DC Pandey Solutions: Circular Motion- 2 | DC Pandey Solutions for JEE Physics PDF Download

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Introductory Exercise 7.3
1. | | v
1
®
= u and | | v
2
®
= v (say)
Dv v v
® ® ®
= + -
2 1
( )
       = -
® ®
v v
2 1
    D D v u u v v u
® ® ® ® ® ®
× = - × - ( ) ( )
2 1 2 1
         = × + ×
® ® ® ®
u v u v
2 2 1 1
 - -
® ®
2
2 1
v v
| | | | v v
2
2
1
2 2 2
® ®
+ = + v u
             = - + u gL u
2 2
2
       | | ( ) Dv
®
= -
2 2
2 u gL
         | | ( ) Dv
®
= - 2
2
u gL
2. Ball motion from A to B :
  0 2 2
2 2
= + - u g R
min
( )( )
Þ       u gR
min
2
4 =
Ball motion from C to A :
v u g h
2 2
2 = + -
min
( )
= - 4 2 gR gh 
Þ v g R h = - 2 2 ( )
3. Decrease in KE of bob
     = Increase in PE of bob
  
L
v
2
– v
1
Di
g
v
1
®
®
®
®
h
v
C
A
2R
B
PE of bob
= K (say)
KE of bob
2
= 1/2 mv
0
v
0
Page 3


            
          
 
          
Introductory Exercise 7.3
1. | | v
1
®
= u and | | v
2
®
= v (say)
Dv v v
® ® ®
= + -
2 1
( )
       = -
® ®
v v
2 1
    D D v u u v v u
® ® ® ® ® ®
× = - × - ( ) ( )
2 1 2 1
         = × + ×
® ® ® ®
u v u v
2 2 1 1
 - -
® ®
2
2 1
v v
| | | | v v
2
2
1
2 2 2
® ®
+ = + v u
             = - + u gL u
2 2
2
       | | ( ) Dv
®
= -
2 2
2 u gL
         | | ( ) Dv
®
= - 2
2
u gL
2. Ball motion from A to B :
  0 2 2
2 2
= + - u g R
min
( )( )
Þ       u gR
min
2
4 =
Ball motion from C to A :
v u g h
2 2
2 = + -
min
( )
= - 4 2 gR gh 
Þ v g R h = - 2 2 ( )
3. Decrease in KE of bob
     = Increase in PE of bob
  
L
v
2
– v
1
Di
g
v
1
®
®
®
®
h
v
C
A
2R
B
PE of bob
= K (say)
KE of bob
2
= 1/2 mv
0
v
0
  
1
2
0
2
mv mgh =
       v gh
0
2
2 =
       v gl
0
2 1 = - ( cos ) q
      = ´ ´ ´ - ° 2 5 1 60 9.8 ( cos )
         =7 ms
-1
        
      
          
      
  
       
 
          
           
         
        
       
          
  
 
  
          
      
        
   
   
h
PE of bob = K + mgh