CEE 271: Applied Mechanics II, Dynamics– Lecture 27: Ch.18, Sec.1–5 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Tuesday, November 27, 2012

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KINETIC ENERGY, WORK, PRINCIPLE OF WORKAND ENERGY

Today’s objectives: Studentswill be able to

1 Define the various ways aforce and couple do work.

2 Apply the principle of workand energy to a rigid body.

In-class activities:• Reading Quiz• Applications• Kinetic Energy• Work of a Force or Couple• Principle of Work and

Energy• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 Kinetic energy due to rotation (only) of the body is definedas(a) 1

2m(vG)2

(b) 12m(vG)2 + (1/2)IGω

2

(c) 12IGω

2

(d) IGω2

ANS: (c)

2 When calculating work done by forces, the work of aninternal force does not have to be considered because

(a) internal forces do not exist(b) the forces act in equal but opposite collinear pairs(c) the body is at rest initially(d) the body can deform

ANS: (b)

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APPLICATIONS

• The work of the torque (or moment) developed by thedriving gears on the two motors on the concrete mixer istransformed into the rotational kinetic energy of the mixingdrum.

• If the motor gear characteristics are known, how would youfind the rotational velocity of the mixing drum?

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APPLICATIONS(continued)

• The work done by the soil compactor’s engine istransformed into (1) the translational kinetic energy of theframe and (2) the translational and rotational kinetic energyof the roller and wheels (excluding the internal kineticenergy developed by the moving parts of the engine anddrive train).

• Are the kinetic energies of the frame and the roller relatedto each other? If so, how?

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KINETIC ENERGY (Section 18.1)

• The kinetic energy of a rigid body can be expressed as thesum of its translational and rotational kinetic energies. Inequation form, a body in general plane motion has kineticenergy given by:

T =1

2m(vG)2 +

1

2IGω

2

• Several simplifications can occur.

1. (Pure) Translation: When a rigidbody is subjected to onlycurvilinear or rectilineartranslation, the rotational kineticenergy is zero (ω = 0).Therefore,

T =1

2m(vG)2

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KINETIC ENERGY (continued)

2. Pure Rotation: When a rigid body isrotating about a fixed axis passing throughpoint O, the body has both translationaland rotational kinetic energy. Thus,

T = 12mv

2G + 1

2IGω2

Since vG = rGω where rG = OG, we canexpress the kinetic energy of the body as:

T =1

2[IG +mr2G]ω2 =

1

2IOω

2

which is the parallel axis theorem.• If the rotation occurs about the mass center, G, then what

is the value of vG? In this case, the velocity of the masscenter is equal to zero. So the kinetic energy equationreduces to: T = 1

2IGω2.

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THE WORK OF A FORCE (Sec. 18.2)

• Recall that the work done by a force can be written as:UF =

∫F · dr =

∫(F cos θ)ds

• When the force is constant, this equation reduces toUFc = (Fc cos θ)s where Fc cos θ represents the componentof the force acting in the direction of the displacement, s.

• Work of a weight (W = mg): As before, thework can be expressed as UW = −W∆y(+ ↑). Remember, if the force andmovement are in the same direction, thework is positive.

• Work of a spring force: For a linear spring,the work is:

Us =

∫ s2

s1

−kxdx = −1

2k[(s2)

2 − (s1)2]

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FORCES THAT DO NO WORK

There are some external forces that do no work.1 For instance, reactions at fixed supports do no work

because the displacement at their point of application iszero.

2 Normal forces and friction forces acting onbodies as they roll without slipping over arough surface also do no work since there isno instantaneous displacement of the pointin contact with ground (it is an instant center,IC).

3 Internal forces do no work because they always act inequal and opposite pairs. Thus, the sum of their work is zero.

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THE WORK OF A COUPLE (Sec. 18.3)

• When a body subjected to a coupleexperiences general plane motion, thetwo couple forces do work only whenthe body undergoes rotation.

• If the body rotates through an angulardisplacement dθ, the work of the couplemoment, M , is:

UM =∫ θ2θ1M · dθ

• If the couple moment, M , is constant, thenUM = M(θ2 − θ1)

• Here, the work is positive, provided M and (θ2 − θ1) are inthe same direction.

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PRINCIPLE OF WORK AND ENERGY (Sec. 18.4)

• Recall the statement of the principle of work and energyused earlier:

T1 + ΣU1−2 = T2

• In the case of general plane motion, this equation statesthat the sum of the initial kinetic energy (T1, bothtranslational and rotational) and the work done (ΣU1−2) byall external forces and couple moments equals the body’sfinal kinetic energy (T2, translational and rotational).

• This equation is a scalar equation. It can be applied to asystem of rigid bodies by summing contributions from allbodies.

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EXAMPLE

• Given: The disk weighs 40 lb andhas a radius of gyration (kG) of0.6 ft. A 15 ft · lb moment isapplied and the spring has aspring constant of 10 lb/ ft.

• Find: The angular velocity of the wheel when point Gmoves 0.5 ft. The wheel starts from rest and rolls withoutslipping. The spring is initially un-stretched.

• Plan: Use the principle of work and energy to solve theproblem since distance is the primary parameter. Draw afree body diagram of the disk and calculate the work of theexternal forces.

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EXAMPLE (Solution)

• Free body diagram of the disk⇒• Since the disk rolls without

slipping on a horizontal surface,only the spring force and couplemoment M do work. Why don’tforces FB and NB do any work?

• Since the spring is attached tothe top of the wheel, it will stretchtwice the amount of displacementof G (or 1 ft).

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EXAMPLE (continued)

• Work:U1−2 = −1

2k[s22 − s21] +M(θ2 − θ1)U1−2 = −1

2(10)(12 − 0) + 15(0.5/0.8)

= 4.375ft · lb

• Kinematic relation: vG = rω = 0.8ω

• Kinetic energy:T1 = 0

T2 = 12m(vG)2 + 1

2IGω2

= 12( 40

32.2)(0.8ω)2 + 12( 40

32.2)(0.6)2ω2

= 0.621ω2

• Work and energy: from T1 + U1−2 = T2, one gets0 + 4.375 = 0.621ω2, i.e., ω = 2.65 rad/s

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CONCEPT QUIZ1 If a rigid body rotates about its center of gravity, its

translational kinetic energy is at all times.(a) constant(b) zero(c) equal to its rotational kinetic energy(d) Cannot be determined

ANS: (b)2 A rigid bar of mass m and length L is released from rest in

the horizontal position. What is the rod’s angular velocitywhen it has rotated through 90◦?

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L

m (a)√g/3L

(b)√

3g/L

(c)√

12g/L

(d)√g/L

ANS: (b)15 / 42

GROUP PROBLEM SOLVING

• Given: The 50 kg pendulum of theCharpy impact machine isreleased from rest when θ = 0.The radius of gyrationkA = 1.75 meter.

• Find: The angular velocity of thependulum when θ = 90◦.

• Plan: Since the problem involves distance, the principle ofwork and energy is an efficient solution method. The onlyforce involved doing work is the weight, so only its workneed be determined.

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GROUP PROBLEM SOLVING (Solution)

• Calculate the vertical distance the mass center moves.Because

∆y = 1.25 sin θ

We can determine the work due to theweight (W = mg) from θ = 0 to π/2.

UW = −W∆y

= 50(9.81)(0.0− 1.25)

= 613.1N ·m

• The mass moment of inertia about A is:IA = m(kA)2 = 50(1.75)2 = 153.1 kg ·m2

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GROUP PROBLEM SOLVING (continued)

• Kinetic energy:

T1 = 0

T2 =1

2IAω

2

=1

2(153.1)ω2

• Now apply the principle of work and energy equation:

T1 + U1−2 = T2

0 + 613.1 = 76.55ω2

ω = 2.83 rad/s

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ATTENTION QUIZ1 A disk and a sphere, each of mass m and radius r, are

released from rest. After 2 full turns, which body has alarger angular velocity? Assume roll without slip.

θ

r

(a) Sphere(b) Disk

(c) Equal.(d) Cannot

ANS: (a)

2 A slender bar of mass m and length L is released from restin a horizontal position. The work done by its weight whenit has rotated through 90◦ is?

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L

m (a) mg(π/2)

(b) mgL

(c) mg(L/2)

(d) −mg(L/2)

ANS: (c)19 / 42

PLANAR KINETICS OF A RIGID BODY:CONSERVATION OF ENERGY

Today’s objectives: Studentswill be able to

1 Determine the potentialenergy of conservativeforces.

2 Apply the principle ofconservation of energy.

In-class activities:• Reading Quiz• Applications• Potential Energy• Conservation of Energy• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 Elastic potential energy is defined as .(a) + 1

2k(s)2

(b) − 12k(s)2

(c) + 12k(v)2

(d) None of the aboveANS: (a)

2 The kinetic energy of a rigid body consists of the kineticenergy due to .(a) translational motion and rotational motion(b) only rotational motion(c) only translational motion(d) the deformation of the body

ANS: (a)

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APPLICATIONS

• The torsion springs located at thetop of the garage door wind up asthe door is lowered.

• When the door is raised, thepotential energy stored in the springis transferred into the gravitationalpotential energy of the door’s weight,thereby making it easy to open.

• Are parameters such as the torsionalspring stiffness and initial rotationangle of the spring important whenyou install a new door?

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APPLICATIONS(continued)

• Two torsional springs are used to assistin opening and closing the hood of thetruck.

• Assuming the springs are uncoiledwhen the hood is opened, can wedetermine the stiffness of each springso that the hood can easily be lifted,i.e., practically no external force appliedto it, when a person is opening it?

• Are the gravitational potential energy ofthe hood and the torsional springstiffness related to each other? If so,how?

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CONSERVATION OF ENERGY (Section 18.5)

• The conservation of energy theorem is a ‘simpler’ energymethod (recall that the principle of work and energy is alsoan energy method) for solving problems.

• Once again, the problem parameter of distance is a keyindicator for when conservation of energy is a goodmethod to solve a problem.

• If it is appropriate for the problem, conservation of energy iseasier to use than the principle of work and energy.

• This is because the calculation of the work of aconservative force is simpler. But, what makes a forceconservative?

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CONSERVATIVE FORCES

• A force F is conservative if the work done by the force isindependent of the path.

• In this case, the work depends only on the initial and finalpositions of the object with the path between the positionsof no consequence.

• Typical conservative forces encountered in dynamics aregravitational forces (i.e., weight) and elastic forces (i.e.,springs).

• What is a common force that is not conservative?ANS: friction and drag

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CONSERVATION OF ENERGY

• When a rigid body is acted upon by a system ofconservative forces, the work done by these forces isconserved. Thus, the sum of kinetic energy and potentialenergy remains constant. This principle is calledconservation of energy and is expressed as:

T1 + V1 = T2 + V2 = Constant

• In other words, as a rigid body moves from one position toanother when acted upon by only conservative forces,kinetic energy is converted to potential energy and viceversa.

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GRAVITATIONAL POTENTIAL ENERGY

• The gravitational potential energy of an object is a functionof the height of the body’s center of gravity above or belowa datum.

• The gravitational potentialenergy of a body is found bythe equation

Vg = WyG

• Gravitational potential energy is positive when yG ispositive, since the weight has the ability to do positive work(why is it positive?) when the body is moved back to thedatum.

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ELASTIC POTENTIAL ENERGY

• Spring forces are also conservative forces.

• The potential energy of a springforce (F = ks) is found by theequation

Ve = +12ks

2

• Notice that the elastic potential energy is always positive.

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PROCEDURE FOR ANALYSIS

• Problems involving velocity, displacement and conservativeforce systems can be solved using the conservation ofenergy equation.

1 Potential energy: Draw two diagrams: one with the bodylocated at its initial position and another at the final position.Compute the potential energy at each position usingV = Vg + Ve, where Vg = WyG and Ve = 1

2ks2.

2 Kinetic energy: Compute the kinetic energy of the rigidbody at each location. Kinetic energy has two components:translational kinetic energy, 1

2mv2G, and rotational kinetic

energy, 12IGω

2.3 Apply the conservation of energy equation.

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EXAMPLE I

• Given: The rod AB has a mass of 10kg. Piston B is attached to a springof constant k = 800 N/m. The springis un-stretched when θ = 0◦.Neglect the mass of the pistons.

• Find: The angular velocity of rod AB at θ = 0◦ if the rod isreleased from rest when θ = 30◦.

• Plan: Use the energy conservation equation since allforces are conservative and distance is a parameter(represented here by θ). The potential energy and kineticenergy of the rod at states 1 and 2 will have to bedetermined.

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EXAMPLE I (Solution): Potential Energy

• Initial Position • Final Position

• Let’s put the datum in line with the rod when θ = 0◦.Then, the gravitational potential energy and the elasticpotential energy will be zero at position 2: V2 = 0

• Gravitational potential energy at 1:Vg1 = −(10)(9.81)(120.4 sin 30◦)

• Elastic potential energy at 1: Ve1 = 12(800)(0.4 sin 30◦)2

• So V1 = Vg1 + Ve1 = −9.81 + 16.0 = 6.19 N ·m

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EXAMPLE I (Solution): Kinetic Energy

• Initial Position • Final Position

• The rod is released from rest from position 1: T1 = 0.• At position 2, the angular velocity is ω2 and the velocity at

the center of mass is vG2:

T2 =1

2(10)(vG2)

2 +1

2

(1

12

)(10)(0.42)(ω2)

2

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EXAMPLE I(continued)

• At position 2, point A is theinstantaneous center of rotation so thatvG2 = rG/ICω = 0.2ω2 and

T2 = 0.2(ω2)2 + 0.067(ω2)

2

= 0.267(ω2)2

• Now apply the conservation of energy equation and solvefor the unknown angular velocity, ω2.

T1 + V1 = T2 + V2

0 + 6.19 = 0.267(ω2)2 + 0

ω2 = 4.82 rad/s

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EXAMPLE II

• Given: The 30 kg rod isreleased from rest whenθ = 0◦. The spring isunstretched when θ = 0◦.

• Find: The angular velocityof the rod when θ = 30◦.

• Plan: Since distance is a parameter and all forces doingwork are conservative, use conservation of energy.Determine the potential energy and kinetic energy of thesystem at both positions and apply the conservation ofenergy equation.

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EXAMPLE II (Solution): Potential Energy

• Let’s put the datum in line with the rodwhen θ = 0◦. Then, the gravitationalpotential energy when θ = 30◦ is

Vg2 = −30(9.81)(121.5 sin 30◦)

= −110.4N ·m

• The elastic potential energy at θ = 0◦ is zero since thespring is un-stretched. The un-stretched length of thespring is 0.5 meter. Elastic potential energy at θ = 30◦ is

Ve2 =1

280(x− 0.5)2 = 11.09 N ·m

x2 = 22 + 1.52 − 2(2)(1.5) cos θ = 1.054 = (1.0266)2

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EXAMPLE II (continued): Kinetic Energy

• The rod is released from rest atθ = 0◦, so vG1 = 0 and ω1 = 0.Thus, the kinetic energy atposition 1 is T1 = 0.

• At θ = 30◦, the angular velocity isω2 and the velocity at the centerof mass is vG2

• Since vG2 = 0.75ω2,

T2 = 12m(vG2)

2 + 12IG(ω2)

2

= 12(30)(vG2)

2 + 12(1/12)30(1.5)2(ω2)

2

= 12(30)(0.75ω2)

2 + 12(1/12)30(1.5)2(ω2)

2

= 11.25(ω2)2

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EXAMPLE II(continued)

• Now all terms in the conservation of energy equation havebeen formulated. Writing the general equation and thensubstituting into it yields:

T1 + V1 = T2 + V2

0 + 0 = 11.25(ω2)2 + (−110.4 + 11.09)

• Solving for ω2 = 2.97 rad/s

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UNDERSTANDING QUIZ

1 At the instant shown, the spring is undeformed. Determinethe change in potential energy if the 20 kg disk(kG = 0.5 meter) rolls 2 revolutions without slipping.

(a) + 12 (200)(1.2π)2 + (20)9.81(1.2π sin 30◦)

(b) − 12 (200)(1.2π)2 − (20)9.81(1.2π sin 30◦)

(c) + 12 (200)(1.2π)2 − (20)9.81(1.2π sin 30◦)

(d) + 12 (200)(1.2π)2

ANS: (c)

2 Determine the kinetic energy of the disk at this instant.

(a) ( 12 )(20)(3)2

(b) 12 (20)(0.52)(10)2

(c) Answer (a) + Answer (b)(d) None of the above

ANS: (c)

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GROUP PROBLEM SOLVING

• Given: The 30 kg pendulum has itsmass center at G and a radius ofgyration about point G ofkG = 0.3 meter. It is released fromrest when θ = 0◦. The spring isun-stretched when θ = 0◦.

• Find: The angular velocity of the pendulum when θ = 90◦.• Plan: Conservative forces and distance (θ) leads to the

use of conservation of energy. First, determine thepotential energy and kinetic energy for both positions.Then apply the conservation of energy equation.

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GROUP PROBLEM SOLVING: Potential Energy

• Let’s put the datum when θ = 0◦.Then, the gravitational potentialenergy and the elastic potentialenergy will be zero. So,

Vg1 = Ve1 = 0Note that the un-stretched lengthof the spring is 0.15 meter.

• Gravitational potential energy at θ = 90◦:

Vg2 = −30(9.81)(0.35) = −103.0 N ·m

• Elastic potential energy at θ = 90◦ is :

Ve2 =1

2300(

√0.62 + 0.452 − 0.15)2 = 54.0 N ·m

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GROUP PROBLEM SOLVING: Kinetic Energy

• When θ = 0◦, the pendulum isreleased from rest. Thus, T1 = 0.

• When θ = 90◦, the pendulum hasa rotational motion about point O.Thus, T2 = 1

2IO(ω2)2

where

IO = IG +m(dOG)2 = (30)0.32 + 30(0.35)2 = 6.375kg ·m2

T2 = 126.375(ω2)

2

• Now, substitute into the conservation of energy equation.

T1 + V1 = T2 + V2

0 + 0 = 126.375(ω2)

2 + (−103 + 54.0)

• Solving for ω yields ω = 3.92 rad/s.41 / 42

ATTENTION QUIZ

1 Blocks A and B are released from rest and the disk turns 2revolutions. The V2 of the system includes a term for ....?

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����������

����������

datum1m

40kg

80kg

(a) only the 40 kgblock

(b) only the 80 kgblock

(c) the disk andboth blocks

(d) only the twoblocksANS: (d)

2 A slender bar is released from rest while in the horizontalposition. The kinetic energy (T2) of the bar when it hasrotated through 90◦ is?

��������������������������������������������������������

L

m

(a) 12m(vG2)2

(b) 12IG(ω2)2

(c) 12k(s1)2 −W (L/2)

(d) 12mv

2G2 + 1

2IGω22

ANS: (d)

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