Motion of Rigid Bodies in Two Dimensions | Mathematics Optional Notes for UPSC PDF Download

 Page 1


Edurev123 
3. Motion of Rigid Bodies in Two 
Dimensions 
3.1 The ends of a heavy rod of length ?? ?? are rigidly attached to two light rings 
which can respectively slide on the thin smooth fixed horizontal and vertical wires 
?? ?? and ?? ?? . The rod starts at an angle ?? to the horizon with an angular velocity 
v[?? ?? (?? -?????? ??? )/?? ?? ] and moves downwards. Show that it will strike the horizontal 
wire at the end of time -?? v?? /(?? ?? )?????? ?[?????? ?(
?? ?? -
?? ?? )?????? ?
?? ?? ]. 
(2011 : 30 Marks) 
Solution: 
Let the rod ???? whose centre of gravity is ?? be inclined at an angle ?? to the horizontal at 
the time ?? . 
Co-ordinates of ?? are (?? cos??? ,?? sin??? ) 
??( Velocity) ?
2
 of ?? =?? 2
???
2
 
So at the time ?? , kinetic energy 
?=
?? 2
[
?? 2
3
???
2
+?? 2
???
2
] 
 
Initial angular velocity 
=
v
3?? 2?? (1-sin??? ) 
? From (i), 
Page 2


Edurev123 
3. Motion of Rigid Bodies in Two 
Dimensions 
3.1 The ends of a heavy rod of length ?? ?? are rigidly attached to two light rings 
which can respectively slide on the thin smooth fixed horizontal and vertical wires 
?? ?? and ?? ?? . The rod starts at an angle ?? to the horizon with an angular velocity 
v[?? ?? (?? -?????? ??? )/?? ?? ] and moves downwards. Show that it will strike the horizontal 
wire at the end of time -?? v?? /(?? ?? )?????? ?[?????? ?(
?? ?? -
?? ?? )?????? ?
?? ?? ]. 
(2011 : 30 Marks) 
Solution: 
Let the rod ???? whose centre of gravity is ?? be inclined at an angle ?? to the horizontal at 
the time ?? . 
Co-ordinates of ?? are (?? cos??? ,?? sin??? ) 
??( Velocity) ?
2
 of ?? =?? 2
???
2
 
So at the time ?? , kinetic energy 
?=
?? 2
[
?? 2
3
???
2
+?? 2
???
2
] 
 
Initial angular velocity 
=
v
3?? 2?? (1-sin??? ) 
? From (i), 
 Initial kinetic energy =
1
2
?? 4?? 2
3
·
3?? 2?? (1-sin??? ) 
Hence, energy equation gives 
1
2
?? 4?? 2
3
???
2
-
1
2
?? 4?? 2
3
·
3?? 2?? (1-sin??? ) =???? (?? sin??? -?? sin??? )
??
2?? 2
3
???
2
-???? (1-sin??? )=???? (sin??? -sin??? )
?? ???
2
=
3?? 2?? (1-sin??? )
??
????
????
=-
v
3?? 2?? (1-sin??? )
 
(negative sign is taken because the motion is towards ?? decreasing) Integrating it 
between the limits of ?? from ?? to 0 , the required time is 
?? ?=-v
2?? 3?? ? ?
0
?? ?
1
v1-sin??? ????
?=v
2?? 3?? ? ?
?? 0
?
????
(cos?
?? 2
-sin?
?? 2
)
=
v
?? 3?? ? ?
?? 0
?
????
sin?(
?? 2
-
?? 2
)
?=v
?? 3?? ? ?
?? 0
?cosec?(
?? 4
-
?? 2
)????
 
?=-2v
?? 3?? [log?tan?(
?? 8
-
?? 4
)]
0
?? ?=+2v
?? 3?? ·log?(
tan?
?? 8
tan?(
?? 8
-
?? 4
)
]
?=2v
?? 3?? ·log?[tan?
?? 8
cot?(
?? 8
-
?? 4
)]
?=-2v
?? 3?? log?[tan?(
?? 8
-
?? 4
)·cot?
?? 8
]