Unit2
2. Kinematics of particles
Contents
2.1. Introduction
2.2. Rectilinear Motion: Position,
Velocity & Acceleration
2.3. Determination of the Motion of
a Particle
Eg. 2.1
Eg.2.3
2.4 Uniform Rectilinear-Motion
2.5 Uniformly Accelerated
Rectilinear-Motion
2.6 Motion of Several Particles:
Relative Motion
Eg.2.4
2.7. Motion of Several Particles:
Dependent Motion
Eg.2.5
2.8. Graphical Solution of RectilinearMotion Problems
2.9 Other Graphical Methods
2.10 Curvilinear Motion: Position,
Velocity & Acceleration
2.11 Derivatives of Vector Functions
2.12 Rectangular Components of
Velocity and Acceleration
2.13 Motion Relative to a Frame in
Translation
2.14.Tangential and Normal
Components
2.15 Radial and Transverse
Components
Eg 2.10
11 - 2
Eg.2.11
2.1.Introduction
 Dynamics includes:
 Kinematics: study of the geometry of motion.
 Kinematics is used to relate displacement, velocity,
acceleration, and time without reference to the cause of motion.
 Kinetics: study of the relations existing between the forces
acting on a body, the mass of the body, and the motion of the
body.
 Kinetics is used to predict the motion caused by given forces or
to determine the forces required to produce a given motion.
11 - 3
 Rectilinear motion: position, velocity, and acceleration
of a particle as it moves along a straight line.
 Curvilinear motion: position, velocity, and
acceleration of a particle as it moves along a curved
line in two or three dimensions.
11 - 4
2.1 Rectilinear Motion: Position, Velocity & Acceleration
 Particle moving along a straight line is
said to be in rectilinear motion.
 Position coordinate of a particle is defined
by positive or negative distance of
particle from a fixed origin on the line.
 The motion of a particle is known if the
position coordinate for particle is known for
every value of time t.
 Motion of the particle may be expressed in
the form of a function, or in the form of a
graph x vs. t. e.g.,
x  6t  t
2
3
11 - 5
 Consider particle which occupies
position P at time t and P’ at t+Dt,
Average velocity 
x
t
x
t 0 t
Instantaneous velocity  v  lim
 Instantaneous velocity may be
positive or negative.
 Magnitude of velocity is referred to as
particle speed.
• From the definition of a derivative , Eg
x dx

dt
t 0 t
2
3
v  lim
x  6t  t
dx
v
 1112- 6t  3t
dt
 Consider particle with velocity v at
time t and v’ at t+t,
Instantaneous acceleration
v
t  0  t
 a  lim
 Instantaneous acceleration may be:
 positive: increasing positive velocity
or decreasing negative velocity
 negative: decreasing positive velocity
or increasing negative velocity.
 From the definition of a derivative,
v dv d 2 x
a  lim

 2

t
dt
t 0
dt
e.g. v  12t  3t 2
a
dv
 12  6t
dt
11 - 7
 Eg 2.1
 Consider particle with motion given by
x  6t 2  t 3
dx
v
 12t  3t 2
dt
dv d 2 x
a
 2  12  6t
dt dt
 Required:-x=? V=? a=?
 Solution
 at t = 0, x = 0, v = 0, a = 12 m/s2
 at t = 2 s,
x = 16 m, v = vmax = 12 m/s, a = 0
 at t = 4 s,
x = xmax = 32 m, v = 0, a = -12 m/s2
 at t = 6 s,
2
x = 0, v = -36 m/s, a = -24 11m/s
-8
 Recall, motion of a particle is known if position is known for all
time t.
 Typically, conditions of motion are specified by the type of
acceleration experienced by the particle.
 Determination of velocity and position requires two successive
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)
11 - 9
 Acceleration given as a function of time, a = f(t):
dv
 a  f t 
dt
dx
 vt 
dt
t
v0
0
 dv   f t  dt
dv  f t  dt
dx  vt  dt
v t 
x t 
t
 dx   vt  dt
x0
t
vt   v0   f t  dt
0
t
xt   x0   vt  dt
0
0
 Acceleration given as a function of position, a = f(x):
dx
dx
v
or dt 
dt
v
v dv  f  x dx
v x 
dv
dv
a
or a  v  f  x 
dt
dx
x
 v dv   f  x dx
v0
x0
1 v x 2
2
x
 12 v02   f  x dx
x0
11 - 10
Eg 2.2
Ball tossed with 10 m/s vertical velocity from window 20 m above
ground.
Determine:
 velocity and elevation above ground at time t, highest elevation
reached by ball and corresponding time, and time when ball will
hit the ground and corresponding velocity.
 Given
v=+10 m/s
y=+20 m
a=-9.81m/s2
Required
t=? @v(t)=0
v(t) =?
y(t)=?
11 - 11
Solution:
 Integrate twice to find v(t) and y(t).
dv
 a  9.81m s 2
dt
v t 
t
vt   v0  9.81t
 dv    9.81dt
v0
0
vt   10
m 
m
  9.81 2  t
s 
s 
dy
 v  10  9.81t
dt
y t 
t
 dy   10  9.81t dt
y0
0
y t   y0  10t  12 9.81t 2
m
 m 
yt   20 m  10 t   4.905 2 t 2
 s 
s 
11 - 12
 Solve for t at which velocity equals zero
and evaluate corresponding altitude.
vt   10
m 
m
  9.81 2  t  0
s 
s 
t  1.019s
m
 m 
y t   20 m  10 t   4.905 2 t 2
 s 
s 
m

 m
y  20 m  10 1.019 s    4.905 2 1.019 s 2
 s

s 
y  25.1m
11 - 13
 Solve for t at which altitude equals zero
and evaluate corresponding velocity.
m
 m 
yt   20 m  10 t   4.905 2 t 2  0
 s 
s 
t  1.243s meaningless 
t  3.28 s
vt   10
m 
m
  9.81 2  t
s 
s 
v3.28 s   10
m 
m
  9.81 2  3.28 s 
s 
s 
v  22.2
m
s
11 - 14
Eg.2.3
 Brake mechanism used to reduce gun recoil consists of piston
attached to barrel moving in fixed cylinder filled with oil. As
barrel recoils with initial velocity v0, piston moves and oil is
forced through orifices in piston, causing piston and cylinder to
decelerate at rate proportional to their velocity. Determine v(t),
x(t), and v(x).
 Given
 a=-kv
 requred
 v(t)=? x(t)=?
 v(x)=?
11 - 15
Solution:
 Integrate a = dv/dt = -kv to
find v(t).
v t 
t
dv
 v  k  dt
v
0
dv
a
 kv
dt
0
ln
vt 
 kt
v0
vt   v0 e kt
 Integrate v(t) = dx/dt to find x(t).
vt  
x t 
dx
 v0 e  kt
dt
t
 dx  v0  e
0
0
 kt
t
dt
 1

xt   v0  e  kt 
 k
0

v
xt   0 1  e  kt
k

11 - 16
 Integrate a = v dv/dx = -kv to find v(x).
dv
a  v  kv
dx
v  v0  kx
dv  k dx
v
x
v0
0
 dv  k  dx
v  v0  kx
 Alternatively, xt   v0 1  e kt 
k
with
and
then
vt 
vt   v0 e kt or e kt 
v0
v  vt  

xt   0 1 
k 
v0 
v  v0  kx
11 - 17
Uniform Rectilinear Motion
 For particle in uniform rectilinear motion, the acceleration is zero
and the velocity is constant.
dx
 v  constant
dt
x
t
x0
0
 dx  v  dt
x  x0  vt
x  x0  vt
11 - 18
Uniformly Accelerated Rectilinear Motion
 For particle in uniformly accelerated rectilinear motion, the
acceleration of the particle is constant.
dv
 a  constant
dt
v
t
v0
0
 dv  a  dt
v  v0  at
v  v0  at
dx
 v0  at
dt
x
t
x0
0
 dx   v0  at dt
x  x0  v0t  12 at 2
x  x0  v0t  12 at 2
dv
v  a  constant
dx
v 2  v02  2a x  x0 
v
x
v0
x0
 v dv  a  dx
1
2
v 2  v02   ax  x0 
11 - 19
Motion Of Several Particles: Relative Motion
 For particles moving along the same
line, time should be recorded from the
same starting instant and displacements
should be measured from the same
origin in the same direction.
xB A  xB  x A  relative position of B
xB  x A  xB A
with respect to A
vB A  vB  v A  relative velocity of B
vB  v A  vB A
with respect to A
a B A  a B  a A  relative acceleration
aB  a A  aB A
of B with respect to
11 - 20
A
Eg .2.4
Ball thrown vertically from 12 m level in elevator shaft with initial velocity of 18
m/s. At same instant, open-platform elevator passes 5 m level moving upward at 2
m/s.
Determine (a) when and where ball hits elevator and (b) relative velocity of ball
and elevator at contact.
Given
yo=12m
Vo=18
A=g=-9.81m/s2
Required
a.
t=?
b. V=?
11 - 21
 Solution:
 Substitute initial position and velocity
and constant acceleration of ball into
general equations for uniformly
accelerated rectilinear motion.
v B  v0  at  18
m 
m
  9.81 2 t
s 
s 
m
 m 
y B  y0  v0t  12 at 2  12 m  18 t   4.905 2 t 2
 s 
s 
 Substitute initial position and constant
velocity of elevator into equation for
uniform rectilinear motion.
vE  2
m
s
 m
y E  y0  v E t  5 m   2 t
 s
11 - 22
 Write equation for relative position of ball
with respect to elevator and solve for zero
relative position, i.e., impact.
yB
E


 12  18t  4.905t 2  5  2t   0
t  0.39 s meaningless 
t  3.65 s
 Substitute impact time into equations for
position of elevator and relative velocity of ball
with respect to elevator.
y E  5  23.65
y E  12.3 m
v B E  18  9.81t   2
 16  9.813.65
m
v B E  19.81
s
11 - 23
Motion Of Several Particles: Dependent Motion
 Position of a particle may depend on position of one or
more other particles.
 Position of block B depends on position of block
A. Since rope is of constant length, it follows that
sum of lengths of segments must be constant.
x A  2 xB 
constant (one degree of freedom)
 Positions of three blocks are dependent.
2 x A  2 xB  xC  constant (two degrees of
freedom)
 For linearly related positions, similar relations
hold between velocities and accelerations.
dx
dx A
dx
 2 B  C  0 or 2v A  2v B  vC  0
dt
dt
dt
dv
dv
dv
2 A  2 B  C  0 or 2a A  2a B  aC  0
dt
dt
dt
2
11 - 24
Eg.2.5
Pulley D is attached to a collar which is pulled down at 3 in./s. At t
= 0, collar A starts moving down from K with constant acceleration
and zero initial velocity. Knowing that velocity of collar A is 12 in./s
as it passes L, determine the change in elevation, velocity, and
acceleration of block B when block A is at L. Collar A has
uniformly accelerated rectilinear motion. Solve for acceleration
and time t to reach L.
11 - 25
Solution:
 Define origin at upper horizontal surface
with positive displacement downward.
v 2A  v A 02  2a A x A   x A 0 
2
 in. 
12   2a A 8 in.
 s
aA  9
in.
s2
 Collar A has uniformly accelerated
rectilinear motion. Solve for acceleration
and time t to reach L.
v A  v A 0  a At
12
in.
in.
9 2t
s
s
t  1.333 s
11 - 26
 Pulley D has uniform rectilinear motion.
Calculate change of position at time t.
x D   x D 0  v D t
 in. 
x D   x D 0   3 1.333s   4 in.
 s 
VD=𝑽𝑫𝒐 + 𝒂𝑫𝒕
3in/s=0in/s+aDt
aD=-3in/s/1.3333s=2.25in/s2
 Block B motion is dependent on
motions of collar A and pulley D.
Write motion relationship and solve for
change of block B position at time t.
Total length of cable remains constant,

x A  2 x D  x B   x A 0  2 x D 0   x B 0
x A   x A 0   2xD   xD 0   xB   xB 0   0
8 in.  24 in.  x B   x B 0   0
x B   x B 0  16 in.
11 - 27
 Differentiate motion relation twice to
develop equations for velocity and
acceleration of block B.
x A  2 x D  x B  constant
v A  2v D  v B  0
 in.   in. 
12   2 3   v B  0
 s   s 
in.
v B  18
s
𝑎𝐴 + 2𝑎𝐷 + aB = 0
9in/s2+2(2.25in/s2)+aB=0
9in/s24.5in/s2+aB=0
aB=-13.5in/s2
11 - 28
Graphical Solution Of Rectilinear-motion Problems
 Given the x-t curve, the v-t curve is equal to the x-t curve
slope.
 Given the v-t curve, the a-t curve is equal to the v-t curve
slope.
11 - 29
 Given the a-t curve, the change in velocity between t1 and t2 is
equal to the area under the a-t curve between t1 and t2.
 Given the v-t curve, the change in position between t1 and t2 is
equal to the area under the v-t curve between t1 and t2.
11 - 30
Other Graphical Methods
 Moment-area method to determine
particle position at time t directly from
the a-t curve:
x1  x0  area under v  t curve
 v0t1 
using dv = a dt ,
v1
 t1  t dv
v0
v1
v1
v0
v0
 t1  t  a dt  x1  x0  v0t1   t1  t  a dt
 First moment of area
under a-t curve with
respect to t = t1 line.
x1  x0  v0t1  area under a-t curvet1  t 
t  abscissa of centroid C
11 - 31
 Method to determine particle
acceleration from v-x curve:
dv
dx
 AB tan 
av
 BC 
subnormal to v-x curve
11 - 32
Curvilinear Motion: Position, Velocity & Acceleration
 Particle moving along a curve other than a
straight line is in curvilinear motion.
 Position vector of a particle at time t is
defined by a vector between origin O of a
fixed reference frame and the position
occupied by particle.
 Consider particle which occupies position P
defined by r at time t and P’ defined by r  at t +


t, 
r dr
v  lim

dt
t 0 t
 instantaneous velocity (vector)
s ds

dt
t 0 t
v  lim
 instantaneous speed (scalar)
11 - 33
Curvilinear Motion: Position, Velocity & Acceleration
 Consider velocity v of particle at time t
and velocity v at t + Dt,


v d v

a  lim

dt
t 0 t
 instantaneous
acceleration (vector)
 In general, acceleration vector is not
tangent to particle path and velocity
vector.
11 - 34
Derivatives Of Vector Functions
 LetP𝒑(𝒖)
u  be a vector function of scalar variable u,




dP
P
Pu  u   Pu 
 lim
 lim
du u 0 u u 0
u
 Derivative of vector
sum,

 

d P  Q  dP dQ


du
du du
 Derivative of product
of scalar
and vector


functions, d  f P   df P  f dP
du
du
du
 Derivative of scalar product and vector
product,

 

d P  Q  dP   dQ

Q  P 
du
du
du

 



d P  Q  dP
dQ

Q  P
du
du
du
11 - 35
Rectangular Components Of Velocity & Acceleration
 vector of particle P is
 position


 When
r  xi  y j  zk
given by its rectangular components,
 Velocity vector,



 dx  dy  dz 
v  i  j  k  xi  y j  zk
dt
dt
dt



 vx i  v y j  vz k
• Acceleration vector,



 d 2 x d 2 y  d 2 z 
a  2 i  2 j  2 k  xi  y j  zk
dt
dt
dt



 ax i  a y j  az k
11 - 36
 Rectangular components particularly
effective when component accelerations
can be integrated independently, e.g.,
motion of a projectile, with initial
conditions,
a x  x  0
a y  y   g
v x  v x 0
v y  v y  gt
0
y  v y y  12 gt 2
a z  z  0
v x 0 , v y 0 , v z 0  0
x0  y 0  z 0  0
Integrating twice yields
x  v x 0 t
 
 0
vz  0
z0
 Motion in horizontal direction is
uniform.
 Motion in vertical direction is uniformly
accelerated.
 Motion of projectile could be replaced by
11 - 37
two independent rectilinear motions.
Motion Relative To A Frame In Translation
 Designate one frame as the fixed frame of reference.
All other frames not rigidly attached to the fixed
reference frame are moving frames of reference.
 Position vectors for particles A and B with respect


rAare
and rB .
to the fixed frame of reference Oxyz

r
 Vector B A joining A and B defines the position
of B with respect to the moving frame Ax’y’z’

 
and
rB  rA  rB A
 Differentiating twice,



vB  v A  vB A



a B  a A  aB A

vB A  velocity of B relative to A.

a B A  acceleration of B
relative to A.
 Absolute motion of B can be obtained by
combining motion of A with relative motion of B
with respect to moving reference frame attached
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to A.
Tangential And Normal Components
 Velocity vector of particle is tangent to path
of particle. In general, acceleration vector is
not. Wish to express acceleration vector in
terms of tangential and normal components.


• et and et are tangential unit vectors for the
particle path at P and P’. When drawn with
  
respect to the same origin, et  et  et and
 is the angle between them.
∆𝜃
∆𝑒𝑡 = 2 sin
2
∆𝜃𝑒𝑡
2 𝑠𝑖𝑛(∆𝜃/2)
= lim
𝑒𝑡 =𝑒𝑛
∆𝜃
∆𝜃
∆𝜃→0
∆𝜃→0
lim
𝑒𝑛 =
𝑑𝑒𝑡
𝑑𝜃
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 With the velocity vector expressed as v  vet
the particle acceleration may be written as



de dv 
de d ds
 dv dv 
a
 et  v
 et  v
dt dt
dt dt
d ds dt
but 
det 
ds
 en
 d  ds
v
d
dt
After substituting,
dv
v2
 dv  v 2 
a  et  en
at 
an 
dt

dt

 Tangential component of acceleration
reflects change of speed and normal
component reflects change of direction.
 Tangential component may be positive or
negative. Normal component always points
toward center of path curvature.
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 Relations for tangential and normal
acceleration also apply for particle
moving along space curve.
 dv  v 2 
a  et  en
dt

dv
at 
dt
an 
v2

 Plane containing tangential and normal unit
vectors is called the osculating plane.
 Normal to the osculating plane is
found from

 
eb  et  en

en  principalnormal

eb  binormal
 Acceleration has no component
along binormal.
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Radial And Transverse Components


r  re r

der 
 e
d
 When particle position is given in polar
coordinates, it is convenient to express velocity
and acceleration with components parallel and
perpendicular to OP. The particle velocity
vector is

d
e
d
dr
dr 
d 



v  rer   er  r r  er  r
e
dt
dt
dt
dt
dt


 r er  r e

de

  er
d


der der d  d

 e
dt
d dt
dt


de de d
 d

  er
dt
d dt
dt
 Similarly, the particle acceleration vector is
d  
 d  dr 
a   er  r
e 
dt  dt
dt 


d 2 r  dr der dr d 
d 2 
d de
 2 er 

e  r 2 e  r
dt dt dt dt
dt dt
dt
dt


 r  r 2 er  r  2r e


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 When particle position is given in
cylindrical coordinates, it is convenient
to express the velocity and acceleration
vectors using the unit vectors

 
eR , e , and k .
 Position vector,



r  R e R z k
 Velocity
vector,


 dr  


v
 R eR  R e  z k
dt
 Acceleration vector,


 dv


2




  R eR  R  2 R  e  z k
a
 R
dt


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Eg.2.6
A motorist is traveling on curved section of highway at 60 mph.
The motorist applies brakes causing a constant deceleration rate.
Knowing that after 8 s the speed has been reduced to 45 mph,
determine the acceleration of the automobile immediately after the
brakes are applied.
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Solution:
 Calculate tangential and normal
components of acceleration.
at 
an 
60 mph  88 ft/s
45 mph  66 ft/s
v 66  88 ft s
ft

 2.75 2
t
8s
s
v2

88 ft s 2
ft

 3.10 2
2500ft
s
 Determine acceleration magnitude
and direction with respect to tangent
to curve.
a  at2  an2   2.752  3.102
ft
a  4.14 2
s
3.10
1 an
  tan
 tan 1
at
2.75
  48.4
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Eg2 .7
Rotation of the arm about O is defined by 𝜽 = 0.15t2 where 𝜽 is in
radians and t in seconds. Collar B slides along the arm such that r
= 0.9 - 0.12t2 where r is in meters.
After the arm has rotated through 30o, determine (a) the total
velocity of the collar, (b) the total acceleration of the collar, and (c)
the relative acceleration of the collar with respect to the arm.
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Solution:
 Evaluate time t for  = 30o.
  0.15t 2
 30  0.524 rad
t  1.869 s
 Evaluate radial and angular positions, and
first and second derivatives at time t.
r  0.9  0.12 t 2  0.481 m
r  0.24 t  0.449 m s
r  0.24 m s 2
  0.15 t 2  0.524 rad
  0.30 t  0.561rad s
  0.30 rad s 2
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 Calculate velocity and acceleration.
vr  r  0.449 m s
v  r  0.481m 0.561rad s   0.270 m s
v  vr2  v2
v
  tan 1 
vr
  31.0
v  0.524 m s
ar  r  r 2
 0.240 m s 2  0.481m 0.561rad s 2
 0.391m s 2
a  r  2r


 0.481m  0.3 rad s 2  2 0.449 m s 0.561rad s 
 0.359 m s 2
a  ar2  a2
a
  tan 1 
a  0.531m s
ar
  42.6
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 Evaluate acceleration with
respect to arm.
 Motion of collar with respect
to arm is rectilinear and
defined by coordinate r.
a B OA  r  0.240 m s 2
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The End