2019-03-01_Physics_Times

2

ext Self 2

GM

W U

R



 

A geostationary satellite goes around the earth

in west-east direction. The time period a

geostationary satellite is 24 hours. The angle

between the equatorial plane and the orbital plane

of geostationary satellite is 0 .

The block shall topple about its edge through

O. The torque FL of the applied force is clockwise.

The torque MgL/2 of the weight is anticlockwise.

Applying condition for rotational equilibrium.

/2 0

2

mg

 FL mgL  or F

From the conservation of energy

1 1

2 2

mgh= mv^2  Iω^2

As the cord does not slip v=r

2 2 2 2

2 2 2

.

mgh mgh mgh

mr mk mr I I mr

   

  

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  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
 Li Lf
mvr mv r I' 
  
mv r'  mr^2 v r'/ 2 ' mv r
'
2 2
v J
v
m
 
Tension in the string when bob passes through
lowest point
2
( )
mv
T mg mg mv v r
r
       
Putting v 2 gh and
2 2
T 2
 
  
we get T m g( 2 ) gh
If v' are the velocities of the block of mass M
and (M+m) while passing from the mean position
when executing SHM.
Using law of conservation of linear momentum, we
have
mv ( ) 'M m v
or v' /( )mv Mm
Also, maximum PE = maximum KE
(^1) '2 (^1) ( ) '
2 2
 kA  Mm v
1/
'
( )
M m mv
A
k M m
  
  
   ( )
mv
M m k


x  x1 2 3x x
  2sin 2 3sin 2cos 3sint     t t t
3 3cost
 (5 2 3)sin (2 3 3)cos  t t
 x Asin( ) t
Where A 68 32 3 and
tan^1 2 3 3
5 2 3
    
 
  
x11.1sin( 31.54) t^0
x5sint and y 5cos 5sint t 2

    
 
So these two are perpendicular SHMs with same
amplitude (and frequency) and their phase
difference is.
2

So the path will be a circle of radius
5 units.
(^) TV^1  constant
7.Sol:
8.Sol:
9.Sol:
10.Sol:
11.Sol:
12.Sol:
13.Sol:
14.Sol:
15.Sol:
5/3 1
5/3 1
2
8
(300)
27
V
V T

   
 
 
2 2/
(300)^328
27
V
V T
 
  
 