5. radial and transverse compo. 2 by-ghumare s m

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
r
r
r
r
ViVj
VeVe
dr d
ere
dt dt
rere











Resultant Velocity : 
22
V
r
VV


(r - )  tan
r
V
V


Radial Comp. of Velocity:
Transverse Comp. of Velocity: ()
r
dr
Vr
dt

Dir. ()
d
Vr r
dt




7. Radial and Transverse Components of Accl.
Accl.
Comp.   
2
22
22
2
2
2
r
r
r
r
r
aaiaj
aaeae
dr d drd d
a r e r e
dt dt dtdtdt
arre rre





 
 


 
    

 
  
Resultant Velocity : 
22
r
aa a


(r - )  tan
r
a
a


Radial Comp. of Accl.:
Tran. Comp. of Accl.:
Dir.  
2
2
2
2r
dr d
a r rr
dt dt




   


 2
2
22
drd d
a r rr
dtdtdt




   


8. Summary
Velocity Components:
Radial Comp. of Velocity:
Transverse Comp. of Accl. :
Acceleration Components:  
2
2
2
2r
dr d
a r rr
dt dt




   


 2
2
22
dr d d
a r r r
dt dt dt




   

Radial Comp. of Accl.: ()
r
dr
Vr
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 dr
ie r r
dt dt
 
Then put in velocity and Accl. Equation
Velocity Comp. will be 0rr  r θ
(V) = r = 0onlyv= rθwillbepresent. .   Accl. Comp will be   
22
r r
a=r-rθ,herer=0,th a=ren-θ θθ
a=2rθ+rθherer=0,then a=rθ

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
dd
ie
dt dt

 
Then put in velocity and Accl. Equation
Velocity Comp. will be 0  r θ
V = r andv= rθ=0 asθ=0 .   Accl. Comp will be   
2
r r
a=r-rθ,thenθ=0,He a=rnce 0
θθ
a=2rθ+rθherer=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
0.40.120.06tt  2
0.80.10.05r t t  2
2
2
0.80.10.05,
0.10.10,
0.1,
r t t
dr
rt
dt
dr
r
dt


 3
2
2
2
2
0.40.120.06,
0.120.18,
0.120.18
tt
d
t
dt
d
t
dt





 



13. Example-1 continue… To find v and a at t = 2secs.
2
2
2
0.80.10.05,
,2sec,
0.10.10,
2sec,0.10.102,
0.1,
2
r =0.4m
r x-0.3m/s
r = -0.1m/s
r t t
Att s
dr
rt
dt
t
dr
r
dt



  
 3
2
2
2
2
0.40.120.06,
0.120.18,
,2sec, ,
0.120.18,
,2sec,
2
θ=0.84rad/s
θ=0.72rad/s
tt
d
t
dt
Att
d
t
dt
Att





 




Velocity Components: 
22
() 0.3/sec, Re.Velo.
v = 0.45m/sec
rr
vv
dr
Vr vm
dt

  () 0.40.840.338/x
d
Vr r ms
dt


  

14. Problem 1 continue….
Transverse Comp.
of Acele. :
Acceleration Components:  
2
2
(0.10.4(0.84))
2
r
a= - 0.38m/s
r
r
arr
a

 2
2(0.3)0.840.40.72
2
θ
a= - 0.246m/
xx
s
arr
a



 
Radial Comp. of Accel. : 
  
22
22
0.380.246
2
a = 0.437m/s
r
aa 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
0.006trad
Given, To find V and a Compo. at t = 4secs. 0
2
2
5.5
400sin 400sin5.50.048
400 (400 )
2
r =398.16m
x r =-1.84m/s
putalltermsr =-1.36
r=400cosθ
7m/s
put
dr
r
dt
dr
r Cos Cos
dt


 

  
  0
2
2
4sec,5.5
0.012
2
2
θ 0.048rad/sec
θ0.012rad/s
θ=0.0
e
06t,
c
Putt
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 = -1.84m/sec,
dt θ
dθ
(V) = r= rθ =398.16x0.048 = 19.11m/s
dt 
  
22
22
1.8419.11
Resultant Velocity
v = 19.19m/sec
r
vv v
v


 
Transverse Compo. of Velocity:

17. Problem 1 continue….
Transverse Comp. of
Acele. :
Acceleration Components:  
2
2
(1.367398.17(0.48))
2
r
a= - 2.28m/s
r
r
arr
a

  2
2(1.84)0.048398.160.012
2
θ
xx
a= 4.6m/s
arr
a



 
Radial Comp. of Accel. : 
 
22
22
2.284.6
2
a = 5.13 m/s
r
aa a
a


 
Total Acceleration :

18. Important Point
Some time some components are directly given 22
22
,,
dθdθ drd
dtdt dtdt
r
or
or components are directly given ,,rrand
Thank You