Rectilinear Kinematics Erratic Motion.pdf

8/14/2019 Rectilinear Kinematics Erratic Motion.pdf

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DynamicsMCB 2043

Rectilinear Kinematics: Erratic Motion

May 2013 Semester

Dereje Engida Woldemichael (PhD, CEng MIMechE)[email protected]


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Lesson OutcomesAt the end of this lecture you should be able to:

Determine position, velocity, and acceleration of aparticle using graphs .


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Overview Rectilinear kinematics: erratic motion

s-t, v-t, a-t, v-s, and a-s diagrams


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Application In many experiments, avelocity versus position (v-s)

profile is obtained.

If we have a v-s graph for thetank truck, how can we

determine its acceleration atposition s = 1500 m?

Rectilinear Kinematics: Erratic Motion


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Erratic Motion

The approach builds on the facts that slope and differentiation arelinked and that integration can be thought of as finding the areaunder a curve .

Graphing provides a good wayto handle complex motions that

would be difficult to describewith formulas.

Graphs also provide a visual

description of motion andreinforce the calculus conceptsof differentiation and integrationas used in dynamics.


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S-t Graph

Plots of position vs. time can beused to find velocity vs. time

curves. Finding the slope of theline tangent to the motion curve atany point is the velocity at thatpoint (or v = ds/dt).

Therefore, the v-t graph can be

constructed by finding the slope atvarious points along the s-t graph.


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V-t Graph

Also, the distance moved(displacement) of the particle is thearea under the v-t graph during time t.

Plots of velocity vs. time can be used tofind acceleration vs. time curves.Finding the slope of the line tangent tothe velocity curve at any point is theacceleration at that point (or a = dv/dt).

Therefore, the acceleration vs. time (ora-t) graph can be constructed byfinding the slope at various pointsalong the v-t graph.


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A-t Graph

Given the acceleration vs. timeor a-t curve, the change invelocity ( v) during a timeperiod is the area under the a-tcurve.

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

graph

(v 1 v o) = = area under the a-ss 2

s 1a ds

A more complex case is presented bythe acceleration versus position or a-sgraph. The area under the a-s curve

represents the change in velocity(recall a ds = v dv ).

This equation can be solved for v 1,allowing you to solve for the velocity

at a point. By doing this repeatedly,you can create a plot of velocityversus distance .


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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

a point on the curve andmultiplying it by the slope of thecurve (dv/ds) at this same point,we can obtain the acceleration atthat point. Recall the formula

a = v (dv/ds).

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


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Summary of Graphical Interpretation

s dt

dsv ========

1

v dt

dv a ========

1

dt

v

t

dt

a

v=slope of s-t curve a=slope of v-t curve

v=area under a-t curve s=area under v-t curve

dt

dsv ====

dt

dv a ====

======== 21

2

112

t

t

s

svdt ds s s ======== 2

1

2

112

t

t

v

v adt dvvv

tt

t1 t2

s

tt1

t

t2

v

tt1 t

2

a


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Summary of Graphical Interpretation (Contd)

ds

a

==== ds

dvv a

(((( )))) ==== 21

)(2

1 21

22

s

s ds s avv ==== 2

1

2

1

)( s

s

v

v ds s a dvv or

ds

dv

1

ss 1 s s 2

v

area under a-s curve

a = v x slope of v-s curve

ss 1 s 2

a

s


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Graphical Interpretation of Constant Acceleration

Assuming t 0 =0 a =Const.

a

tO

a t Curve

v=v0+atv

tO

v t Curve

tO

s t Curve

s

200 2

1 at tv s s ++++++++====

t

s0

v0

s s0

20 2

1 at tv ++++

at

v0v


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Example #1

What is your plan of attack for the problem?

Given: The s-t graph for a sports car moving along a straight road.

Find: The v-t graph and a-t graph over the time interval shown.


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EXAMPLE #1 (continued)

Solution: The v-t graph can be constructed by finding the slopeof the s-t graph at key points. What are those?

when 0 < t < 5 s; v 0-5 = ds/dt = d(3t 2)/dt = 6 t m/s

when 5 < t < 10 s; v 5-10 = ds/dt = d(30t 75)/dt = 30 m/s

v-t graphv(m/s)

t(s)

30

5 10


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Example #1 (continued)

Similarly, the a-t graph can be constructed by finding the slope at various pointsalong the v-t graph.

when 0 < t < 5 s; a 0-5 = dv/dt = d(6t)/dt = 6 m/s 2

when 5 < t < 10 s; a 5-10 = dv/dt = d(30)/dt = 0 m/s 2

a-t graph

a(m/s 2)

t(s)

6

5 10


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Example #2

Find slopes of the v-t curve and draw the a-t graph.Find the area under the curve. It is the distance traveled.Finally, calculate average speed (using basic definitions!).

Given: The v-t graph shown.

Find: The a-t graph, averagespeed, and distancetraveled for the 0 - 90 s

interval.Plan:


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Solution:

Find the at graph:

For 0 t 30 a = dv/dt = 1.0 m/s

For 30 t 90 a = dv/dt = -0.5 m/s

a-t graph

-0.5

1

a(m/s)

30 90t(s)


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Now find the distance traveled:

s0-30 = v dt = (1/2) (30) 2 = 450 m

s30-90 = v dt= (1/2) (-0.5)(90) 2 + 45(90) (1/2) (-0.5)(30) 2 45(30)

= 900 m

s0-90 = 450 + 900 = 1350 m

vavg(0-90)

= total distance / time

= 1350 / 90

= 15 m/s


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Example #3

A motorcycle starts from rest and travels on a straight road with a constantacceleration of 5 m/s 2 for 8 sec, after which it maintains a constant speed for 2sec. Finally it decelerates at 7 m/s 2 until it stops. Plot a-t, v-t diagrams for theentire motion.Determine the total distance travelled.

Sketch a-t diagram from the known accelerations, thus

5

-7


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When v=0 (i.e. motorcycle stops)

110'70 ++++==== t s t 71.15' ====

Thus, the velocity as the function of time canbe expressed as

++++


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Example #4

A test projectile is fired horizontally into a viscous liquid with a velocityv0.The retarding force is proportional to the square of the velocity, sothat the acceleration becomes a=-kv 2 . Derive expressions for distanceD travelling in the liquid and the corresponding time t required to reducethe velocity to v0 /2 .Neglect any vertical motion.

Note the acceleration a is non-constant .

Using dskvadsvdv 2==

=

= 22

20

0

0

0

0

v

v

v

v

D

kv

dv

kv

vdvds

k kv

v

k k

v D

v

v

693.02ln2ln1ln

0

020

0

================

Using 2 kv dt

dv a ========

==== t

v

v dt

kv

dv

0

22

0

00

2 1110

0 kvv k

t

v

v

========


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Example #5

The acceleration of a particle which moves in thepositive s-direction varies with its position asshown. If the velocity of the particle is 0.8 m/swhen s=0 , determine the velocities v of the particlewhen s=0.6 and 1.4 m .

a x (m/s 2 )

s (m)

0.4

0.2

0.4 0.8 1.2Using

22

20

22

00

0

vvvvdvads

v

v

v

v

s ===

smadsvv / 17.102.04.04.0)4.02.0(2

1)4.04.0(28.02 2

4.1

0

20

=+++++=+= For x=1.4m

Where v 0 =0.8 m/s Area under a x -x curve

(0 x 1.4)

For x=0.6m

1.40.6

smadsvv / 05.12.0)4.03.0(2

1)4.04.0(28.02 2

6.0

0

20

=+++=+= Area under a x -x curve

(0 x 0.6)


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Example #6

The v-s diagram for a testing vehicle travelling on astraight road is shown. Determine the accelerationof the vehicle at s=50 m and s=150 m. Draw thea-s diagram.

v (m/s)

s (m)100 200

8

Since the equations for segments of v-s diagram are given,we can use ads=vdv to determine a-s diagram.

m s 1000


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1. The slope of a v-t graph at any instant represents instantaneous

A) velocity. B) acceleration.C) position. D) jerk.

2. Displacement of a particle in a given time interval equals thearea under the ___ graph during that time.

A) a-t B) a-s

C) v-t D) s-t

Summary Questions


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3. If a particle starts from rest andaccelerates according to the graphshown, the particles velocity att = 20 s is

A) 200 m/s B) 100 m/sC) 0 D) 20 m/s

4. The particle in Problem 3 stops moving at t = _______.

A) 10 s B) 20 s

C) 30 s D) 40 s

Summary Questions (continued)


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5. If a car has the velocity curve shown, determine the time tnecessary for the car to travel 100 meters.

A) 8 s B) 4 sC) 10 s D) 6 s

t

v

6 s

75

t

v

6. Select the correct a-t graph for the velocity curve shown.

A) B)

C) D)

a

t

a

ta

t

a

t

Summary Questions (continued)


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References:

R.C. Hibbeler, Engineering Mechanics: Dynamics,

SI 13th

Edition, Prentice-Hall, 2012.

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Rectilinear Kinematics Erratic Motion.pdf