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
