lecture 10 engg mech. dyanmics

8/7/2019 Lecture 10 Engg Mech. Dyanmics

http://slidepdf.com/reader/full/lecture-10-engg-mech-dyanmics 1/57

Engineering Mechanics

(C) 2005 Pearson Education South Asia Pte Ltd. 1

DYNAMICS




Chapter Outline:

Position, Velocity and Acceleration

Straight-Line Motion

Curvilinear Motion

Relative Motion

Chapter Summary




Chapter 13: Motion of a Point

Study of motion involves describing and analyzing

the motion of a point in space (either along an

arbitrary path or trajectory) from its position,

velocity and acceleration using various coordinate

systems.




13.1 Position, Velocity and

Acceleration

A reference frame is a coordinate system

specifying the positions of points by specifying the

components of the position vector r relative to the

origin. Shown in Fig 13.1(a) and (b) are examplesfor specifying positions of objects in a room and

an airplane.





Acceleration

We can describe the position of a point P relative

to a given reference frame with origin O by the

position vector r from O to P as seen in Fig

13.2(a).





Acceleration

Suppose P is in a motion relative to the chosen

reference frame, so the r is a function of time t

(see Fig 13.2(b)).

Thus, we can express this

by notation

)(r r t !





Acceleration

The velocity of P relative to the given reference

frame at time t is defined by

)1.13()(r )(r limr v0 t

t t t dt d

t (

(!!p(

where the vector r(t+t) ± r(t) is the change in

position, or displacement, of P during the interval

of time t (refer to Fig 13.2(c)).





Acceleration

Thus, the velocity v is the rate of change of

position of P.





Acceleration

Time derivative of the sum of two vector functions

u and w is

d t

d

d t

d

d t

d wu

wu !

Time derivative of the product of a scalar function

f and a vector function u is

dt

d f

dt

df

dt

f d u

u

u

!





Acceleration

The acceleration of P relative to the given

reference frame at time t is defined by

)2.13()(v)(vlimva0 t

t t t dt d

t (

(!!

p(

where v(t+t) ± v(t) is the change in velocity of Pduring the interval of time t. (see Fig 13.3)





Acceleration

Acceleration is the rate of change of the velocity of

P at time t i.e.

)/(r r 2

2

2

sm

d t

d or

d t

d

d t

d ¹ º

¸©ª

¨





Acceleration

Let O¶ be an arbitrary fixed point and r¶ be the

position vector from O¶ to P (refer to Fig 13.4(a)).

Velocity of P relative to O¶is

d t

d 'r 'v !

Velocity of P relative toorigin O is

dt

d r v !





Acceleration

To show v¶ = v, let R

be the vector from O

to O¶ (refer to Fig

13.4(b)), such that

R r r' !





Acceleration

Since the vector R is constant, the velocity of P

relative to O¶ is

vr R r 'r 'v !!!!d t d

d t d

d t d

d t d

and it is given that from definition,

d t

d and

d t

d va

'v'a !!





Acceleration

Therefore, v¶ = v and a¶ = a.

Thus the velocity and acceleration of a point P

relative to a given reference frame do not dependon the location of the fixed reference point used to

specify the position of P




13.2 Straight-Line Motion

Description of the Motion

Consider a straight line through the origin O

of a given reference frame. Assuming the

direction of the line relative to the referenceframe is fixed, we can specify the position of

a point P relative to O by a coordinate s

measured along the line from O to P (Fig

13.5(a)).





Since s is defined to be positive to the right, thus

1.s is positive when P is to the right of O

2.s is negative when P is to the left of O

The displacement of P during an interval of time

from to to t is the change in the position s(t) ±

s(to), where s(t) denotes the position at time t.





From Fig 13.5(b), by introducing a unit vector e

that is parallel to the line in positive s direction, the

position vector of P relative to O is

er s!

Because the magnitude and direction of e are

constant, such that , therefore the velocity

of P relative to O is

0

e

!d t

d

er

vd t

ds

d t

d !!





Writing the velocity vector as , we can

obtain the scalar equation

ev v!

d t dsv !

where the velocity of point P along the straight

line is the rate of change of position s.

v





From Fig 13.6, is equal to the slope at time t of

the line tangent to the graph of s as a function of

time. Therefore the acceleration of P relative to O

is

v

eev

ad t

d vv

d t

d

d t

d !!!





By introducing the unit vector e, the position is

specified by the coordinate s and the velocity and

acceleration are governed by the equations

)3.13(d t

dsv !

)4.13(dt

d va !





Applying the chain rule of differential calculus, the

derivation of velocity w.r.t time is

d t ds

dsd v

d t d v !

An alternative expression for the acceleration is

)5.13(vd s

d va !





Analysis of the Motion

In some cases, the position s of a point of an

object is known as a function of time. Hence

methods such as radar and laser-Doppler interferometry are used. In this case, we can

obtain the velocity and acceleration as

functions of time from Eqn (13.3) & (13.4) by

differentiation.





For example in Fig 13.8, the position of the truck

from t = 2 s to t = 4 s is given by the eqn

m316 3t s !

Thus, the velocity and acceleration of the truck

during that interval of time are

22 m/s2andm/s t d t

d vat

d t

dsv !!!!





Acceleration specified as a function of time

if the acceleration is a known function of time a(t ),

we can integrate the relation

)6.13()(t ad t

d v!

w.r.t time to determine the velocity as a function of

time such that

´ ! Adt t av )(





Given A is an integration constant, we can

integrate the relation

)7.13(vd t

ds

!

to determine the position as a function of time,

´ ! Bd t v s

where B is another integration constant.





To determine the constants A and B, we can write

Eqn (13.6) asd t t ad v )(!

and integrate in terms of definite integrals:

)8.13()(0 0

´ ´!v

v

t

t d t t ad v

where is the velocity at time t0

is the velocity at an arbitrary time t

0v

v





Evaluating the integral on the left side of Eqn

(13.8), we obtain an expression for the velocity as

a function of time:

´!t

t d t t avv

0

)9.13()(0

Writing Eqn (13.7) as , and integrating it to

obtain:

d t vds !

´ ´! s

s

t

t dt vd s

0 0





Once again, by evaluating the integral on the left

side, we obtain the position as a function of time:

´! t

t d t t v s s

0

)10.13()(0

It is recommended that straight-line motion

problems be solved by using Eqns (13.3)-(13.5).It will be demonstrated in later examples.





From Eqns (13.9) & (13.10), we can observed that

The area defined by the acceleration graph of P

from t0

to t is equal to the change in velocity from t0to t. (Fig 13.9(a))





The area defined by the velocity graph of P

from t0 to t is equal to the change in position

from t0 to t. (Fig 13.9(b))





Constant Acceleration

Let the acceleration be a known constant a0. From

Eqns (13.9) & (13.10), the velocity and position asfunctions of time are

)11.13()( 000 t t avv !

)12.13()(2

1)( 2

00000 t t at t v s s !





Notice that s0 and v0 are the position and velocity

at time t0. If the acceleration is constant, the

velocity is a linear function of time. From Eqn

(13.5), we can write the acceleration as

vd s

d va !0

Rewriting as , and integrating,d sad vv 0!

´ ´!v

v

s

sd sad vv

0 00





We can obtain an equation for the velocity as a

function of position:

)13.13()(2 0020

2

s savv !

Eqns (13.11) to (13.13) are suitable to be used

only when the acceleration is constant.




Example 13.1 Straight-Line Motion

with Constant Acceleration

Question

Engineers testing a vehicle that will be

dropped by parachute estimate that the

vertical velocity of the vehicle when it reachesthe ground will be 6 m/s. If they drop the

vehicle from the test

rig in Fig 13.10, from

what height h shouldthey drop it to match the

impact velocity of the

parachute drop?






S trategy

If only the significant force acting on an object

near the earth¶s surface is its weight, the

acceleration of the object is approximatelyconstant and equal to the acceleration due to

gravity at sea level. Therefore we can assume

that the vehicle¶s acceleration during its short fall

is g = 9.81 m/s2. Integrating Eqns (13.3) & (13.4)to obtain the vehicle¶s velocity and position, and

hence its position when its velocity is 6 m/s.






S ol ution

let t = 0 when the vehicle is dropped, and let s

be the position of the bottom of the cushioning

material beneath the vehicle relative to itsposition at t = 0 (Fig a). The vehicle¶s

acceleration is a = 9.81 m/s2.

From Eqn (13.4),

2m/s81.9!! adt

d v






Integrating, we obtain

At v ! 81.9

where A is an integration constant.

Because the vehicle is at rest when it is released,

v = 0 at t = 0. Therefore A = 0 and hence,

m/s81.9 t v !






Substitute the result into Eqn (13.3), we have

t vd t

ds81.9!!

and integrating it to obtain,

Bt s ! 291.4

When t = 0, s = 0 and B = 0, hence

291.4 t s !






The time necessary for the vehicle to reach 6 m/s

as it falls is

s612.0m/s81.9

m/s6

m/s81.9 !!!

v

t

Hence, the required height h is

m83.1)612.0(91.491.4 22 !!! t h




Example 13.2 Graphical Solution of


Question

The cheetah can run as fast as 120 km/h. If

assuming the animal¶s acceleration is constant

and that it reaches top speed in 4 s, whatdistance can the cheetah cover in 10 s?






S trategy

the acceleration has a constant value for the

first 4 s and is then zero. We can determine

the distance traveled during each of these³phases´ of the motion and sum them to

obtain the total distance covered, either

analytically or graphically.






S ol ution

the top speed in terms of feet per second is

m/s33.33s3600

m1000120km/h6.1 !v!

1.Analytical Method

let a0 be the acceleration during the first 4 s.we integrate Eqn (13.4) to get






obtaining the velocity as a function of time during

the first 4 s:

? A ? A

)0(0 0

000

0 00

!

!

!´ ´

t avt

av

d t ad v

t v

v t

m/s0 t av !






When t = 4 s, v = 33.33 m/s and a0 = 8.33 m/s2.

since the velocity during the first 4 s is v = 8.33t

m/s, integrating Eqn (13.3) we have

? A

)02

(33.80

233.8

33.8

2

0

2

0

0 0

!

¼½

»

¬-

«

!

!´ ´

t s

t

s

d t t ds

t

s

s t






Obtaining the position as a function of time during

the first 4 s:

m17.4 2t s !

At t = 4 s, the position is s = 4.17(4)2 = 66.7 m.

From t = 4 s to t = 10 s, v = 33.33 m/s. Rewriting

Eqn (13.3) asdt dt vd s 33.33!!






Integrate to determine the distance traveled during

the second phase of the motion,

? A ? A

m200

)410(33.330

33.33

33.331040

0

10

4

!

!

!

!

´ ´

s

s

t s

dt d s

s

s

Hence the total distance traveled in 10 s is 66.7 m

+ 200 m = 266.7 m or 0.267 km.






2.Graphical method

By drawing a graph of the cheetah¶s velocity

as a function of time in Fig (a), the totaldistance covered is the sum of the areas

during the two phases of motion. Acceleration

is constant during the first 4 s of motion, so the

graph is linear from v = 0 at t = 0 to v = 33.33m/s at t = 4 s. Velocity is constant during the

last 6 s.






Total distance covered by the cheetah is

m7.266

m200m7.66m/s33.33s6m/s33.33s4

2

1

!

!




Example 13.3 Acceleration that is a

Function of Time

Question

Suppose that the acceleration of the train in Fig.

13.12 during the interval of time from t = 2 s to t

= 4 s is a = 2t m/s2, and at t = 2 s its velocity is v = 180 km/h. what is the train¶s velocity at t = 4 s,

and what is its displacement (change in position)

from t = 2 s to t = 4 s?





Function of Time

S trategy

We can integrate Eqns (13.3) & (13.4) to

determine the train¶s velocity and position as

functions of time.





Function of Time

S ol ution

The velocity at t = 2 s is

m/s50s3600

h1km1

m1000km/h180 !¹ º ¸©

ª¨¹

º ¸©

ª¨

Rewrite Eqn (13.4) as

dt t dt ad v 2!!





Function of Time

Given the condition v = 50 m/s at t = 2 s,

? A ? A

m/s46

250

2

2

22

22

50

50 2

!

!

!

!´ ´

t v

t v

t v

dt t d v

t v

v t





Function of Time

Rewriting Eqn (13.3), we have

d t t d t vds 462 !!

Integrating at t = 2 s, s = 0:

? A

t s

s t

t t

s

dt t d s

2

3

0

0 2

2

463

)46(

¼½

»¬-

«!

!´ ´





Function of Time

m7.9446

3

)2(463

2463

0

3

33

!

!

t t

s

t t s

Hence yielding the position as a function of time,




Function of Time

Therefore, the displacement from t = 2 s to t = 4 s

is

m11107.94)4(4643

1 3!¼½

»¬-«

m/s624642 !!v

Using our equations for the velocity and position,

the velocity at t = 4 s is

lecture 10 engg mech. dyanmics

Documents