The conservation of linear

momentum equation can be used if the linear

impulses are small or non-impulsive.

CONSERVATION OF ANGULAR MOMENTUM

If the initial condition of the rigid body (or

system) is known, conservation of momentum is

often used to determine the final linear or

angular velocity of a body just after an event

occurs.

EXAMPLE

Given A 10 kg wheel (IG 0.156 kgm2) rolls

without slipping and does not bounce at A.

Find The minimum velocity, vG, of the wheel to

just roll over the obstruction at A.

Note Since no slipping or bouncing

occurs, the wheel pivots about point A. The force

at A is much greater than the weight, and since

the time of impact is very short, the weight can

be considered non-impulsive. The reaction force

at A, since we dont know either its direction or

magnitude, can be eliminated by applying the

conservation of angular momentum equation about A.

EXAMPLE (solution)

y

Impulse-momentum diagram

x

Conservation of angular momentum (HA)1

(HA)2 r ' (mvG)1 IG w1 r (mvG)2 IG w2 (0.2

- 0.03)(10)(vG)1 0.156 w1 0.2(10)(vG)2

0.156 w2

Kinematics no slip, w vG / r 5 vG

.Substituting and solving

(vG)2 0.892 (vG)1

EXAMPLE (continued)

To complete the solution, conservation of energy

can be used. Since it cannot be used for the

impact (why?), it is applied just after the

impact. In order to roll over

the bump, the wheel must go to position 3 from 2.

When (vG)2 is a minimum, (vG)3 is zero. Why?

T2 V2 T3 V3

½(10)(vG)22 ½(0.156)w22 0 0 98.1

(0.03)

Substituting w2 5 (vG)2 and (vG)2 0.892

(vG)1 and solving yields (vG)1 0.729 m/s,

(vG)2 0.65 m/s

GROUP PROBLEM SOLVING

Given A slender rod (Wr 5 N) has a wood block

(Ww 10 N) attached. A bullet (Wb 0.2 N) is

fired into the center of the block at 1000 m/s.

Assume the pendulum is initially at rest and the

bullet embeds itself into the block.

Find The angular velocity of the pendulum just

after impact.

GROUP PROBLEM SOLVING (continued)

Solution

To use conservation of angular momentum,

First draw a FBD.

IA ? IA 7.2kg.m2

GROUP PROBLEM SOLVING (continued)

Apply the conservation of angular momentum

equation

- Solving yields
- w2 7.0 rad/s

End of 19.3

Let Learning Continue

CHAPTER 22 VIBRATIONS (section 22.1)

Objectives Discuss Undamped single degree of

freedom (SDOF) vibration of a rigid body

Vibration definition

- Vibration is the periodic motion of a body or

system of connected bodies displaced from a

position of equilibrium - Types of vibrations
- Free vibrations motion is maintained by

conservative forces (like gravitational or

elastic restoring forces). - Forced vibrations motion is caused by external

periodic or intermittent forces.

Undamped free response for SDOF systems

- If you displace mass m a distance u from

equilibrium position and release it, vibration

occurs

Undamped free response for SDOF systems (cont)

- Equilibrium eq
- A linear, constant coefficients, homogeneous,

second order ordinary differential equation

- A and B are found from initial conditions

Tn natural period of vibration

Undamped free response for SDOF systems

- ?n natural circular frequency of vibration

fn 1/Tn natural cyclic frequency of vibration

in hertz (cycles per second)

Solving initial conditions

Example 1

Example 1 (continued)

Example 1 (continued)

- By analogy

By analogy

Example 2

Example 2 (continued)

From equilibrium

Example 2 (continued)

- Thus

Group Solving problem

- Draw FBD
- Write equation of motion

- Use analogy

End of 22.1

Let Learning Continue

Energy method Example 1

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Energy methods Example 2

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Undamped forced vibrations

Free Body Diagram

Equilibrium Equation

Solution for constant loading

Solving initial conditions

Sine curve

90

Three cases

Case 1 ? gt 1 (overdamping) ?1

and ?2 lt 0

91

92

Case 3 ?lt 1 (underdamping)

Define damped natural frequency

Period ?d 2?/?d

93

It is because ? c/(2m?n), measurement of ?

determines the viscous damping constant c.

94

B. Viscous Damped Forced vibration

The equation of motion

95

96

Maximum M occurs at

The resonance frequency is

97

tan ?

Consider the following regions (1) ? is small,

tan? gt 0, ? ? 0, xp in phase with the driving

force

(2) ? is large, tan? lt 0, ? ? 0-, ? ?, xp lags

the driving force by 90o

(3) ? ? ?n-, tan? ??, ? ? ?/2(-) ? ? ?n,

tan? ?-?, ? ? ?/2()

98

If the driving force is not applied to the mass,

but is applied to the base of the system

If b?2 is replaced by Fo/m

This can be used as a device to detect

earthquake.

99

Example

m 45 kg, k 35 kN/m, c 1250 N.s/m, p 4000

sin (30 t) Pa, A 50 x 10-3 m2.

Determine (a) steady-state displacement

(b) max. force transmitted to the

base.

100

The amplitude of the steady-state vibration is

101

The force transmitted to the base is

For max Ftr

102

End of 22.2

Let Learning Continue

Newton Laws, Energy Theorems and Momentum

Principles

- Integration Principles

Group Working Problems Problem 1 Chapters 17, 18

and 19

- Given The 30-kg disk is pin connected at its

center. If it starts from rest, - Find
- the number of revolutions it must make and the

time needed to attain an angular velocity of

20rad/s?

Problem 1 Newton Laws

- Find angular acceleration (11.7rad/s2)
- Find angle (17.1 radians)
- Find number of revolutions (2.73)
- Find time (1.71 sec)

Problem 1 Energy Theorems

- Find angle (17.1 radians)
- Find number of revolutions (2.73)
- Find angular acceleration (11.7rad/s2)
- Find time (1.71 sec)

Problem 1 Momentum Principle

- Find time (1.71 sec)
- Find angular acceleration (11.7rad/s2)
- Find angle (17.1 radians)
- Find number of revolutions (2.73)

Problem 2 Chapter 22 in view of Chapters 17, 18

and 19

- Given a 10-kg block suspended from a cord

wrapped around a 5kg disk. - Find natural period of vibration

Problem 2 Newton Laws

- Find differential equation
- Find period (1.57s)

Problem 2 Energy Theorems use conservation of

energy

- Write conservation of energy equation
- Differentiate equation
- Find period (1.57s)

Problem 2 Momentum Principle use Md(HG)/dt

- Write Md(HG)/dt for a displacement ? from

equilibrium position - Differentiate equation
- Find period (1.57s)

End of Integration Principles

Let Learning Continue