chapter 18 planar kinetics of a rigid body: work and energy (sections 18.1-18.4) objectives:...

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 ω2

Several simplifications can occur.1. Pure Translation: the

rotational kinetic energy is zero (ω = 0): T = 0.5 m (vG)2

KINETIC ENERGY (continued)

If the rotation occurs about the mass center (pure rotation), G, then vG=0

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 + 0.5IG2

Since

vG = rG, T = 0.5(IG + m(rG)2)2 = 0.5IO2

T = 0.5 IG 2

WORK OF A FORCE

The work done by a force is

UF = F•dr = (F cos θ) ds

Constant force : UFc = (Fc cos θ)s

s

Work of a weight: Uw = -WΔy.

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.

3. Internal forces because they always act in equal and opposite pairs.

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.

THE WORK OF A COUPLE

A body subjected to a couple produces work only when it undergoes rotation.

constant M: UM = M (2 – 1)

If the body rotates through an angular displacement d, the work of the couple moment, M, is

2

1

M dUM

PRINCIPLE OF WORK AND ENERGY

The principle states: U1-2 = T12

i.e. the work done by all external forces and couple moments equals the change in body’s 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.

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

Given:The 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.

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)

Work: U1-2 = -0.5k[(s2)2 – (s1)2] + M(2 – 1)

U1-2 = -0.5(10)(12 – 0) + 15(0.5/0.8) = 4.4N.m

Kinematic relation: vG = r = 0.8Kinetic energy: T1 = 0

T2 = 0.5m (vG)2 + 0.5 IG 2

T2 = 0.5(40/9.8)(0.8)2 + 0.5(40/9.8)(0.6)22

T2 = 2.0 2

Work and energy: U1-2 = T12

4.4 = 2.0 2

= 1.5 rad/s

GROUP PROBLEM SOLVING

Find: The angular velocity of the sphere when = 45° if the sphere rolls without slipping.

Given: A sphere weighing 10N rolls along a semicircular hoop. Its equals 0 when = 0.

GROUP PROBLEM SOLVING (continued)

Now calculate the _______________ :

Solution: Draw a FBD and calculate the vertical distance the mass center moves.


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:

= 13.1 rad/s

PLANAR KINETICS OF A RIGID BODY: CONSERVATION OF ENERGY (Section 18.5)

Objectives:a) Determine the potential

energy of conservative forces.

b) 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 door’s 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 body’s center above or below a datum.

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.

The gravitational potential energy is found by:

Vg = W yG

ELASTIC POTENTIAL ENERGY

Spring forces are also conservative forces.

Notice that the elastic potential energy is always +ve

The potential energy of a spring force (F = ks) is found by the equation

Ve = ½ ks2

Find: The angular velocity of rod AB at = 0° if the rod is released from rest when = 30°.

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.

EXAMPLE 1 (solution)

Potential Energy: put datum in line with rod when =0°. Gravitational and elastic energy=0 at 2. => 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

Initial Position Final Position

EXAMPLE 1 (continued)

The rod released from rest (vG1=0, 1=0). Thus, T1 = 0. At 2, the angular velocity is 2 and the velocity is vG2 .

Kinetic Energy:

Initial Position Final Position


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

At 2, vA=0. Hence, vG2 = r = 0.2 2 .

Then, T2 = 0.2 22 + 0.067 2

2 = 0.267 22

conservation of energy:

T1 + V1 = T2 + V2

0 + 6.19 = 0.26722 + 0 => 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.


Kinetic Energy:

Released from rest: vG1=1=0, T1=0

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


conservation of energy:

T1 + V1 = T2 + V2

0 +16.0 = 1.41 22 + 9

Solving , 2 = 2.23 rad/s


Find: The angle (measured down from the horizontal) to which the bar rotates before it stops its initial downward movement.

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

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 =


At 1 (when = 0):

Kinetic Energy:

At 2:


conservation of energy

Thus, = 49.9 deg.

CHAPTER 19PLANAR KINETICS: IMPULSE AND

MOMENTUM (Sections 19.1-19.2)

Today’s 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

LINEAR AND ANGULAR MOMENTUM (continued)

Translation.When a rigid body undergoes rectilinear or curvilinear translation, its angular momentum is zero because ω = 0.

Therefore:

L = m vG

HG = 0

Rotation about a fixed axis.

HO = ( rG x mvG) + IG = IO

The body’s linear momentum and angular momentum are:

L = m vG

HG = IG


General plane motion. The linear and angular

momentum computed

about G are required.

L = m vG

HG = IGω

The angular momentum about point A is HA = IGω + (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.

Linear impulse-linear momentum equation:

L1 + F dt = L2 or (mvG)1 + F dt = (mvG)2

t2

t1

t2

t1

Angular impulse-angular momentum equation:

(HG)1 + MG dt = (HG)2 or IG1 + MG dt = IG2

t2

t1

t2

t1

PRINCIPLE OF IMPULSE AND MOMENTUM (continued)

The previous relations can be represented graphically by drawing the impulse-momentum diagram.

+ =

Find: The angular velocity of the disk after 4 seconds if it starts from rest and rolls without slipping.

EXAMPLE

Given: A disk weighing 50 N has a rope wrapped around it. The rope is pulled with a force P equaling 2 N.

EXAMPLE (solution)

Impulse-momentum diagram:

Kinematics: (vG)2 = r 2t2

t1

Impulse & Momentum: (HA)1 + MA dt = (HA)2

(P t) 2 r = (mvG)2 r + IG2 = m r2 2 + 0.5 m r2 2 = 1.5 m r2 2

3.5 rad/s3(50/9.81)(0.6)

4(2)(4)

3mr

4 Pt2

(m vG)1 (m vG)2G G

IG1 IG 2

P t

F t

N t

W t

r+ =A

y

x


Find: The angular velocity of gear A after 5 seconds if the gears start turning from rest.

Plan: Time is a parameter, thus

Given: A gear set with:WA = 15 NWB = 10 NkA = 0.5 mkB = 0.35 mM = 2(1 – e-0.5t) N.m


Impulse-momentum diagrams:

Gear A: y

x

Gear B:

Solution:


Angular impulse & momentum relation for gear A about point A yields:

For gear B:

Kinematics:


and A = 14.4 rad/s

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 skater’s 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

Recall that the linear impulse and momentum relationship is

L1 + F dt = L2 or (m vG)1 + F dt = (m vG)2

t2

t1

t2

t1

If the sum of all the linear impulses acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, the linear momentum for a rigid body (or system) is constant, or conserved. So L1 = L2.

00

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

The angular impulse-angular momentum relationship is:

(HG)1 + MG dt = (HG)2 or IG1 + MG dt = IG2

t2

t1

t2

t1

00

Similarly, if the sum of all the angular impulses due to external forces acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, angular momentum is conserved . The resulting equation is referred to as the conservation of angular momentum or (HG)1 = (HG)2 .

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.

Given: A 10 kg wheel(IG = 0.156 kg·m2) 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.

EXAMPLE

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 don’t know either its direction or magnitude, can be eliminated by applying the conservation of angular momentum equation about A.

Conservation of angular momentum: (HA)1 = (HA)2

r ' (mvG)1 + IG 1 = r (mvG)2 + IG 2

(0.2 - 0.03)(10)(vG)1 + 0.156 1 = 0.2(10)(vG)2 + 0.156 2

Kinematics: no slip, = vG / r = 5 vG .Substituting and

solving: (vG)2 = 0.892 (vG)1

Impulse-momentum diagram:

+ =

x

y

EXAMPLE (solution)

EXAMPLE (continued)

T2 + V2 = T3 + V3

{½(10)(vG)22 + ½(0.156)2

2 } + 0 = 0 + 98.1 (0.03)

Substituting 2 = 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

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 overthe bump, the wheel must go to position 3 from 2. When (vG)2 is a minimum, (vG)3 is zero. Why?


Find: The angular velocity of the pendulum just after impact.

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.


To use conservation of angular momentum,

IA = ?

IA = 7.2kg.m2

Solution:

First draw a FBD.


Solving yields:2 = 7.0 rad/s

Apply the conservation of angular momentum equation:

CHAPTER 22VIBRATIONS (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

2

2

2

20

uKu m

t

um Ku

t

2

2

22

2

0

0

cos sin

n

n n

um ku

t

uu

tu A t t

A and B are found from initial conditions

Tn =natural period of vibration

ωn =natural circular frequency of vibration

n

k

m

2n

n

T

fn =1/Tn= natural cyclic frequency of vibration in hertz (cycles per second)

Undamped free response for SDOF systems:

Solving initial conditions

0

. .

0

.

0

0

( 0)

( 0)

cos sin

n

n nn

u t u A

u t u B

uu u t t

cos sinn nu A t t

Example 1

Example 1 (continued)


By analogy

By analogy

Example 2


From equilibrium

Thus:


Group Solving problem

Draw FBD

Write equation of motion

Use analogy

Energy methodExample 1

Energy methods: Example 2

Undamped forced vibrations

Free Body Diagram

Equilibrium Equation

2

2

2

2

( )

( )

uF t ku m

t

um ku F t

t

Solution for constant loading

2

2

22

2/

cos sin /

n

n n

um ku P

t

uu P m

tu A t t P k

Solving initial conditions

.

0

0

max

0, 0 0

0, 0 /

(1 cos )

2 / 2

n

static

t u B

t u A P k

Pu t

Ku P K u

cos sin /u A t t P k

Sine curve

..

2

2

sin

1cos sin sin

1 ( )

1

1 ( )

o

n n oo

n

o

n

mu ku F t

Fu C t D t t

k

DMF

90

Viscous Damped free vibration :

xmxckxF

0 kxxcxm

02 2 xxx nn

k

c

m

x0x

)(dashpot

nm

czeta

2)(

c : viscous damping constant (in N s/m)

Define: viscous damping factor (no unit)

02 22 nn Use a trial solution x = A et :

Obtain two possible ’s :

)1( 21 n )1( 2

2 nand

tt eAeAx 2121

te 1λ te 2λ are two solutions. General solution is a linear combination

and

91

Case 1 : > 1 (overdamping) : 1 and 2 < 0

tt eAeAx 21

2

1

x decays to zero without oscillation

t

x

1,

overdamped1Critical

damped

1 underdamping

Three cases:

92

Case 2: = 1 (critical damping) : 1 = 2 = -n

A1 exp(-nt) is a solution

A2 t exp(-nt) is also a solution (prove !)

The general solution is x = (A1 + A2 t )exp(-nt)

x approaches zero quickly without oscillation.

93

Case 3: < 1 (underdamping) :

ttiti nnn eeAeAx )(22 1

21

121 ndDefine damped natural frequency

d

1x 2x

1t 2t

x

t

)sin()(

]e2

2 and e2

1Let

conjucate.complex are 2 and 1 that so real, is [

)( 21

tdtnCex

iCAiCA

AAx

ttiti ndd eeAeAx

Period d = 2/d

94

22

21

1

2

12

/decrement clogarithmi Define

n

ndn

xxn

It is because = c/(2mn), measurement of determines the viscous damping constant c.

dn

dn

n

eCe

Cexx

t

t

)(2

1

1

1

cycles successive twoof ntsdisplaceme theof ratio The

95

xmtFxckxF o sin

B. Viscous Damped Forced vibration

The equation of motion :

tFkxxcxm o sin t

mFxxx o

nn sin2 2

kcm

x0x

)(dashpot

tFF O sin

)(spring

96

2/1222

2

2/1222

/2/1/1/

factor ion magnificator ratio amplitude Define

/1

/2tan

/2/1

/

nno

n

n

nn

o

kF

CM

kFC

)sin( and )cos( then),sin( Let

2

tCxtCxtCx

pp

p

tmFxxx o

nn sin2 2

97

M

n /

1

1

01.0

2.01

Maximum M occurs at: 0 })2

(])(1{[d

d 222 nn

nres 21 2The resonance frequency is

98

n /1)(resonance

02.0

1

0

2/

Consider the following regions:

(1) is small, tan > 0, 0+, xp in phase with the driving force(2) is large, tan < 0, 0-, = , xp lags the driving force by 90o

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

n+, tan -, /2(+)

tan = 2n

n

)(

99

If the driving force is not applied to the mass, but is applied to the base of the system:

tsin2 22nn bxxx

If b2 is replaced by Fo/m:

)( Bxxmkxxc

k

c

x0x

tbxB sin

m

tsin2 2nn

m

Fxxx o

This can be used as a device to detect earthquake.

Bxmkxxcxm

100

Determine : (a) steady-state displacement (b) max. force transmitted to the base.

k c

m

pA

m =45 kg, k = 35 kN/m, c = 1250 N.s/m, p = 4000 sin (30 t) Pa, A= 50 x 10-3 m2.

Example

)(498.0)9.27452/(1250)2/(/9.2745/1035/ 3

dunderdampemcsradmk

n

n

101

The amplitude of the steady-state vibration is:

m

nn

koFX

00528.02/12

79.2/30498.0222

9.27/301

31035/

310504000

2/12

/222

/1

/

rad

n

n716.12

9.27/301

9.27/30498.021tan2

/1

/21tan

#)716.130cos(00528.0)cos( mttXxP

102

The force transmitted to the base is :

)(cos)(sin

tXctkXxckxF pptr

For max Ftr :

kct

tddFtr

1tan0

o

3

31 9.46

10351030351250tan

t

#

o

o3

max

2719.46sin00528.0301250

9.46cos00528.01035

N

Ftr

Newton Laws, Energy Theorems and Momentum Principles

Integration Principles

Group Working ProblemsProblem 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





Problem 1: Momentum Principle





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 M=d(HG)/dt

Write M=d(HG)/dt for a displacement θ from equilibrium position

Differentiate equation

Find period (1.57s)

chapter 18 planar kinetics of a rigid body: work and energy (sections 18.1-18.4) objectives:...

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