Rectilinear Kinematics Erratic Motion.pdf 4w1h6m

Dynamics MCB 2043 Rectilinear Kinematics: Erratic Motion

Dereje Engida Woldemichael (PhD, CEng MIMechE) [email protected] May 2013 Semester

Lesson Outcomes

At the end of this lecture you should be able to:

Determine position, velocity, and acceleration of a particle using graphs.

Overview



Rectilinear kinematics: erratic motion s-t, v-t, a-t, v-s, and a-s diagrams

Rectilinear Kinematics: Erratic Motion Application

In many experiments, a velocity versus position (v-s) profile is obtained. If we have a v-s graph for the tank truck, how can we determine its acceleration at position s = 1500 m?

Erratic Motion Graphing provides a good way to handle complex motions that would be difficult to describe with formulas. Graphs also provide a visual description of motion and reinforce the calculus concepts of differentiation and integration as used in dynamics. The approach builds on the facts that slope and differentiation are linked and that integration can be thought of as finding the area under a curve.

S-t Graph Plots of position vs. time can be used to find velocity vs. time curves. Finding the slope of the line tangent to the motion curve at any point is the velocity at that point (or v = ds/dt).

Therefore, the v-t graph can be constructed by finding the slope at various points along the s-t graph.

V-t Graph Plots of velocity vs. time can be used to find acceleration vs. time curves. Finding the slope of the line tangent to the velocity curve at any point is the acceleration at that point (or a = dv/dt). Therefore, the acceleration vs. time (or a-t) graph can be constructed by finding the slope at various points along the v-t graph. Also, the distance moved (displacement) of the particle is the area under the v-t graph during time ∆t.

A-t Graph Given the acceleration vs. time or a-t curve, the change in velocity (∆v) during a time period is the area under the a-t curve. So we can construct a v-t graph from an a-t graph if we know the initial velocity of the particle.

a-s Graph A more complex case is presented by the acceleration versus position or a-s graph. The area under the a-s curve represents the change in velocity (recall ∫ a ds = ∫ v dv ). s2

½ (v1² – vo²) =

∫ s1

a ds=

area under the a-s graph

This equation can be solved for v1, allowing you to solve for the velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance.

v-s graph Another complex case is presented by the velocity vs. distance or v-s graph. By reading the velocity v at a point on the curve and multiplying it by the slope of the curve (dv/ds) at this same point, we can obtain the acceleration at that point. Recall the formula a = v (dv/ds). Thus, we can obtain an a-s plot from the v-s curve.

Summary of Graphical Interpretation a=slope of v-t curve

v=slope of s-t curve

v= s

ds dt

a=

v

a

a=

ds v = = s& dt

t

dv = v& dt

1

a

v

1

t1

dv dt

t

t

t1

t

dt

s2

t2

s1

t1

s2 − s1 = ∫ ds = ∫ vdt

s=area under v-t curve

t1

t 2

2

t

t v2

t2

v1

t1

v 2 − v1 = ∫ dv = ∫ adt

v=area under a-t curve

t dt

t 2

Summary of Graphical Interpretation (Cont’d) area under a-s curve

∫

v2

v1

vdv =

∫

s2

s1

a ( s )ds

or

s2 1 2 v 2 − v12 = ∫ a ( s )ds s1 2

(

)

v

a

1

dv ds

a s s1

s ds

s2

s1

 dv  a = v   ds  a = v x slope of v-s curve

s

s2

s

Graphical Interpretation of Constant Acceleration a

Assuming t0=0

a =Const. a –t Curve

O

t v v=v0+at at

v –t Curve v0

v v0

O

t s 1 s = s0 + v0 t + at 2 2

s –t Curve

s0 O

1 v 0 t + at 2 2

s0 t

s t

Example #1 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.

What is your plan of attack for the problem?

EXAMPLE #1 (continued) Solution: The v-t graph can be constructed by finding the slope of the s-t graph at key points. What are those? when 0 < t < 5 s;

v0-5 = ds/dt = d(3t2)/dt = 6 t m/s

when 5 < t < 10 s; v5-10 = ds/dt = d(30t−75)/dt = 30 m/s v(m/s) v-t graph 30

t(s) 5

10

Example #1 (continued) Similarly, the a-t graph can be constructed by finding the slope at various points along the v-t graph. when 0 < t < 5 s;

a0-5 = dv/dt = d(6t)/dt = 6 m/s2

when 5 < t < 10 s; a5-10 = dv/dt = d(30)/dt = 0 m/s2 a-t graph

a(m/s2) 6

t(s) 5

10

Example #2

Given: The v-t graph shown. Find: The a-t graph, average speed, and distance traveled for the 0 - 90 s interval. Plan: 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!).

Example #2 (continued) Solution:

Find the a–t graph: For 0 ≤ t ≤ 30

a = dv/dt = 1.0 m/s²

For 30 ≤ t ≤ 90

a = dv/dt = -0.5 m/s² a(m/s²)

a-t graph

1 30 -0.5

90 t(s)

Example #2 (continued) 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

Example #3 A motorcycle starts from rest and travels on a straight road with a constant acceleration of 5 m/s2 for 8 sec, after which it maintains a constant speed for 2 sec. Finally it decelerates at 7 m/s2 until it stops. Plot a-t, v-t diagrams for the entire motion.Determine the total distance travelled. a (m/s2)

Sketch a-t diagram from the known accelerations, thus

 5 (0 ≤ t < 8 s )  a =  0 (8 ≤ t < 10s) − 7 (10 ≤ t ≤ t ' ) 

5

(segment I) (segment II)

8

(segment III)

v

∫

0

t

∫

dv =

0

5 dt

v = 5t

When t =8 s, v =5× ×8= 40m/s. Using this as the initial condition for segment II, thus

∫

8 ≤ t < 10 s

v

40

∫

dv =

t

8

0 dt

v = 40m / s

Similarly, for segment III

10 ≤ t ≤ t '

∫

v

40

dv =

∫

t

10

( − 7 ) dt

10

-7

Since dv=adt, the v-t diagram is determined by integrating the straight line segments of a-t diagram. Using the initial condition t=0, v=0 for segment I, we have

0 ≤ t < 8s

t (s)

v = −7 t + 110

a-t Diagram

t' (=15.71)

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

0 = −7 t '+110

t ' = 15 .71 s

v (m/s) 40

Thus, the velocity as the function of time can be expressed as

s1

s2 8

5t (0 ≤ t < 8 s )   v= 40 (8 ≤ t < 10s ) − 7t + 110 (10 ≤ t ≤ 15.71s ) 

s3 10

t (s) 15.71

v-t Diagram

The total distance travelled (using the area under v-t diagram)

1 1   s = s1 + s 2 + s 3 =  × 8 × 40  + (2 × 40 ) +  × 5 .71 × 40  = 354 .2 m 2  2 

Example #4 A test projectile is fired horizontally into a viscous liquid with a velocity v0.The retarding force is proportional to the square of the velocity, so that the acceleration becomes a=-kv2. Derive expressions for distance D travelling in the liquid and the corresponding time t required to reduce the velocity to v0/2.Neglect any vertical motion. Note the acceleration a is non-constant.

vdv = ads = −kv 2 ds

Using

∫

D

0

ds =

v0 2 v0

∫

v0 2

 ln v  D = −   k  v0 Using

a= v0 2 v0

∫

v

0 vdv dv 2 = − ∫v0 kv − kv 2

v0 1 ln 2 0 .693 = − ln 2 = = k v0 k k

dv = − kv 2 dt

dv = − kv 2

∫

t

0

dt

t=

1 k

v0 2

1 1 = v  kv 0   v0

Example #5 ax (m/s2) The acceleration of a particle which moves in the positive s-direction varies with its position as 0.4 shown. If the velocity of the particle is 0.8 m/s when s=0, determine the velocities v of the particle 0.2 when s=0.6 and 1.4 m.

Using

∫

s

0

v

v  v −v ads = ∫ vdv =   = v0 2  2  v0 v

2

2

2 0

0.4 0.6 0.8

s (m) 1.2 1.4

For x=0.6m 0 .6 1   v = v02 + 2 ∫ ads = 0 .8 2 + 2 ×  ( 0 .4 × 0 .4 ) + ( 0 .3 + 0 .4 ) × 0 .2  = 1 .05 m / s 0 2  

Area under ax-x curve (0≤x ≤ 0.6) For x=1.4m 1.4 1   v = v02 + 2 ∫ ads = 0.8 2 + 2 × (0.4 × 0.4) + (0.2 + 0.4) × 0.4 + 0.4 × 0.2 + 0 = 1.17 m / s 0 2  

Area under ax-x curve (0≤x ≤ 1.4)

Where v0=0.8 m/s

Example #6 The v-s diagram for a testing vehicle travelling on a v (m/s) straight road is shown. Determine the acceleration of the vehicle at s=50 m and s=150 m. Draw the 8 a-s diagram. s (m) Since the equations for segments of v-s diagram are given, we can use ads=vdv to determine a-s diagram.

0 ≤ s < 100 m a=v

100

v = 0.08 s

dv d = ( 0 . 08 s ) ( 0 . 08 s ) = 0 . 0064 s ds ds

200

a (m/s2) 0.64

100 ≤ s ≤ 200 m v = −0.08 s + 16 d a = ( − 0 . 08 s + 16 ) ( − 0 . 08 s + 16 ) = 0 .0064 s − 1 .28 ds -0.64 When s=50 m, then

100

a = 0.0064× 50 = 0.32m / s 2 (acceleration in segment I)

2 When s=150 m, then a = 0.0064× 150− 1.28 = −0.32m / s (deceleration in segment II)

200 s (m)

Summary Questions 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 the area under the ___ graph during that time. A) a-t

B) a-s

C) v-t

D) s-t

Summary Questions (continued) 3. If a particle starts from rest and accelerates according to the graph shown, the particle’s velocity at t = 20 s is A) 200 m/s

B) 100 m/s

C) 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) 5. If a car has the velocity curve shown, determine the time t necessary for the car to travel 100 meters. v A) 8 s

B) 4 s

C) 10 s

D) 6 s

75

6s

t

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

a t

A)

C)

t

B)

a

v

a t

D)

t t

References:

R.C. Hibbeler, Engineering Mechanics: Dynamics, SI 13th Edition, Prentice-Hall, 2012.