Physics Times – July 2019

6. An impulse J is exerted on a rod of mass m and
   length L at a distance x from the C.M. Find the
   value of x at which the net velocity of point P is
   zero.


7. A cockroach of mass m is moving on the rim of a
   disc with velocity v in the anticlockwise direction.
   The moment of inertia of the disc about its own
   axis is I and it is rotating in the clockwise direction
   with angular speed

(a)

L

12

(b)

L

8 (c)

L

3 (d)

L

6

. If the cockroach stops

moving then the angular speed of the disc will be

(a)

I mvR

I



(b) 2

I mvR

I mR





(c) 2

I mvR

I mR





(d) 2

I

I mR





8. 
   

   The two uniform discs rotate separately on parallel
   axles. The upper disc (radius a and momentum of
   inertia

   
   


9. 
   


10. 
    


11. 
    

I 1 ) is given an angular velocity  0 and the

lower disc of (radius b and momentum of inertia

I 2 ) is at rest. Now the two discs are moved together

so that their rims touch. Final angular velocity of

the upper disc is

(a)

 

 

1 0

2 2

1 2 /

I

I a I b



   (b)

 

 

1 0

2 2

1 2 /

I

I b I a



  

(c)

 

 

2 0

2 2

2 1 /

I

I b I a



   (d)

 

 

2 0

2 2

2 1 /

I

I a I b



  

1. c 2. d 3. c 4. a 5. a


2. d 7. c 8. a


3. 
   
   
   
   2. 
      
      
      
      3. 
         
         
         
         4. 
            
            
            
            5. 
               
            
            
            
            
         
         
         
         
      
      
      
      
   
   
   
   


4. 
   
   
   
   7. 
      
      
      
      8. 
         
      
      
      
      
   
   
   
   

1.Sol: Let l is the length of the rod.

2.Sol: Let v be the velocity of COM of ring just after

the impulse is applied and v’ its velocity when pure

rolling starts. Angular velocity

2

v J

m



2 2

2 2 2

l l l

J m m 

      

       

      

J

ml



(^) A 2 2 2
l J J J
v v
m m m
     
 of the ring at
this instant will be
v'
r
.
From impulse = change in linear momentum, we
have
J = mv v = J/m
Between the two positions shown in the figure,
force of friction on the ring acts backwards.
Angular momentum of the ring about bottom most
point will remain conserved
 L Li f
(^) mvr mv r I ' 
mv r mr v r mv r' ^2  '/  2 '
' 2 2
v v J
m
 
3.Sol: Impulse provided by the edge in the horizontal
direction:
1.Sol:
2.Sol:
3.Sol: