Physics Times – July 2019

(Darren Dugan) #1
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      
   
W WOAB OB
The given force is conservative.
(Method-II)
The given force F Fi F j Fk  x y zˆ ˆ ˆ is said to
be conservative if it satisfies the following three
equations,
Fx Fy
y x

 

  (1)

Fy FZ
z y

 

  (2)

Fz Fx
x z

 
  (3)
For a two dimensional force
F Fi F j x yˆ ˆ then the force is conservative if
it satisfies eq (1)
Given F xy i yx j ^2 ˆ ˆ^2
2
F xyx
2
F x yy

Fx ( )xy 2
y y

 

 

x y( )^2
y




 x y xy(2 ) 2

Fy ( )yx 2
x x

 

 

y x( )^2
y

 
  y x xy(2 ) 2
The given force is conservative


  1. Consider the gravitation force between two
    masses m and M


2 ˆ 3
F r rGmM GmM
r r

 

Where r is the radial vector drawn from the
centre of M to the centre of m. Verify whether
it is a conservative or not.

2.Sol : 2 ˆ
F rGmM
r


(Method-II)





2.Sol :

r xi yj ˆ ˆ

r x y ^22

r i jˆ   cos sinˆ ˆ

r i jˆ x yˆ ˆ
r r

 

 

2 2 2 ˆ ˆ

F C x yi j
x y r r

  
  

(where C = GmM)

   
2 2 3/2ˆ 2 2 3/2ˆ
F Cx i Cy j
x y x y

 
 

x   2 2 3/

Cx
F
x y


^  

y 2 2 3/

Cy
F
x y



 
2 2 3/

Fx Cx
y y x y

  
  
    
 

 

2 2 3/
Cx x y
y

 
 

 

3 2 2 5/
(2 )
2

Cx x y y


 

 

Fx 3 Cxy x y 2 2 5/
y

    

 
2 2 3/

Fx Cy
y x x y

  
  
    
 

 
2 2 3/
Cy x y
x

 
 
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