Kinematics of a Particle All

Chapter TwoKinematics of particles

IntroductionKinematics: is the branch of dynamics which describes

the motion of bodies without reference to the

forces that either causes the motion or are

generated as a result of the motion.

Kinematics is often referred to as the “geometry of

motion”

Examples of kinematics problems that engage

the attention of engineers.

The design of cams, gears, linkages, and other

machine elements to control or produce

certain desired motions, and

The calculation of flight trajectory for

aircraft, rockets and spacecraft.

• If the particle is confined to a specified path, as with a bead sliding along a fixed wire, its motion is said to be Constrained.

Example 1. - A small rock tied to the end of a string and whirled in a circle undergoes constrained motion until the string breaks

• If there are no physical guides, the motion is

said to be unconstrained.

Example 2. - Airplane, rocket

• The position of particle P at any time t can be described by specifying its:

- Rectangular coordinates; X,Y,Z

- Cylindrical coordinates; r,θ,z

- Spherical coordinates; R, θ,Ф

- Also described by measurements along the tangent t and normal n to the curve(path variable).

• The motion of particles(or rigid bodies) may be described by using coordinates measured from fixed reference axis (absolute motion analysis) or by using coordinates measured from moving reference axis (relative motion analysis).

Rectilinear motion• Is a motion in which a particle moving along a

straight line(one-dimensional motion)

• Consider a particle P moving along a straight

line.

Average velocity: for the time interval Δt, it is defined as

the ratio of the displacement Δs to the time interval Δt.

2.1

• As Δt becomes smaller and approaches zero in the limit, the average velocity approaches the instantaneous velocity of the particle.

2.2

ts =Vav ∆

∆

•

→∆→∆==

∆∆

== Sdtds

ts

tt 0av0limVlimV

Average acceleration

For the time interval Δt, it is defined as the ratio of the change in velocity Δv to the time interval Δt.

2.3

Instantaneous acceleration

2.4(a)

2.4(b)

tvaav ∆

∆=

•

→∆==

∆∆

= vdtdv

tv

t 0lima

sdt

sddtds

dtd

dtdv

tv

t==

==

∆∆

=→∆ 2

2

0lima

• Note:-The acceleration is positive or negative

depending on whether the velocity increasing

or decreasing.

• Considering equation 2.2 and 2.4(a) , we have

dsssdss

sd

s

dsdt

adsvdvadv

••••

••

•

• =⇒==

=⇒

==vdsdt

General representation of Relationship among s, v, a & t.

1. Graph of s Vs t

• By constructing tangent to the curve at any time t,we obtain the slope, which is the velocity v=ds/dt

2. Graph of v Vs t

• The slope dv/dt of the v-t curve at any instant gives the acceleration at that instant.

• The area under the v-t curve during time dt is vdt which is the displacement ds

• The area under the v-t curve is the net

displacement of the particle during the

interval from t1 to t2.

(area under v-t curve )

∫ ∫=2

1

2

1

s

s

t

t

vdtds

=− 22 ss

3. Graph of a vs t

• The area under the a-t curve during time dt is the net change in velocity of the particle between t1 and t2.

v2 - v1=(area under a-t curve) ∫ ∫=2

1

2

1

v

v

t

t

adtdv

4. Graph of a Vs s

• The net area under the curve b/n position

coordinates s1 and s2 is

(areas under a-s curve)

∫ ∫=2

1

2

1

v

v

s

s

adsvdv

=− )(21 2

12

2 vv

5. Graph of v vs. s

dsCBvdvv

CBdsdv

vCBds

dv

=⇒=

==1

tanθ

• The graphical representations described are useful for:-

visualizing the relationships among the several motion quantities.

approximating results by graphical integration or differentiation

when a lack of knowledge of the matimatical relationship prevents

its expression as an explicit mathematical function .

experimental data and motions that involve discontinuous

relationship b/n variables.

Methods for determining the velocity and displacement

functions

a) Constant acceleration, (a=const.)- boundary conditions

at t=0 , s=s0 and v=v0

using integratingdvadt

dtdva =⇒=

atvv

atvvadtdv

o

t

o

v

vo

+=⇒

=−⇒= ∫∫0

• Using

• Using

( )

)sa(svv

ssavvasv

adsvdvadsvdv

o

oo

s

s

s

v

v

s

o

v

ov

oo

022

222

222

−+=⇒

−=−

⇒=⇒

=⇒= ∫∫

2

2

212

)(0

attvss

attvss

dtatvds

vdtdsdtdsv

oo

oo

t

oo

s

s

++=⇒

+=−⇒

+=⇒

=⇒=

∫∫

• These relations are necessarily restrictedto the special case where the accelerationis constant.

• The integration limits depend on the initialand final conditions and for a givenproblem may be different from those usedhere.

• Typically, conditions of motion are specified by the typeof acceleration experienced by the particle.Determination of velocity and position requires twosuccessive integrations.

• Three classes of motion may be defined for:

- acceleration given as a function of time, a = f(t)

- acceleration given as a function of position, a = f(x)

- acceleration given as a function of velocity, a = f(v)

b) Acceleration given as a function of time, a=f(t)

( ) ( )( )

( ) ( ) ( )

( ) ( )( )

( ) ( ) ( )∫∫∫

∫∫∫

=−===

=−====

tttx

x

tttv

v

dttvxtxdttvdxdttvdxtvdtdx

dttfvtvdttfdvdttfdvtfadtdv

00

0

00

0

0

0

c) Acceleration given as a function of position, a = f(x)

( )

( ) ( ) ( )∫∫∫ =−==

=====

x

x

x

x

v

v

dxxfvvdxxfdvvdxxfdvv

xfdxdvva

dtdva

vdxdt

dtdxv

000

202

1221

or or

d) Acceleration given as a function of velocity, a = f(v)

( ) ( ) ( )

( )

( ) ( ) ( )

( )∫

∫∫

∫

∫∫

=−

====

=

====

v

v

v

v

x

x

v

v

tv

v

vfdvvxx

vfdvvdx

vfdvvdxvfa

dxdvv

tvf

dv

dtvf

dvdtvf

dvvfadtdv

0

00

0

0

0

0

Example 1• Consider a particle moving in a straight line, and assuming

that its position is defined by the equation

• Where, t is express in seconds and s is in meters.Determine the velocity and acceleration of the particles atany time t

326 tts −=

Example 2• The acceleration of a particle is given by ,

where a is in meters per secondsquared and t is in seconds. Determine thevelocity and displacement as function time. Theinitial displacement at t=0 is so=-5m, and theinitial velocity is vo=30m/s.

304 −= ta

Example 1• The position of a particle which moves along a straight

line is defined by the relation ,where x is expressed in m and t in

second.Determine:a) The time at which the velocity will be zero.b) The position and distance traveled by the particle at

that time.c) The acceleration of the particle at that time.d) The distance traveled by the particle between 4s and

6s.

40156 23 +−−= tttx

Example 2• A particle moves in a straight line with velocity

shown in the figure. Knowing that x=-12m at t=0

• Draw the a-t and x-t graphs, and • Determine:a) The total distance traveled by the particle when

t=12s.b) The two values of t for which the particle passes

the origin.c) The max. value of the position coordinate of the

particle.d) The value of t for which the particle is at a

distance of 15m from the origin.

Example 3• the rocket car starts from rest and

subjected to a constant acceleration ofuntil t=15sec. The brakes are then

applied which causes a decelerated at arate shown in the figure until the carstops. Determine the max. speed of thecar and the time when the car stops.

26m/sa =

Example 4• A motorcycle patrolman starts from rest at A two seconds

after a car, speeding at the constant rate of 120km/h, passes point A. if the patrolman accelerate at the rate of 6m/s2 until he reaches his maximum permissible speed of 150km/h, which he maintains, calculate the distance s from point A to the point at which he overtakes the car.

Example 5• The preliminary design for a rapid transient system calls

for the train velocity to vary with time as shown in the plot as the train runs the 3.2km between stations A and B.

• The slopes of the cubic transition curves(which are of form a+bt+ct2+dt3) are zero at the end points.

• Determine the total run time t between the stations and the maximum acceleration.

Plane curvilinear motion

Curvilinear motion of a particle• When a particle moves along a curve other

than a straight line, we say that the particle is in curvilinear motion.

Plane curvilinear motion• The analysis of motion of a particle along a

curved path that lies on a single plane.

• Consider the continuous motion of a particle along a plane curve.

- At time t, the particle is at position P, which islocated by the position vector r measured fromsome convenient fixed origin o.

- At time , the particle is at P’ located by the

position vector .

- The vector Δr joining p and p’ represents the

change in the position vector during the time

interval Δt (displacement) .

tt ∆+

rr ∆+

• The distance traveled by the particle as it

moves along the path from P to P’ is the scalar

length Δs measured along the path.

• The displacement of the particle, that

represents the vector change of position and is

clearly independent of the choice of origin.

• The average velocity of the particle between P and P’ defined as:

• which is a vector whose direction is that of .

• The instantaneous velocity,

trV av

∆∆

=

r∆v

•

→∆→∆

==∆∆

== rdt

rdtrvv

tav

tlimlim

00

Note: As ∆t approaches zero, the direction ofapproaches to the tangent of the path. Hence the

velocity V is always a vector tangent to the path.

• The derivative of a vector

is itself a vector having both

a magnitude and a direction.

r∆

•

=== sdtdsvv

Note: there is a clear distinction between the

magnitude of the derivative and the derivative

of the magnitude.

- The magnitude of the derivative.

- The derivative of the magnitude

speedvvrdt

rd⇒===

•

•

== rdtdr

dt

rd

- The rate at which the length of the position vector is changing.

• The magnitude of the vector v is called the speed of the particle.

ts

tppv

tt ∆∆

=∆

=→∆→∆

limlim0

'

0

dtdsv =

•

rr

Consider the following figure

- let the velocity at p be- let the velocity at p’ be

v

v′

• Let us draw both vectors v and v’ from the same origin o’. The vector ∆v joining Q and Q’ represents the change in the velocity of the particle during the time interval ∆t.

v’=∆v+v

• Average acceleration, of the particle between

P and P’ is defined as , which is a vector and

whose direction is that of ∆v.

• Instantaneous acceleration,

tv

∆∆

tvaav ∆

∆=

a

•••

===∆∆

== rvdtvd

tvaa av limlim

Note: The direction of the acceleration of aparticle in curvilinear motion is neithertangent to the path nor normal to thepath.

• Suppose we take the set of velocity vectors and trace out a continuous curve; such a curve is called a hodograph.

• The acceleration vector is tangent to the hodograph, but this does not produce vectors tangent to the pathof the particle.

Rectangular co-ordinates (x-y-z)

• This is particularly useful for describing

motions where the x,y and z-components of

acceleration are independently generated.

• When the position of a particle P is defined at any instant by its rectangular coordinate x,y and z, it is convenient to resolve the velocity v and the acceleration a of the particle into rectangular components.

• Resolving the position vector r of the particle into rectangular components,

r=xi+yj+zk• Differentiating

)ˆˆˆ( kzjyixdtd

dtrdv ++==

kzjyixv•••

++=

• All of the following are equivalent:

)ˆˆˆ( kzjyixdtd

dtrdv ++==

kdtdzj

dtdyi

dtdx ˆˆˆ ++=

kvjviv zyxˆˆˆ ++=

kzjyix ˆˆˆ ++=

• Since the speed is defined as the magnitude of the velocity, we have:

222zyx vvvv ++=

Similarly,

)ˆˆˆ( kvjvivdtd

dtvda zyx ++==

kdt

dvjdt

dvi

dtdv zyx ˆˆˆ ++=

kvjviv zyxˆˆˆ ++=

kzjyix ˆˆˆ ++=

• The magnitude of the acceleration vector is:

222zyx aaaa ++=

• From the above equations that the scalar components of the velocity and acceleration are

••

•

=

=

xa

xv

x

x••

•

=

=

ya

yv

y

y

••

•

=

=

za

zv

z

z

• The use of rectangular components to describe the position, the velocity and the acceleration of a particle is particularly effective when the component ax of the acceleration depends only upon t,x and/or vx,similarly for ay and az.

• The motion of the particle in the x direction, its motion in the y direction, and its motion in the z direction can be considered separately.

Projectile motion

• An important application of two – dimensional

kinematic theory is the problem of projectile

motion.

Assumptions• Neglect the aerodynamic drag, the earth

curvature and rotation,

• The altitude range is so small enough sothat the acceleration due to gravity can beconsidered constant, therefore;

• Rectangular coordinates are useful for the trajectory analysis.

• In the case of the motion of a projectile, it can be shown that the components of the acceleration are

0==••

xax gyay −==••

0==••

zaz

Boundary conditions

at t=0 ; x=x0 ,y=y0; vx=vxo and vy=vy0

Position

Velocitytvzz

gttvyy

tvxx

ozo

y

x

+=

−+=

+=

200

00

21

)(222

0

0

oyoy

zoz

yy

xx

yygvv

vzv

gtvyv

vxv

−−=

==

−==

==

•

•

•

• In all these expressions, the subscript zero denotes initial conditions

• But for two dimensional motion of the projectile,

200

00

21 gttvyy

tvxx

y

x

−+=

+=

)(2220

0

oyoy

yy

xx

yygvv

gtvyv

vxv

−−=

−==

==•

•

• If the projectile is fired from the origin O, we have xo=yo=0 and the equation of motion reduced to

20

0

21 gttvy

tvx

y

x

−=

=

gtvvvv

yy

xx

−==

0

0

ExampleA projectile is fired from the edge of a 150m cliff with an initial

velocity of 180m/s at angle of 300 with the horizontal. Neglect air resistance, find

a) the horizontal distance from gun to the point where the projectile strikes the ground.

b) the greatest elevation above the ground reached by the projectile.

ExampleA projectile is launched from point A with the initial conditions shown in the figure. Determine the slant distance s that locates the point B of impact and calculate the time of flight.

ExampleThe muzzle velocity of a long-range rifle at A is u=400m/s. Determine the two angles of elevation θ that will permit the projectile to hit the mountain target B.

Curvilinear motion Normal and tangential coordinates

Normal and tangential coordinate

• When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian.

• When the path of motion is known, normal (n) and tangential (t) coordinates are often used.

• They are path variables, which are measurements made along the tangent t and normal n to the path of the particle.

• The coordinates are considered to move along the path with the particle.

• In the n-t coordinate system, the origin is located on the particle (the origin moves with the particle).

• The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion.

• The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature of the curve.

• The coordinate n and t will now be used to describe the velocity v and acceleration a.

• Similarly to the unit vectors i and j introduced for rectangular coordinate system, unit vectors for t-n coordinate system can be used.

• For this purpose we introduce unit vector

• et in the t-direction• en in the n-direction.

• et - directed toward the direction of motion.

• en-directed toward the center of curvature of the path.

• During the differential increment of time dt, the particle moves a differential distance ds along the curve from A to A’.

• With the radius of curvatureof the path at this position designated by ρ, we see that

ds = ρdβ

velocity• The magnitude of the velocity is:-

• Since it is unnecessary to consider the

differential change in between A and A’,

dtd

dtd

dtdsvv βρβρ

====

ρ

)1....(....................tt eevv βρ==

Acceleration• The acceleration a of the particle was defined

by:

• Now differentiate the velocity by applying the ordinary rule (chain rule) for the differentiation of the product of a scalar and a vector.

( )tvedtd

dtdva ==

( )••

+=

+===

tt

ttt

eveva

dtdeve

dtdvve

dtd

dtdva

• Where the unit vector et now has a derivative because its direction changes.

. . . . . . . . . . . (1) dtdeve

dtdva t

t +=

• To find the derivative of consider the following figure

• Using vector additione’t = et + ∆et

• Since the magnitude | e’t |= | et | = 1

dtdet

• The magnitude of ∆et

| ∆et |= 2 sin ∆ѳ/2 • Dividing both sides by ∆ѳ

• As ∆ѳ→0, is tangent to the path;i.e,

perpendicular to et .

θ

θ

θ ∆

∆=

∆∆ 2sin2

te

θ∆∆ te

• Taking the limit as ∆ѳ→0

• The vector obtained in the limit is a unit vector along the normal to the path of the particle.

12

2sinlimlim

00=

∆

∆=

∆∆

→∆→∆ θ

θ

θ θθte

1lim0

==∆∆

→∆ θθθ ddee tt

• But

• Dividing both sides by dtBut dѳ = ds/ρ

• Then

ntt

n

nt

eddeddee

edde

.

.1

θθ

θ

=⇒=∴

=

nt e

dtd

dtde .θ

=

nt

nt ev

dtdee

dtds

dtde

ρρ=⇒= .1

• Equation (1) becomes

• We can write

where, and

tntt e

dtdvevae

dtdv

dtdeva ..

2

+=⇒+=ρ

ttnn eaeaa +=22 •

== θρρvan

•••

== θρvat

22tn aaaa +==

Note: • an is always directed towards the center of

curvature of the path.• at is directed towards the positive t-direction of

the motion if the speed v is increasing and towards the negative t-direction if the speed v is decreasing.

• At the inflection point in the curve, the normal acceleration, goes to zero since ρ becomes infinity. ρ

2v

Special case of motion• Circular motion

but ρ=r and ρ

2van =•

= θrv

2•

= θran

nt

t

t

erera

ra

dtdrr

dtd

dtdva

2•••

••

••

+=

=

=

==

θθ

θ

θθ

• The particle moves along a path expressed as y = f(x). The radius of curvature, ρ, at any point on the path can be calculated from

2

2

23

2)(1

dxyddxdy

xy

+

=ρ

APPLICATIONSCars traveling along a

clover-leaf interchange

experience an acceleration

due to a change in speed as

well as due to a change in

direction of the velocity.

Example 1• Starting from rest, a motorboat

travels around a circular path of

r = 50 m at a speed that increases with time, v = (0.2 t2) m/s.

Find the magnitudes of the boat’s velocity and acceleration at the instant t = 3 s.

Example 2• A jet plane travels along a vertical

parabolic path defined by the equation

y = 0. 4x2. At point A, the jet has a

speed of 200 m/s, which is increasing at

the rate of 0. 8 m/s2. Find the magnitudeof the plane’s acceleration when it is at

point A.

Example 3• A race traveling at a speed of 250km/h on the

straightway applies his brakes at point A and reduce his speed at a uniform rate to 200km/h at C in a distance of 300m.

• Calculate the magnitude of the total accelerationof the race car an instant after it passes point B.

Example 4• The motion of pin A in the fixed circular slot

is controlled by a guide B, which is being elevated by its lead screw with a constant upward velocity vo=2m/s for the interval of its motion.

• Calculate both the normal and tangentialcomponents of acceleration of pin A as it passes the position for which .

Curvilinear motion Polar coordinate system (r- ѳ)

Polar coordinate(r- ѳ)• The third description for plane curvilinear

motion.

• Where the particle is located by the radialdistance r from a fixed pole and by an angularmeasurement ѳ to the radial line.

• Polar coordinates are particularly useful when amotion is constrained through the control of aradial distance and an angular position,

• or when an unconstrained motion is observed bymeasurements of a radial distance and an angularposition.

• An arbitrary fixed line,such as the x-axis, isused as a reference forthe measurement ѳ.

• Unit vectors er and eѳare established in thepositive r and ѳ

directions, respectively.

• The position vector to the particle at A has a

magnitude equal to the radial distance r and a

direction specified by the unit vector er.

• We express the location of the particle at A by

the vector

rer.r • =

r

Velocity• The velocity is obtained by differentiating the vector r.

• Where the unit vector er now has a derivative because

its direction changes.

• We obtain the derivation in exactly the same way that

we derived for et.

••

+=

+===

rr

rr

r

ererv

dtedre

dtdr

dtedr

dtrdv

..

.

• To find the derivative of consider the following figure

• Using vector additione’r = er + ∆er

e’ѳ = eѳ + ∆eѳ• Since the magnitude

|e’r| = |er| = |e’ѳ|= |eѳ| = 1

dtder

• The magnitude of ∆er and ∆eѳ| ∆er|= |∆eѳ| =2 sin ∆ѳ/2

• Dividing both sides by ∆ѳ

• As ∆ѳ→0, is perpendicular to er .

θ

θ

θθθ

∆

∆=

∆∆

=∆∆ 2sin2eer

θ∆∆ re

Note: As ∆ѳ→0,

1. is directed towards

the positive eѳ direction.

2. is directed towards

the negative er direction.

• Then,

θ∆∆ re

θθ

∆∆e

12

2sinlimlimlim

000=

∆

∆=

∆∆

=∆∆

→∆→∆→∆ θ

θ

θθ θθ

θθ

eer

• Therefore;

1lim

1lim

0

0

=∆

=∆∆

=∆

=∆∆

→∆

→∆

θθ

θθ

θθθ

θ

dee

dee rr

rrr

rr

eddeeedde

eddeeedde

..1

..1

θθ

θθ

θθ

θθθ

−=⇒−=−=

=⇒==

• Dividing both sides by dt, we have

• Therefore the velocity equation becomes;

rr

rr

edtede

dtd

dted

edtede

dtd

dted

.

.

•

•

−=⇒−=

=⇒=

θθ

θθ

θθ

θθ

θθ ererdtedre

dtdrv r

rr

••

+=+= ..

• Whereand•

= rvr

•

= θθ .rv

=

+=

−

r

r

vv

vvv

θ

θ

α 1

22

tan

• The r-component of v is merely the rate at which

the vector r stretches.

• The ѳ-component of v is due to the rotation of r.

Acceleration• Differentiating the expression for v to obtain

the acceleration a.

• But from the previous derivation

dtedre

dtdre

dtdr

dtedre

dtrda

ererdtd

dtrd

dtvda

rr

r

θθθ

θ

θθθ

θ

••

•••

••

++++=

+=== 2

2

rr e

dtdeande

dtde .,.

••

−== θθ θθ

• Substituting the above and simplifying

• Where

θ

θθθ

θθθ

θθθθθ

errerra

ererererera

r

rr

++

−=∴

−++++=

•••••••

••••••••••

22

+=

−=

••••

•••

θθ

θ

θ rra

rrar

2

2

=

+=

−

r

r

aa

aaa

θ

θ

φ 1

22

tan

• For motion in a circular path• Velocity

Where, because r=constant

• Acceleration where,

θθ ererv r

••

+= ..

0=•

r

θθ erv•

=∴ .

0==•••

rr

θθθ erera r

+

−=∴

••• 2

Kinematics of particles

Relative motion

Relative motion• Relative motion analysis : is the motion analysis of

a particle using moving reference systemcoordinate in reference to fixed reference

system.

• In this portion we will confine ourattention to:-– moving reference systems that translate but

do not rotate.

– The relative motion analysis is limited to planemotion.

• Note: in this section we need

1. Inertial(fixed) frame of reference.

2. Translating(not rotating) frame of reference.

• Consider two particles A and B that may have

separate curvilinear motion in a given plane or in

parallel planes.

• X,Y : inertial frame of reference

• X,y : translating coordinate system

• Using vector addition:• position vector of particle B is

Where: rA, rB – absolute position vectors

rB/A – relative position vector of particle B (B

relative to A or B with respect to A)

ABAB rrr /+=

• Differentiating the above position vector once we

obtain the velocities and twice to obtain

accelerations. Thus,

- Velocity - Acceleration

ABAB

ABAB

vvvdtrd

dtrd

dtrd

/

/

+=

+=

ABAB

ABAB

aaadtvd

dtvd

dtvd

/

/

+=

+=

• Note: In relative motion analysis, we employed

the following two methods,

1. Trigonometric(vector diagram) – A sketch of

the vector triangle is made to reveal the

trigonometry

2. Vector algebra – using unit vector i and j,

express each of the vectors in vector form.

Example 1• A 350m long train travelling at a constant speed of 40m/s crosses

over a road as shown below. If an automobile A is traveling at

45m/s and is 400m from the crossing at the instant the front of the

train reaches the crossing, determine

a) The relative velocity of the train with respect to the automobile, and

b) The distance from the automobile to the end of the last car of the train at the instant.

Example 2• For the instant represented, car A has a speed of

100km/h, which is increasing at the rate of 8km/h

each second. Simultaneously, car B also has a speed of

100km/h as it rounds the turn and is slowing down at

the rate of 8km/h each second. Determine the

acceleration that car B appear to have an observer in

car A.

Example 3• For the instant represented, car A has an acceleration in

the direction of its motion and car B has a speed of

72km/h which is increasing. If the acceleration of B as

observed from A is zero for this instant,

• Determine the acceleration of A and the rate at which the

speed of B is changing.

Example 4• Airplane A is flying horizontally with a constant speed

of 200km/h and is towing the glider B, which is

gaining altitude. If the tow cable has a length r=60m

and is increasing at the constant rate of 5 degrees

per second, determine the velocity and acceleration

of the glider for the instant when =15

Constrained motion of connected particles

Constrained motion(dependent motion)

• Sometimes the position of a particle will depend

upon the position of another or of several particles.

• If the particles are connected together by an

inextensible ropes, the resulting motion is called

constrained motion

• Considering the figure, cable AB is subdivided into three segments:

• the length in contact with the pulley, CD

• the length CA• the length DB

• It is assumed that, no matter how A and B move, the length in contact with the pulley is constant.

• We could write:

constant==++ ABBCDA lsls

• Differentiating with respect to time,

• Differentiating the velocity equation0vv

0

BA =+

=+dtds

dtds BA

0aa BA =+

Important points in this technique:

• Each datum must be defined from a fixed position.

• In many problems, there may be multiple lengths like lCD that don’t

change as the system moves. Instead of giving each of these

lengths a separate label, we may just incorporate them into an

effective length:

• where it’s understood that

l = cable length less the length in contact with the pulley = lAB – lCD.

constant==+ lss BA

constant2 ==++ lshs BA

• considering the fig, we could write:

• where l is the length of the cable less

the red segments that remain

unchanged in length as A and B move.

Differentiating,

0202

=+=+

BA

BA

aavv

constant)(2 ==−++ lshhs BA

• we could also write the length of the cable by taking another datum:

• Differentiating,

0202

=+=+

BA

BA

aavv

• Consider the fig.,

• Since L, r2, r1 and b is

constant, the first and

second time derivatives

are:-

bryrxL ++++= 12 2

2ππ

yxyx

2020

+=+=

• Consider the

following figure

.)(.2

constyyyyLconstyyL

DCCBB

DAA

+−++=++=

• NB. Clearly, it is impossible for the

signs of all three terms to be positive

simultaneously.

Example 1• Cylinder B has a downward velocity of 0.6m/s and an

upward acceleration of 0.15m/s2.

• Calculate the velocity and acceleration of block A.

•

= xvA

Example 2

• Collars A and B slides along the fixed rods are

connected by a cord length L. If collar A has a

velocity to the right, express the velocity

of B in terms of x, vA, and s. •

−= svB

Part III

Kinetics of particles

Kinetics of particles

• It is the study of the relations existing between

the forces acting on body, the mass of the body,

and the motion of the body.

• It is the study of the relation between

unbalanced forces and the resulting motion.

• Newton ’s first law and third law are sufficient

for studying bodies at rest (statics) or bodies in

motion with no acceleration.

• When a body accelerates ( change in velocity

magnitude or direction) Newton ’s second law is

required to relate the motion of the body to the

forces acting on it.

Kinetics problems

• Force-mass-acceleration method

• Work and energy principles

• Impulse and momentum method

Force, mass and acceleration

• Newton ’s Second Law: If the resultant force

acting on a particle is not zero the particle will

have an acceleration proportional to the

magnitude of resultant and in the direction of the

resultant.

• The basic relation between force and

acceleration is found in Newton's second law of

motion and its verification is entirely

experimental.

• Consider a particle subjected to constant forces

• We conclude that the constant is a measure of

some property of the particle that does not

change.

constaF

aF

aF

==== ...2

2

1

1

• This property is the inertia of the particle

which is its resistance to rate of change of

velocity.

• The mass m is used as a quantitative measure

of inertia, and therefore the experimental

relation becomes,F=ma

• The above relation provides a complete

formulation of Newton's second law; it expresses

not only that the magnitude F and a are

proportional but also that the vector F and a have

the same direction.

Types of dynamics problems• Acceleration is known from kinematics conditions

Determine the corresponding forces• Forces acting on the particle are specified

(Forces are constant or functions F( t, s, v, …)Determine the resulting motion

Equation of motion and solution of problems

• When a particle of mass m acted upon by severalforces. The Newton’s second law can beexpressed by the equation

• To determine the acceleration we must use theanalysis used in kinematics, i.e

• Rectilinear motion• Curvilinear motion

∑ = maF

Rectilinear Motion• If we choose the x-direction, as the direction

of the rectilinear motion of a particle of mass m, the acceleration in the y and z direction will be zero, i.e

∑∑∑

=

=

=

0

0

z

y

xx

F

FmaF

• Generally,

• Where the acceleration and resultant force are given by

ZZ

yy

xx

maFmaFmaF

=

=

=

∑∑∑

222zyx

zyx

aaaa

kajaiaa

++=

++=

( ) 222 )()( ∑∑∑∑

∑++=

++=

zyx

zyx

FFFF

kFjFiFF

Curvilinear motion

• In applying Newton's second law, we shall make

use of the three coordinate descriptions of

acceleration in curvilinear motion.

Rectangular coordinates

Where and

∑∑

=

=

yy

xx

maFmaF

••

= xax

••

= yay

Normal and tangential coordinate

• Where∑∑

=

=

tt

nn

maF

maF

••

=== vava tn ,22

ρβρ

Polar coordinates

• Where and ∑∑

=

=

θθ maF

maF rr

2•••

−= θrrar

••••

+= θθ rran 2

Examples

Example 1• Block A has a mass of 30kg and block B has a mass of

15kg. The coefficient of friction between all plane

surfaces of contact are and .

Knowing that Ѳ=300 and that the magnitude of the

force P applied to block A is 250N, determine

a) The acceleration of block A ,and

b) The tension in the cord

15.0=sµ 10.0=kµ

Example 2• A small vehicle enters the top A of the circular path

with a horizontal velocity vo and gathers speed as itmoves down the path.

• Determine an expression for the angle β to theposition where the vehicle leaves the path andbecomes a projectile. Evaluate your expression forvo=0. Neglect friction and treat the vehicle as aparticle

Exercise(problem 3/69)• The slotted arm revolves in the horizontal plane about

the fixed vertical axis through point O. the 2kg slider C is drawn toward O at the constant rate of 50mm/s by pulling the cord S. at the instant for which r=225mm, the arm has a counterclocke wise angular velocity w=6rad/s and is slowing down at the rate of 2rad/s2. For this instant, determine the tension T in the cord and the magnitude N of the force exerted on the slider by the sides of the smooth radial slot. Indicate which side, A or B of the slot contacts the slider.

Exercise (problem 3/43)• The sliders A and B are connected by a light rigid

bar and move with negligible friction in the slots,

both of which lie in a horizontal plane. For the

positions shown, the velocity of A is 0.4m/s to

the right. Determine the acceleration of each

slider and the force in the bar at this instant.

Exercise (problem 3/36)

• Determine the accelerations of bodies A and B

and the tension in the cable due to the application

of the 250N force. Neglect all friction and the

masses of the pulleys.