1.   Using the preceding unit vector relations, obtain an expression for the acceleration.
2.A particle  of mass m moves in a plane under the influence of a force F = - kr, directed toward the origin.  Sketch a polar coordinate system (r, q ) to describe the motion of the particle and thereby obtain the Lagrangian (L = T - V, i.e. difference in kinetic and potential energy).
Solutions:  
1) We differentiate:  v 
=   dr/ dt 
n   +  
r  dq  /dt l
With respect to t to obtain:
a =     ( dr 2/ dt 2) n   +   dr/
dt (dn /dt )  +   dr/ dt (d q /dt ) l   + r
(dq  /dt ) 2 l   
+ r  dq  /dt  (dl/dt)
We next use the 2 equations:
a) :  dn
/dt   =   
dq  /dt l
And:
b) dl /dt  
=   -  dq  /dt n
To separate the components :
a =    [ dr 2/ dt 2   -  r
(dq  /dt
) 2] n   +  [r (d2 q  /dt 2)   + 2 dr/dt (dq  /dt
)] l
Where the first term is just the radial acceleration:  a r
And the second term is the angular acceleration:     a q
2) We make use here of the sketch below:
From which:  r  =  r n
dr/ dt =   ( dr/ dt  n   +   r
dn /dt)
dn /dt  
=   dn / dq   (dq 
/dt) =   l  (dq /dt) 
(Since:  dn / dq   =  l )
Then:    dr/ dt =  ( dr/ dt)  n  +  r  dq  /dt l
L  = T   - V
Where:  T  =    ½ m [( dr 2/ dt 2 )   +  r 2 (d2 q /dt 2) ]  
V   =  ò r o     kr dr =     ½ k r 2
 Then: 
L  = T   - V   =   
½ m [( dr 2/ dt 2 )   +  r 2 (d2 q /dt 2) ]    -   ½ k r 2
This can also be written in more concise notation:
L = T - V = ½ m[ r" + r 2 ( q " ) ] - ½ k r 2

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