Radial and Transverse coordinate system

1. Coordinate System in Curvilinear Motion Part-2
Sub- Engg. Mechanics
By: Mr. Ghumare S. M.
Radial and Transverse Coordinate Systems
( Polar coordinate System )
(r - )

2. Coordinate Systems
The motion along a curved path is expressed using following three
Coordinate systems. (velocity and acceleration of particle)
• Coordinate System in Curvilinear Motion:
1. Rectangular / Cartesian Coordinate Systems (x-y system)
2. Normal and Tangential Coordinate Systems / Path variables
(n-t system)
In some engg. problems, it is convenient to describe the motion of the
particle in polar coordinates. It is called cylindrical coordinates. i.e.
3. Radial and Transverse Coordinate Systems / Polar coordinates
(r - )

3. Practical Examples
1. Motion of Caller or Slider along the rod in radial direction
2. Motion along the slotted arms
3. Motion of water through nozzles of sprinkler
The position of particle P is defined by its polar coordinates r and

4. Coordinate Systems

5. Radial and Transverse Coordinate Systems / Polar
coordinates (r - )

6. Radial and Transverse Components of Velocity
Velocity Components:
V
V
V
V
r
rr
r
r
V i V j
V e V e
dr d
e r e
dt dt
r e r e






 
 
 
 
Resultant Velocity :    
2 2
V rV V 
(r - )
tan rV
V
 
Radial Comp. of Velocity:
Transverse Comp. of Velocity:
( )r
dr
V r
dt
 
Dir.
( )
d
V r r
dt


 

7. Radial and Transverse Components of Accl.
Accl.
Comp.
   
22 2
2 2
2
2
2
r
rr
r
r
a a i a j
a a e a e
d r d dr d d
a r e r e
dt dt dt dt dt
a r r e r r e




  
  
 
 
    
            
   
Resultant Velocity :    
2 2
ra a a 
(r - )
tan
r
a
a

 
Radial Comp. of Accl.:
Tran. Comp. of Accl.:
Dir.
 
22
2
2r
d r d
a r r r
dt dt


  
        
2
2
2 2
dr d d
a r r r
dt dt dt

 
 
 
    
 

8. Summary
Velocity Components:
Radial Comp. of Velocity:
Transverse Comp. of Accl. :
Acceleration Components:
  
22
2
2r
d r d
a r r r
dt dt


  
        
2
2
2 2
dr d d
a r r r
dt dt dt

 
 
 
    
 
Radial Comp. of Accl.:
( )r
dr
V r
dt
 
Transverse Comp. of Velocity : ( )
d
V r r
dt


 

9. Important Points
Case – I: If ‘ r ’ is constant means only changes, i.e.
motion will be curvilinear or angular. See Fig.next page
If ‘ r ’ is constant
2
2
. . 0, 0
dr d r
i e r r
dt dt
   
Then put in velocity and Accl. Equation
Velocity Comp. will be
0r r 

r θ(V ) = r = 0 only v = r θ will be present.
.Accl. Comp will be
    
22
r ra = r -r θ ,here r =0, th a = ren - θ
θθa = 2 r θ+r θ here r=0, then a = r θ

10. Important Points
Case – I: If ‘ ’ is constant means only r changes, i.e.
Motion will be rectilinear along radial dir.
If ‘ ’ is constant
2
2
. . 0, 0
d d
i e
dt dt
 
    
Then put in velocity and Accl. Equation
Velocity Comp. will be
0  

r θV = r and v = r θ = 0 as θ = 0
.Accl. Comp will be
    
2
r ra = r -r θ , then θ =0, He a = rnce
0θθa = 2 r θ+r θ here r=θ, θ, Hence a =


11. Figure 1

12. Numerical Example-1
Ex.1. A slider block P along bar OA and rotate about pivot O. The
angular position of bar is given as and the
position of the slider is given as . Determine the
velocity and acceleration of the slider at t=2secs. Where r in meters, t
is in seconds and is in radians.
Given, To find V and a at t = 2secs.
3
0.4 0.12 0.06t t   
2
0.8 0.1 0.05r t t  

2
2
2
0.8 0.1 0.05 ,
0.1 0.10 ,
0.1,
r t t
dr
r t
dt
d r
r
dt
  
   
  
3
2
2
2
2
0.4 0.12 0.06 ,
0.12 0.18 ,
0.12 0.18
t t
d
t
dt
d
t
dt





  
  
  

13. Example-1 continue… To find v and a at t = 2secs.
2
2
2
0.8 0.1 0.05 ,
, 2sec ,
0.1 0.10 ,
2sec, 0.1 0.10 2,
0.1, 2
r =0.4m
r x -0.3m/s
r = -0.1m/s
r t t
At t s
dr
r t
dt
t
d r
r
dt
  

   
    
  
3
2
2
2
2
0.4 0.12 0.06 ,
0.12 0.18 ,
, 2sec, ,
0.12 0.18 ,
, 2sec, 2
θ=0.84 rad/s
θ=0.72 rad/s
t t
d
t
dt
At t
d
t
dt
At t





  
  

  

Velocity Components:
   
2 2
( ) 0.3 / sec, Re. Velo.
v = 0.45 m/sec
rr v v
dr
V r vm
dt
    
( ) 0.4 0.84 0.338 /x
d
V r r m s
dt


   

14. Problem 1 continue….
Transverse Comp.
of Acele. :
Acceleration Components:
  2
2
( 0.1 0.4(0.84) )
2
ra = - 0.38 m/s
r
r
a r r
a
 
  
2
2 ( 0.3) 0.84 0.4 0.72
2
θa = - 0.246m/
x x
s
a r r
a


  
  
Radial Comp. of Accel. :
   
   
2 2
2 2
0.38 0.246
2
a = 0.437m/s
ra a a
a
 
  
Total Acceleration :

15. Numerical Example-2
Ex.2. A car travels around a circular track such that its transverse
component is . Determine the radial and transverse
components of Velocity and acceleration at t=4secs. (Refer Fig.)
2
0.006t rad 
Given, To find V and a Compo. at t = 4secs.
0
2
2
5.5
400sin 400sin5.5 0.048
400 ( 400 ) 2
r =398.16m
x r =-1.84m/s
put all terms r =-1.36
r=400cosθ
7m/s
put
dr
r
dt
d r
r Cos Cos
dt

 
     

     
    
0
2
2
4sec, 5.5
0.012
2
2
θ 0.048 rad/sec
θ 0.012 rad/s
θ=0.0
e
06t ,
c
Put t
d
t
dt
d
dt



 
  
 

16. Example-2 continue… To find v and a at t = 4secs.
Radial Compo. of Velocity:
r
dr
(V )= = r = -1.84 m/sec,
dt
θ
dθ
(V ) = r = r θ =398.16 x 0.048 = 19.11m/s
dt
   
   
2 2
2 2
1.84 19.11
Resultant Velocity
v = 19.19 m/sec
rv v v
v
 
  
Transverse Compo. of Velocity:

17. Problem 1 continue….
Transverse Comp. of
Acele. :
Acceleration Components:
  2
2
( 1.367 398.17(0.48) )
2
ra = - 2.28 m/s
r
r
a r r
a
 
  
2
2 ( 1.84) 0.048 398.16 0.012
2
θ
x x
a = 4.6m/s
a r r
a


  
  
Radial Comp. of Accel. :
   
   
2 2
2 2
2.28 4.6
2
a = 5.13 m/s
ra a a
a
 
  
Total Acceleration :

18. Important Point
Some time some components are directly given
2 2
2 2
, ,
dθ d θ dr d
dt dt dt dt
r
or
or components are directly given
, ,r r and  
Thank You