2019-03-01_Physics_Times

At x distance from end

T

v





xg

v







v xg  v x

a 1 5,a 2  10

 

 

27.Sol:

28.Sol:

29.Sol:
30.Sol:

(^22)
max 1 2
2
min 1 2
5 10 9
5 10 1
I a a
I a a
   
    
   
Let x 0 be the static deflection of the pulley. It
is slightly pulled downward and released at t=0.
At time=t, its displacement from equilibrium
position is x and linear acceleration is ‘a’.
 2 Kx mg 0  (i)
Let T T 1 & 2 are tensions in the strings
(T T mg ma 1    2 ) (ii)
For rotation,  1 2 
a
T T R I
R
 
   
 
( 1 1 ) 2
Ia
T T
R
   (iii)
From (ii) and (iii)
(^212)
Ia
T ma mg
R
  
2( ) 2
Ia
kx kx ma mg
 R
   
2
2 K
a x
I
m
R
  
  
 
 
2
2
2
I
m
T R
K

  
x x x x   1 2 3
  2 sin t 2 3 sin t 2 cos  t 3sin t
(^) 3 3 cost
 (5 2 3) sin  t (2 3 3) cost
 x Asin(  t )
Where A 68 32 3 and
tan^1 2 3 3
5 2 3
    
  
 
x11.1sin( t 31.54)^0