Thursday, February 23, 2023

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 2n   +   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 ) 2n   +  [r (d2 q  /dt 2)   + 2 dr/dt (dq  /dt )] l

Where the first term is just the radial acceleration:  r

And the second term is the angular acceleration:     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 n

dr/ dt =   ( dr/ dt  n   +   r dn /dt)

dn /dt   =   dn / dq   (dq  /dt) =   l  (dq /dt) 

(Since:  dn / dq   =  )

Then:    dr/ dt =  ( dr/ dt)  n  +  r  dq  /dt l

L  = T   - V

Where:  T  =    ½ m [( dr 2/ dt 2 )   +  (d2 q /dt 2) ]  


V   =  ò o     kr dr =     ½ k 2

 Then: 

L  = T   - V   =   

½ m [( dr 2/ dt 2 )   +  (d2 q /dt 2) ]    -   ½ k 2

This can also be written in more concise notation:

L =    T   - V  =  ½ m[ r"   +    2 (  " ) ]    -   ½ k 2

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