Solutions To Theoretical Mechanics Problems

 1.   Using  one or more of the preceding unit vector relations, obtain an expression for the acceleration in two dimensions of polar coordinates.

Solution:

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 following two 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.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).

Solution:

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