-2-
Content
 Two basic quantities of Particle Kinematics
• Velocity and Acceleration of a Particle
• Relativity of motion of a Particle – Velocity addition
 Particle Kinematics – Cartesian Coordinates
KINEMATICS OF PARTICLES
• Velocity and acceleration of a Particle in Cartesian coordinate
system
• Rectilinear Kinematics
 Method of natural Coordinates - General Curvilinear
Motion
• Geometric Properties of a Curved Path
• Natural Coordinates System
• Velocity and Acceleration of a Particle
 Method of polar Coordinates / cylindrical coordinates
/ spherical coordinates
Applied Mechanics - Department of Mechatronics - SME
Applied Mechanics - Department of Mechatronics - SME
-3-
1. Velocity and Acceleration of a Particle
Considering a Particle P in space, the
position of P relative to fixed point O:
 
r = r (t ) – position vector
Trajectory of P is a line that the point P
draws in the space.
Rectilinear Trajec.
Rectilinear Motion
Curvilinear Trajec.
Curvilinear Motion
Applied Mechanics - Department of Mechatronics - SME
-4-
1. Velocity and Acceleration of a Particle
Velocity vector: characterizes the position change of
the particle over time.
trajectory
O
P

r (t )


Dr
vaverage =
Dt
r P’
P
Assume that motion of P during time t is
r (from P to P’), average velocity of the
particle in this period time t :
r(t)
r+r
O
- unit [m/s]
P
v
Velocity at time point t



Dr
dr

v = lim
=
= r
Dt  0 Dt
dt
Applied Mechanics - Department of Mechatronics - SME
r(t)
O
Velocity tangent
to the trajectory
-5-
1. Velocity and Acceleration of a Particle
Acceleration vector: characterizes the
velocity change of the particle over time.
P
Assume that the change of velocity in time
period t is v, average acceleration of
the particle in this period time t :

aavg .



Dv
v (t + Dt ) - v (t )
=
=
Dt
Dt
v(t)
O
v
[m/s2]
 
d 2
v = 2v ⋅ a
dt
ì
ï
ï> 0
í
ï<
0
ï
î

ey
y
x
Faster
Slow down
a =
x2 + y2 + z2 .
Faster or slow down motion
ìï> 0 Faster
 
 + yy
 + zz
 ïí
v ⋅ a = xx
ï<
ïî 0 Slow down
Applied Mechanics - Department of Mechatronics - SME
x (t ) = v 0t cos a
-7-
x (t ) = v0t cos a
y

e y v0
Determine: trajectory, maximum
height, and maximum distance.
Ans.
O


ex
x
Eliminating the time variable t in the two motion equations of x and
y, we get
y=
sin a
g
x- 2
x2
2
cos a
2v0 cos a
vx =
vy =
d
dt
d
dt
y
P

ey v0
1
y(t ) = v0t sin a - gt 2
2
P
v 0 , a, g = const
-8-
Example (cont’)
Determine velocity and acceleration
2
Applied Mechanics - Department of Mechatronics - SME
r(t)
x 2 + y 2 + z 2 .




 z
a = xe x + ye y + ze
Motion of a bullet in vertical plane (subjected to gravity)
x

v0 cos a
P
O

ex
- Velocity vector
- Acceleration vector
Example (parabol motion of a particle)
t=




r  xe x  ye y  ze z
v =
Applied Mechanics - Department of Mechatronics - SME
y(t ) = v0t sin a - gt

ez




 z
v = xe x + y e y + ze


 
Dv
dv

a = lim
=
= v = r
Dt  0 Dt
dt
1
2
x = x (t ), y = y(t ), z = z (t )
- Position vector
Accel. vector toward
the concave side of
the trajectory.
acceleration at time point t
Motion: faster or slow down
z
- Equations of motion
v(t+t)
r(t)
-6-
2. Method of Cartesian Coordinates
O
ax = 0
x = v 0 cos a
y = v0 sin a - gt ay = -g


ex
y=
xmax
sin a
g
x- 2
x2
cos a
2v0 cos2 a
Determine: maximum height and distance from trajec. equation:
y(x max ) = 0  x max = 2v02
y max = y( 21 x max ) =
2
sin a cos a v 0
= sin 2a
g
g
sin a 1
g
( 2 x max ) - 2
( 21 x max )2 = ...
2
cos a
2v0 cos a
Applied Mechanics - Department of Mechatronics - SME
-9-
2. Method of Cartesian Coordinates: rectilinear motion
- Equations of motion
x


r (t ) = x (t )ex
x = x (t ),


v = xe x
M
O
- Velocity
-10-
Example 1.
Given v(t) of a car. Determine the
acceleration
and
traveled
distance after 3 second.
x
v  3t 2  2t [m/s]
v = x
- Acceleration


a = xe x a = x.
d
v(t )  6t  2, m/s2
dt
v(t ) 
d
s(t )  (3t 2  2t ),
dt
Uniform motion
v = const,
Faster / slow down motion
ì
Faster
ï> 0
 
 ï
v ⋅ a = xx
í
ï<
0
Slow down
ï
î
a(t ) 
x = x 0 + vt
Motion with constant acceleration
a = const,
3
3
0
0
v = v 0 + at
s(t  3)  27  9  36 m
Applied Mechanics - Department of Mechatronics - SME
Applied Mechanics - Department of Mechatronics - SME
-11-
Example 2
Find the trajectory equation, velocity, and acceleration from
the equations of motion given in the Cartesian coordinate
system as
x  t3  2,
y  3  t3.
Ans.:
Eliminating the time variable t in the two motion equations of
x and y, we get
Rectilinear motion
Velocity and acceleration:
vx  x  3t 2,
vy  y  3t 2
 v  vx2  vy2  3 2t 2
ax  x  6t,
ay  y  6t

Applied Mechanics - Department of Mechatronics - SME
ds(t )  v(t )dt  (3t 2  2t )dt
s(t )  s 0   v(t )dt   (3t 2  2t )dt  t 3  t 2  s(t )  t 3  t 2
x = x 0 + v 0t + 21 at 2
x  y  5
a(t  3)  20, m/s2
a  ax2  ay2  6 2t
-12-
3. Method of natural coordinates: curvilinear motion
Concepts related to curved path: osculating plane
at point P, the curvature, the radius of curvature,
the natural coordinate system.
The osculating plane at point P

e
n
Consider a curved path in space. Let P
and P’ are two points on the path.

en
s

e
 is small enough,
If the distance s  PP
 can be considered as a planar arc.
the arc PP
The plane containing this arc is the osculating
plane of the trajectory at P.
For a planar curve: The osculating plane is the plane containing
the curve.
Applied Mechanics - Department of Mechatronics - SME

-13-
3. Method of natural coordinates: curvilinear motion
The curvature at P
k  lim
s 0

d

ds
s
n

On the osculating plane at P:
en

• Tangential axis  (unit vector e )

• Normal axis n (unit vector en )
(in osculating plane, toward in concave side)
• Binormal axis b (unit vector eb )

e

e
1
k

e
s


eb

 
eb  et  en
r
E.g., a circle trajectory has a
constant curvature
k

e
Natural coordinates system
The radius of curvature at P:

-14-
3. Method of natural coordinates: curvilinear motion
1
 const
r
Applied Mechanics - Department of Mechatronics - SME
Applied Mechanics - Department of Mechatronics - SME
3. Method of natural coordinates: curvilinear motion

eb
Equation of motion of P
s  s(t )
s(t)
Velocity of P

 dr ds 

 t,
v 
 et  se
dt dt
P0
v  s
O
Acceleration of P


det
d 
 dv

 )  se
 t  s
a 
 (se
dt dt t
dt

det
dt
Applied Mechanics - Department of Mechatronics - SME
P
 ???

en

et
P ds
r
r+dr

en
 
dr  etds


dr / ds  et

et

P‘ e
t'
-15-
3. Method of natural coordinates: curvilinear motion
Acceleration of P (cont.)


 t  s
a  se

det
dt



 t  (s2 /  )en
a  se
  
a  at  a n


 t ,
at  se


an  (s2 /  )en
an  s / ,

et
P
 ???
dt
P‘ 
et '
 d
   en

det  et   et
det

 1d en

det d 
1 
d ds 


en 
en  se
 n
dt
dt
ds dt
P
at  s  v,
2

det
ab  0
Applied Mechanics - Department of Mechatronics - SME
-16-

an

a

at
-17-
Example

et
The motion of point P, moving on an arc
of a circle of radius R, governed by s(t)
= at2/2. Determine the velocity and
acceleration of P at t = 2 s.
v
The equations of screw motion of P is written in the Cartesian
coordinate system as
P
x  r cos t,
R
s(t)



 t  atet ,
v  se


v (t  2)  2aet m/s
P0
P
R
P

an
P0
x
-19-
x  r  sin t,
y  r  cos t,
z  p
v  x  y  z
2
2
2
Coordinates
r  r (t ),   (t )
Unit vectors
 
er , e
2
x
Acceleration:
x  r  2 cos t,
Position vector
P
2
y  r  2 sin t,
y
z  0
a  x2  y2  z2  r  2  const
at  v  0  an  a  v 2 / 
Radius of curvature:
Applied Mechanics - Department of Mechatronics - SME
 a 2  at2  an2
  v 2 / an 
r 2 2  p 2
r2
-20-
4. Polar coordinate system
z
 r   p  const
2
y
Applied Mechanics - Department of Mechatronics - SME
Example
Velocity:
z
• velocity, tangential acceleration,
normal acceleration of P
• radius of curvature
Applied Mechanics - Department of Mechatronics - SME
Solution
z  pt
Determine:

at
 
 s2 
 (at )2 
 t  en  aet 
a  v  se
e ,
R
R n

 4a 2 
a (t  2)  aet 
en , m/s2
R
y  r sin t,
r , , p  const

en
Solution
-18-
Example


r  r (t )er
Velocity vector
 d  d

v  r  [r (t )er ]
dt
dt
de

 r r r
 re
dt

 


 r  re  vr  v ,
v  re
Applied Mechanics - Department of Mechatronics - SME
M

er

e
r

x
O


de  1der


der  1de

e
d

der
dt
O

d

er
d 

e  e
dt
-21-
4. Polar coordinate system
 d 

er 
e  e
dt
d 


e  
er  er
dt


de  1der
Consider a circular motion r  const  r  0,


der  1de

e
d
y
y

v
Acceleration vector
  d 

 r  r e )
a  v  (re

dt
a





 r  re
 r  r e  re  r e e  r
 re
er




 (r  r  2 )er  (r   2r )e
M

aj

ar
M

O

x
O
M

O
x
x



a  (r  r  2 )er  (r   2r )e



a  (r  2 )er  (r )e


 a n  at
x
Applied Mechanics - Department of Mechatronics - SME
Applied Mechanics - Department of Mechatronics - SME
-23-
4. Polar coordinate system
Relationship btw. Polar and Cartesian coordinate systems


r  r (t )er
r  r (t ),
z  z (t )
  (t ),

er

e



r  rer  zez
M
r
Velocity vector

O
xM
x
zM
M

ez
Position vector
yM
-24-
5. Cylindrical Coordinate System
Coordinates
y
xM  (r  r 2 ) cos   (r  2r ) sin 
yM  (r  r 2 ) sin   (r  2r ) cos 
Applied Mechanics - Department of Mechatronics - SME

rM
M
O
xM  r cos   r sin 
yM  r sin   r cos 

er

e
y

v
O
x M  r cos 
yM  r sin 
r  0




 r  r e  r e
v  re

er
d
-22-
4. Polar coordinate system
xM
O
r

e
 r
 er
z

yM
 
v  r




 r  rer  ze
 z  zez
 re



 r  r e  ze
z
 re
Acceleration vector
 



 z
a  v  (r  r  2 )er  (r   2r )e  ze
Applied Mechanics - Department of Mechatronics - SME


er  e ,


e  er ,

ez  0
-25-
5. Cylindrical Coordinate System
Relationship btw. Cylindrical and Cartesian coordinate systems
x M  r cos 
yM  r sin 
zM  z

e
O

 er
r
yM

x
y
x 
 M
 yM 
zM 

Applied Mechanics - Department of Mechatronics - SME
-27-
Coordinates
M

O
xM

-28-
Problems
Problems
zM
   (t )
Determine the equation of the path, velocity, and acceleration of the
point if the equation of motion of the particle in Cartesian coordinates is
given as follows:
r
yM
x  t 3  2,
y  3  t3
Velocity components
xM  r sin  cos   r cos  cos   r sin  sin 
yM  r sin  sin   r cos  sin   r sin  cos 
zM  r cos   r sin 
xM  x(r , ,  , r, , , r, , )
Acceleration components
yM  y(r , ,  , r, , , r, , )
zM  z(r , ,  , r, , , r, , )
Applied Mechanics - Department of Mechatronics - SME
r
xM  r sin  cos   r cos  cos   r sin  sin 
yM  r sin  sin   r cos  sin   r sin  cos 
zM  r cos   r sin 
6. Spherical coordinate system
Conversion to Cartersian coordinates
x M  r sin  cos 
yM  r sin  sin 
z M  r cos 

Velocity components
Applied Mechanics - Department of Mechatronics - SME
   (t ),
   (t )
x M  r sin  cos 
yM  r sin  sin 
z M  r cos 
z
xM  (r  r  2 ) cos   (r   2r ) sin 
yM  (r  r  2 ) sin   (r   2r ) cos 
zM  z
r  r (t ),
   (t ),
Conversion to Cartesian coordinates
r

ez
xM
z
P
M
xM  r cos   r  sin 
yM  r sin   r  cos 
zM  z
Coordinates
r  r (t ),
zM
-26-
6. Spherical coordinate system
x  10 cos
2
t,
5
in which coordinates are x, y in cm, and time t in second.
Applied Mechanics - Department of Mechatronics - SME
y  10 sin
2
t
5
Problems
-29-
Applied Mechanics - Department of Mechatronics - SME
Applied Mechanics - Department of Mechatronics - SME
From experimental data, the motion of a jet plane while traveling along
a runway is defined by the v - t graph shown, v(t). Construct the a(t)
and s(t) graphs for the motion. Also determine the traveled distance
after 30 s.
-31-
Problems
The roller coaster car travels down the helical path at constant speed
such that the parametric equations that define its position are
x = c sin kt, y = c cos kt, z = h − bt,
where c, h, and b are constants. Determine the magnitudes of its
velocity and acceleration.
Applied Mechanics - Department of Mechatronics - SME
-30-
A car travels along a straight road with the speed shown by the v - t graph.
Determine the expression and draw the graph of acceleration and traveled
distance as function of time: a(t) and s(t); velocity and acceleration as
function of traveled distance: v(s), a(s).
A car moves on a straight line with velocity v =(12 – 3t2) m/s, time t in
seconds. When t = 1 s, the car is 10 m on the left from the origin.
Determine the acceleration of the car when t = 4 s and the distance
traveled from when t = 0 to t = 10 s.
Problems
Problems
Applied Mechanics - Department of Mechatronics - SME
-32-
Problems
-33-
Applied Mechanics - Department of Mechatronics - SME
Applied Mechanics - Department of Mechatronics - SME
The crank OA of a slider-crank mechanism rotates about axis O with
the law  = t, where  = const. Givens OA = r and AB = L,
determine:
a) equation of motion and velocity of the slider B, x(t).
b) equation of motion and velocity of a point M on the connecting link
AB, AM = b.
Applied Mechanics - Department of Mechatronics - SME
-34-
Because of telescopic action, the end of the industrial robotic arm
extends along the path of the limacon r = (1 + 0.5 cos) m. At the
instant  = /4, the arm has an angular rotation d/dt = 0.6 rad/s, which
is increasing at d2/dt2 = 0.25 rad/s2. Determine the radial and
transverse components of the velocity and acceleration of the object
held in its grip at this instant, using unit vectors er and e.
The AB rod of length L rotates around the fixed axis A according to the
law  = t2 (rad). On the rod there is a slider G moving according to
the law r = L(1-t2). Find the velocity and acceleration of the point G
when  = /4. Determine the angle of rotation E when G touches the
axis A.
Problems
Problems
-35-
Problems
A block moves outward along the slot in the platform with a speed of
dr/dt = (4t) m/s, where t is in seconds. The platform rotates at a
constant rate of  = 6 rad/s. If the block starts from rest at the center,
determine the magnitudes of its velocity and acceleration when t = 1 s.
Applied Mechanics - Department of Mechatronics - SME
-36-
Problems
-37-
Applied Mechanics - Department of Mechatronics - SME
Applied Mechanics - Department of Mechatronics - SME
-39-
Problems
In each case, determine the speed of the cabine if the end of the rope is
pulled up by the motor M with a velocity v. The cable is always taut and
unstretched.
Applied Mechanics - Department of Mechatronics - SME
-38-
Determine the speed of block A if block B has an upward velocity of 3 m/s.
For a short time the arm of the robot is extending such that dr/dt = 0.5
m/s when r = 0.3 m, z = (0.4t2) m, and  = 0.5t rad, where t is in
seconds. Determine the magnitudes of the velocity and acceleration of
the grip A when t = 3 s.
Problems
Problems
Applied Mechanics - Department of Mechatronics - SME
-40-