Upload
aychiluhimhailu
View
856
Download
9
Embed Size (px)
Citation preview
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.