Coordinate System-Rect and normal,tangent components

1. Coordinate System in Curvilinear Motion Part-1
Sub- Engg. Mechanics
By: Mr. Ghumare S. M.
When particle moves along the curved path, a motion
is called as Curvilinear motion.
In curvilinear motion velocity is always tangential to
the curved path.

2. Coordinate Systems
The Motion of the particle in curvilinear motion is expressed
using following three Coordinate systems.
i.e. To find the velocity and acceleration along a curve.
• Coordinate System in Curvilinear Motion:
1. Rectangular / Cartesian Coordinate Systems (x-y system)
2. Normal and Tangential Coordinate Systems / Path variables
(n-t system)
3. Radial and Transverse Coordinate Systems / Polar coordinates
(r - )

3. Coordinate Systems

4. Rectangular / Cartesian Coordinate Systems (x-y)
Position Vector (1)r x i y j  

5. Rectangular / Cartesian Coordinate Systems (x-y)
Position Vector (1)r x i y j       
Differentiate Eq.(1) w.r.t. t
(2)V x y
dr dx dy
i j
dt dt dt
V i V j
 
     
Resultant Velocity :    
22
V x yV V 
Differentiate Eq.(2) w.r.t. t
(3)
yx
x y
dvdvdv
i j
dt dt dt
a a i a j
 
     
Direction: tan y
x
V
V
 
Direction: tan y
x
a
a
 

6. Example on Rectangular Coordinate System
Ex.1 The motion of a particle is defined by the equation
and where in meters. t is in seconds.
Determine the velocity and acceleration when t =4 seconds
Given
       
3 3
2 2
22 2 2 0
16 16 (4) 1018 / ,
18 4 18 (4) 4 4 280 /
1018 280 1055 / 15.38
V
V x
V
x
y
x y
t m s
t t m s
V V m s 
  
    
     
2
4 6x t t 
3 2
6 2y t t  andx y
4
3
2
2
2
4 6
16 6
48
x
x
x t t
dx
V x t
dt
d x
a x t
dt
 
   
  
&
&&
3 2
2
2
2
6 2
18 4 (1)
36 4 (2)
y
y
y t t
dy
V y t t
dt
d y
a y t
dt
 
         
         
&
&&
Find v & a
at t = 4 Sec.
Put t= 4 secs in Eq. (1) and find Velocity Comp. Vx and Vy

7. Example 1 Continue….
• To find acceleration (a)
   
   
2 2
22
2 2
2
0
48 48(4) 768 / ,
36 4 36(4) 4 142 /
768 142
781 /
tan 10.48
x
y
x y
y
x
a t m s
a t m s
a a a
a
a m s
a
a

  
    
 
 

 

8. Example -2
Ex.2 The motion of a particle is defined by the relation
and , where in meters. t is in seconds.
Determine the velocity and acceleration when t =2 seconds.
Given
       
2 2
22 2 2 0
2 5 2 2 5 9 / ,
3 16 3 (2) 16 2 20 / ,
9 20 21.9
tan
65.32 / 77
V x
V x
V
y
x
x
y
x y
t m s
t t m s
V
v
V m s
v


    
     
 

   
2
5x t t 
3 2
8 2y t t   andx y
2
2
2
5
2 5
2
x
x
x
x t t
dx
V t
dt
dVd x
a
dt dt
 
  
  
3 2
2
2
2
8 2
3 16 (1)
6 16 (2)
y
y
y
y t t
dy
V t t
dt
dVd y
a t
dt dt
  
        
     
Find v & a at t = 2 Sec.
Put t= 4 secs in Eq. (1) and find Velocity Comp. Vx and Vy

9. Example 1 Continue….
• To find acceleration (a) Put t = 2 Sec
   
   
2
2
22
2 2
2
0
2 2 / ,
6 16 6(2) 16 4 /
2 4
4.472 /
tan 63.44
x
y
x y
y
x
a m s
a t m s
a a a
a
a m s
a
a

 
     
 
  

 

10. Rectangular Coordinate system
x yV and V are given by differntiating it you
have calculate a and ax y
 Some times in the problem only velocity
Components are given
 Some times in the problem only Acceleration
Components are given
x yand are given by integrating given equation
it you have calculate v andx y
a a
v

11. Tangential and Normal Comp. of Velocity and
Acceleration (n-t System)
Consider the motion of particle along the curved path.
The velocity vector is always tangential to the path but
acceleration vector is at some other angle.
Let,
ˆ
ˆ
t te =e = Unit vector in
Tangential Dirrction
e =e = Unit vector in
Normal Dirrction
n n

12. Tangential and Normal Comp. of Velocity (n-t )
Velocity
ˆ ˆ
t n
t t n n
V= V i +V j
V= V e +V e
Here, Tangential Comp.
of Velocity (Vt)= V itself
and
Normal Comp. of
Velocity (Vn)= 0
Resultant Velocity (V) =
Velocity Components:
2 2
)t nV= (V ) +(V

13. Normal and Tangential Acceleration Compo. (n-t)
Acceleration Components:
ˆ ˆ
t n
t t n n
a a i a j
a a e a e
 
 
t
dv
a
dt

2
n
v
a


Here, Tangential Comp. of
Accl.
and
Normal Comp. of
Acceleration
Acceleration
2 2
)t nAccl. a= (a ) +(a

14. Important Points
Velocity Components: Tangential Comp. of Velocity Vt =V
Normal Comp. of Velocity Vn = 0
Accl. Components:
Tangential Comp. of Accl. at = = Rate of Change Velocity
Normal Comp. of Accl. an = = Directed towards Centre of
Curvature
dv
dt
2
v

To find Radius of Curvature ( )
1. an

2 2
/ , / nv v a  
2
2 2
1 ( / )
( )
( / )
2. if Eq is expressed as
dy dx
y f x
d y dx

   

15. Important Points
1. If the magnitude of velocity / speed is constant then
velocity comp. Vt =0. only Vn will present.
2. If the direction of the particle does not change i.e.
Motion along straight line, then an=0, only at will
present.
3. If the speed in increasing along the curve then at will be
+ ve and directed in the direction of motion.
4. If the speed in decreasing along the curve then at will
be -ve and directed opposite to the direction of motion.
  

16. Numerical Example-1
Ex.1. An outdoor track is 126m in diameter. A runner increases his
speed at a constant rate from 4.2 to 7.2 m/s over a distance of 28.5m.
Determine the total acceleration of the runner 2 sec after he began to
increase his speed.
Given, Diameter = 126m Hence Radius( )= 63m, t= 2 secs
Find a=?
As speed increases at constant rate, u=4.2 m/s and v=7.2 m/s S=28.5 m
Using
, 4.2 0.6 2
Find Velocity t = 2secs
x x v = 5.4m/sectv u a t v   

2 2 2 2
2
2 , 7.5 4.5 2 28.5
0.6 /
x xt t
t
v u a s a
a m s
   


17. Example 1 continue….
Find normal component of Acceleration (an)
2 2
25.4
0.463 /
63
n
v
a m s r

   
Total Acceleration (a)
       
2 2 2 2
2
0.6 0.463
0.758 /Total Accl.
t na a a
a m s
   


18. Numerical Example-2
Ex.2. A train runs at a speed of 120km/hr in a curved track of
radius 900m. The application of brakes suddenly, causes the train to
slow down at a constant rate. After 6 secs, the speed has been
reduced to 72km/hr. Determine the acceleration immediately after
the brakes are applied.
Given, Radius ( )= 900m, t= 6 secs Find a=?
Initial Speed , u= 120km/hr. =120 x = 33.33 m/s and
Final velocity v= 72km/hr= 72 x = 20 m/s S=28.5 m
Since the train slows down the constant rate, find (at )

5
18
5
18
220 33.33
2.222 /
6
t
v u
a m s
t
 
   

19. Example 2 Continue….
Immediately after the brakes are applied (an)
2 2
233.33
1.235 /
900
Here, velocity v = 33.33 m/sec
n
v
a m s
r
  
Total Acceleration (a)
       
2 2 2 2
2
2.222 1.235
2.542 /Total Accl.
t na a a
a m s
    


20. Numerical Example-3
Ex.3. The automobile is originally at rest at x = 0, It then starts to
increase its speed v according to the equation. It then starts to
increase its speed v according to the equation
where t in seconds and s in meters. Determine magnitude of velocity
and acceleration when t =18 seconds. Find v and a = ? at t=18 secs
2 2
/ 0.015 /dv dt t m s
2 2
3
1
1
3
'
4
2
2
0.015
(3)
3 4
0 0 0
0.0
0.015 / (1)
.(1)
0.015
(2)
3
0 0 0
0.015
(2 )
3
1
Integrate Eq. (2) w.r.t.t to get disIntegrate w.r.t.t to get Velocity p.
x
Given
t
S C
dv
a t m s
dt
At t
Eq
t
v C
At t v
t
C
S
C
v
S
   
     
  

     
  
      
4 4
'5 0.015
(3 )
3 4 12x
t t
    

21. 4 4
2 2
'
3
0.015 0.015(18)
12 12
. 131.22 90
.(1)
0.015 0.015 (18)
.(2 ) 18sec
0.015 0
3
t
2
t=18 Sec Find displacement (S)-
131.22mts
i.e. vehicle is on the Curved path.
Find a at t=18secs
4.86 m/st
t
S
See Fig m
From Eq
dv
a t
dt
From Eq t s
t
v
  

   

 
 
3
22
.015(18)
3
:
26.16
72
n
2
26.16 m/sec
To Find a
9.5m/sn
v
a


  
Example 3 Continue….

22. Also see the next video on Coordinate System
Part-2 for third coordinate system i.e. Radial and
Transverse Component System
Thank You