CEE 271: Applied Mechanics II, Dynamics– Lecture 31: Ch.18, Sec.1–4 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Monday, November 7, 2011

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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 of the body is defined as(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 = 12m(vG)2 + 1

2IGω2

• Several simplifications can occur.

1. Pure Translation: When a rigid body issubjected to only curvilinear or rectilineartranslation, the rotational kinetic energy iszero (ω = 0). Therefore,

T = 12m(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 hasboth translational and rotational kineticenergy. Thus,

T = 12mv

2G + 1

2IGω2

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

T = 12 [IG +mr2G]ω2 = 1

2IOω2

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 = 12IGω

2

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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 =∫ s2s1−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 / 19

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 = 0.5IAω2

= 0.5(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 / 19