cee 271: applied mechanics ii, dynamics lecture 1: ch.12

12.1 Introduction 12.2 Rectilinear Kinematics 12.3 Rectilinear Motion

CEE 271: Applied Mechanics II, Dynamics– Lecture 1: Ch.12, Sec.1-3h –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Tuesday, August 21, 2012

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INTRODUCTION (Sec. 12.1) & RECTILINEARKINEMATICS (Sec. 12.2): CONTINUOUS MOTION

Today’s objectives: Studentswill be able to

1 Find the kinematicquantities (position,displacement, velocity, andacceleration) of a particletraveling along a straightpath.

In-class activities:• Relations between s(t),v(t), and a(t) for generalrectilinear motion.

• Relations between s(t),v(t), and a(t) whenacceleration is constant.

• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 In dynamics, a particle is assumed to have(A) both translation and rotational motions(B) only a mass(C) a mass but the size and shape cannot be neglected(D) no mass or size or shape, it is just a point

ANS: [B]

2 The average speed in a rectilinear system is defined as(A) ∆r/∆t (∆r = displacement)(B) ∆s/∆t (∆s = length change)(C) sT /∆t (sT = total length)(D) None of the above.

ANS: [C]

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APPLICATIONS

• The motion of large objects, suchas rockets, airplanes, or cars,can often be analyzed as if theywere particles. Why?

• If we measure the altitude of thisrocket as a function of time, howcan we determine its velocity andacceleration?

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APPLICATIONS(continued)

• A sports car travels along a straight road.• Can we treat the car as a particle?

ANS: Yes

• If the car accelerates at a constant rate, how can wedetermine its position and velocity at some instant?

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An Overview of Mechanics

Mechanics: The study of how bodies react to forces acting onthem.

1 Statics: The study of bodies in equilibrium.2 Dynamics

1. Kinematics - concerned with the geometric aspects of motion2. Kinetics - concerned with the forces causing the motion

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RECTILINEAR KINEMATICS: CONTINIOUS MOTION(Section 12.2)

• A particle travels along a straight-line pathdefined by the coordinate axis s.

• The position of the particle at any instant,relative to the origin, O, is defined by theposition vector r, or the scalar s. Scalar scan be positive or negative. Typical unitsfor r and s are meters (m) or feet (ft).

• The displacement of the particle is defined as its change inposition. (Vector form: ∆r = r′ − r and Scalar form:∆s = s′ − s)

• The total distance traveled by the particle, sT , is a positivescalar that represents the total length of the path over whichthe particle travels.

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VELOCITY

• Velocity is a measure of therate of change in the position of aparticle. It is a vector quantity havingboth magnitude and direction.

• The magnitude of the velocity is calledspeed, with units of m/s or ft/s.

• The average velocity of a particle during a time interval (orelapsed time) ∆t is

vavg =∆s

∆t(1)

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VELOCITY (Cont’d)

• The instantaneous velocity is the time-derivative of position:

v =dr

dt(2)

• Speed is the magnitude of velocity:

v =ds

dt=√v · v (3)

• Average speed is the total distance traveled divided byelapsed time:

(vsp)avg =sT∆t

(4)

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ACCELERATION

• Acceleration is the rate of change in thevelocity of a particle. It is a vectorquantity. Typical units are m/s2 or ft/s2.

• The instantaneous acceleration is thetime derivative of velocity.

• Vector form:

a =dv

dt=

d2r

dt2(5)

• Scalar form:

a =dv

dt=

d2s

dt2=√a · a (6)

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ACCELERATION (Cont’d)

• Acceleration can be1 positive (speed increasing) or2 negative (speed decreasing).

• As the book indicates, the derivative equations for velocityand acceleration can be manipulated to get

ads = vdv (7)

• Derive Eq. (7) using a = dv/dt and v = ds/dt.

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SUMMARY OF KINEMATIC RELATIONS:RECTILINEAR MOTION

• Differentiate position to get velocity and acceleration.

v =dr

dt(8)

a =dv

dt(9)

a = vdv

ds(10)

• Integrate acceleration for velocity and position.∫ s

s0

ds =

∫ t

0vdt (11)∫ v

v0

dv =

∫ t

0adt (12)∫ v

v0

vdv =

∫ s

s0

ads (13)

where so and vo are the initial position and velocity at t = 0. 12 / 30


CONSTANT ACCELERATION

• The three kinematic equations can be integrated for thespecial case when acceleration is constant (a = ac) toobtain very useful equations.

• A common example of constant acceleration is gravity; i.e.,a body freely falling toward earth.

• In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2 downward.These equations are:

1∫ vv0dv =

∫ t0 acdt yields v = vo + act

2∫ ss0ds =

∫ t0 vdt yields s = so + vot + 1

2act2

3∫ vv0vdv =

∫ ss0acds yields v2 = v2o + 2ac(s− so)

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EXAMPLE

• Given: A particle travels along a straight line to the rightwith a velocity of v = (4t− 3t2) m/s where t is in seconds.Also, s = 0 when t = 0.

• Find: The position and acceleration of the particle whent = 4 s.

• Plan:1 Establish the positive coordinate, s, in the direction the

particle is traveling.2 Since the velocity is given as a function of time, take a

derivative of it to calculate the acceleration.3 Conversely, integrate the velocity function to calculate the

position.

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EXAMPLE (continued)

Solution:1 Take a derivative of the velocity to determine the

acceleration.

a =dv

dt=

d(4t− 3t2)

dt= 4− 6t (14)

a = −20 m/s2 (15)

(or in the← direction) when t = 4 s2 Calculate the distance traveled in 4s by integrating the

velocity using s = o:

v =ds

dt⇒ ds = vdt∫ s

sods =

∫ t

0(4t− 3t2)dt ⇒ s− so = 2t2 − t3

⇒ s− 0 = 2(4)2 − (4)3 ⇒ s = −32 m (or←)

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CONCEPT QUIZ

1 A particle moves along a horizontal path with its velocityvarying with time: v = 3 m/s at t = 2 s and v = −5 m/s att = 7 s. The average acceleration of the particle is(A) 0.4 m/s2 →(B) 0.4 m/s2 ←(C) 1.6 m/s2 →(D) 1.6 m/s2 ←

ANS: (D)2 A particle has an initial velocity of 30 ft/s to the left. If it

then passes through the same location 5 seconds laterwith a velocity of 50 ft/s to the right, the average velocity ofthe particle during the 5 s time interval is .(A) 10 ft/s→(B) 40 ft/s→(C) 16 ft/s→(D) 0 ft/s

ANS: (D)

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GROUP PROBLEM SOLVING

• Given: Ball A is released fromrest at a height of 40 ft at thesame time that ball B is thrownupward, 5 ft from the ground. Theballs pass one another at aheight of 20 ft.

• Find: The speed at which ball Bwas thrown upward.

• Plan: Both balls experience aconstant downward accelerationof 32.2 ft/s2 due to gravity. Applythe formulas for constantacceleration, withac = −32.2 ft/s2.

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GROUP PROBLEM SOLVING (continued)

Solution:(1) First consider ball A. With the origin defined at the ground,

ball A is released from rest ((vA)o = 0) at a height of 40 ft((sA)o = 40 ft). Calculate the time required for ball A todrop to 20 ft (sA= 20 ft) using a position equation.

sA = (sA)o + (vA)ot + 12(ac)t

2 (16)20ft = 40ft + (0)(t) + 1

2(−32.2)(t2) (17)t = 1.115 s (18)

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GROUP PROBLEM SOLVING (continued)

Solution:

(2) Now consider ball B. It is throw upward from a height of 5ft ((sB)o = 5 ft). It must reach a height of 20 ft (sB = 20 ft) atthe same time ball A reaches this height (t = 1.115 s).Apply the position equation again to ball B usingt = 1.115 s.

sB = (sB)o + (vB)ot + 12(ac)t

2 (19)20ft = 5 + (vB)o(1.115) + 1

2(−32.2)(1.115)2 (20)(vB)o = 31.4 ft/s (21)

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ATTENTION QUIZ

1 A particle has an initial velocity of 3 ft/s to the left ats0 = 0 ft. Determine its position when t = 3 s if theacceleration is 2 ft/s2 to the right.(A) 0 ft(B) 6 ft←(C) 18 ft→(D) 9 ft→

ANS: (A)2 A particle is moving with an initial velocity of v = 12 ft/s and

constant acceleration of 3.78 ft/s2 in the same direction asthe velocity. Determine the distance the particle hastraveled when the velocity reaches 30 ft/s.(A) 50 ft(B) 100 ft(C) 150 ft(D) 200 ft

ANS: (B)

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RECTILINEAR KINEMATICS: ERRATIC MOTION(Section 12.3)

Today’s objectives: Studentswill be able to

1 Determine position,velocity, and acceleration ofa particle using graphs.

In-class activities:• Reading Quiz• Applications• s− t, v − t, a− t, v − s,

and a− s diagrams• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 The slope of a v − t graph at any instant representsinstantaneous(A) velocity.(B) acceleration.(C) position.(D) jerk.

ANS: (B)2 Displacement of a particle in a given time interval equals

the area under the graph during that time.(A) a− t(B) a− s(C) v − t(D) s− t

ANS: (C)

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APPLICATIONS

• In many experiments, avelocity versus position(v − s) profile is obtained.

• If we have a v − s graph forthe tank truck, how can wedetermine its accelerationat position s = 1500 feet?

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ERRATIC MOTION (Section 12.3)

• Graphing provides a good way tohandle complex motions thatwould be difficult to describe withformulas.

• Graphs also provide avisual description of motion andreinforce the calculus concepts ofdifferentiation and integration asused in dynamics.

• The approach builds on the facts that slope anddifferentiation are linked and that integration can bethought of as finding the area under a curve.

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S − T GRAPH

• Plots of position vs. time can beused to find velocity vs. timecurves. Finding the slope of theline tangent to the motion curveat any point is the velocity at thatpoint, or

v =ds

dt

• Therefore, the v − t graph can beconstructed by finding the slopeat various points along the s− tgraph.

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V − T GRAPH

• Plots of velocity vs. time can be used tofind acceleration vs. time curves. Findingthe slope of the line tangent to the velocitycurve at any point is the acceleration atthat point, or

a =dv

dt

• Therefore, a− t graph can be constructedby finding the slope at various points alongthe v − t graph.

• Also, the distance moved (displacement) ofthe particle is the area under the v − tgraph during time ∆t.

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A− T GRAPH

• Given the acceleration vs. time ora− t curve, the change in velocity(∆v) during a time period is the areaunder the a− t curve.

• So we can construct a v − t graphfrom an a− t graph if we know theinitial velocity of the particle.

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A− S GRAPH

• A more complex case is presentedby the acceleration versus positionor a− s graph. The area under thea− s curve represents the change invelocity∫ s1

s0

ads =

∫ v1

v0

vdv =1

2(v21 − v20)

In other words, v1 = v1 (s1)

• This equation can be solved for v1,allowing you to solve for the velocityat a point. By doing this repeatedly,you can create a plot of velocityversus distance (v − s).

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V − S GRAPH

• Another complex case is presentedby the velocity vs. distance or v − sgraph. By reading the velocity v at apoint on the curve and multiplying itby the slope of the curve (dv/ds) atthis same point, we can obtain theacceleration at that point.

a = vdv

ds

• Thus, we can obtain an a− s plotfrom the v − s curve.

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End of the Lecture

Let Learning Continue

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cee 271: applied mechanics ii, dynamics lecture 1: ch.12

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