Moment of Inertia | Mathematics Optional Notes for UPSC PDF Download

 Page 1


Edurev123 
2. Moment of Inertia 
2.1 The flat surface of a hemisphere of radius ' ?? ' is cemented to one flat surface 
of a cylinder of the same radius and of the same material. If the length of the 
cylinder be ?? and the total mass be ?? , show tiat the moment of inertia of the 
combination about the axis of the cylinder is given by : 
?? ?? ?? (
?? ?? +
?? ????
?? )
(?? +
?? ?? ?? )
 
(2009 : 12 Marks) 
Solution: 
The combined moment of inertia about the axis of cylinder will be given by adding their 
moment of inertia separately about this axis, i.e., 
?? ?? Tolal 
=?? ?? ?? +?? ?? ?? (1)
 
Now, let ?? be the density of material and ?? ?? ,?? ?? be the masses of hemisphere and 
cylinder respectively. So, 
?? ?=?? ?? +?? ?? ?? ?=?? ×
2
3
?? ?? 3
+?? ×?? ?? 2
?? 
 
Now, we know that moment of inertia of a spherical body (solid) or hemispherical body 
(solid) is given by 
2
5
?? ?? 2
 
Page 2


Edurev123 
2. Moment of Inertia 
2.1 The flat surface of a hemisphere of radius ' ?? ' is cemented to one flat surface 
of a cylinder of the same radius and of the same material. If the length of the 
cylinder be ?? and the total mass be ?? , show tiat the moment of inertia of the 
combination about the axis of the cylinder is given by : 
?? ?? ?? (
?? ?? +
?? ????
?? )
(?? +
?? ?? ?? )
 
(2009 : 12 Marks) 
Solution: 
The combined moment of inertia about the axis of cylinder will be given by adding their 
moment of inertia separately about this axis, i.e., 
?? ?? Tolal 
=?? ?? ?? +?? ?? ?? (1)
 
Now, let ?? be the density of material and ?? ?? ,?? ?? be the masses of hemisphere and 
cylinder respectively. So, 
?? ?=?? ?? +?? ?? ?? ?=?? ×
2
3
?? ?? 3
+?? ×?? ?? 2
?? 
 
Now, we know that moment of inertia of a spherical body (solid) or hemispherical body 
(solid) is given by 
2
5
?? ?? 2
 
So, 
?? ?? ?=
2
5
???? ?? 2
=
2
5
×?? ×
2
3
?? ?? 3
?? 2
 
Similarly, for calculating moment of inertia of cylinder (solid) can be calculated by 
assuming a disc of length/ width ???? 
Now 
???? ?? ?? =
???? ?? 2
2
 
(????? of a disc about its axis is 
?? ?? 2
2
 ) 
Now, 
 mass of disc =???? =?? ×?? ?? 2
???? 
So, 
???? ?? ?? ?=
???? ?? 2
·?? 2
????
2
 
 
From(1), 
?? ?? Total 
?=?? ?? ?? +?? ?? ?? ?=
4???? ?? 5
15
+
???? ?? 4
?? 2
 
[from (4) and (5)] 
From (3), put value of ?? in (6) 
?? ?? Total 
=
?? ?? 4
×?? ?? ?? 2
(?? +
2?? 3
)
(
?? 2
+
4?? 15
)=
?? ?? 2
(
?? 2
+
4?? 15
)
?? +
2?? 3
 
Page 3


Edurev123 
2. Moment of Inertia 
2.1 The flat surface of a hemisphere of radius ' ?? ' is cemented to one flat surface 
of a cylinder of the same radius and of the same material. If the length of the 
cylinder be ?? and the total mass be ?? , show tiat the moment of inertia of the 
combination about the axis of the cylinder is given by : 
?? ?? ?? (
?? ?? +
?? ????
?? )
(?? +
?? ?? ?? )
 
(2009 : 12 Marks) 
Solution: 
The combined moment of inertia about the axis of cylinder will be given by adding their 
moment of inertia separately about this axis, i.e., 
?? ?? Tolal 
=?? ?? ?? +?? ?? ?? (1)
 
Now, let ?? be the density of material and ?? ?? ,?? ?? be the masses of hemisphere and 
cylinder respectively. So, 
?? ?=?? ?? +?? ?? ?? ?=?? ×
2
3
?? ?? 3
+?? ×?? ?? 2
?? 
 
Now, we know that moment of inertia of a spherical body (solid) or hemispherical body 
(solid) is given by 
2
5
?? ?? 2
 
So, 
?? ?? ?=
2
5
???? ?? 2
=
2
5
×?? ×
2
3
?? ?? 3
?? 2
 
Similarly, for calculating moment of inertia of cylinder (solid) can be calculated by 
assuming a disc of length/ width ???? 
Now 
???? ?? ?? =
???? ?? 2
2
 
(????? of a disc about its axis is 
?? ?? 2
2
 ) 
Now, 
 mass of disc =???? =?? ×?? ?? 2
???? 
So, 
???? ?? ?? ?=
???? ?? 2
·?? 2
????
2
 
 
From(1), 
?? ?? Total 
?=?? ?? ?? +?? ?? ?? ?=
4???? ?? 5
15
+
???? ?? 4
?? 2
 
[from (4) and (5)] 
From (3), put value of ?? in (6) 
?? ?? Total 
=
?? ?? 4
×?? ?? ?? 2
(?? +
2?? 3
)
(
?? 2
+
4?? 15
)=
?? ?? 2
(
?? 2
+
4?? 15
)
?? +
2?? 3
 
Hence, moment of inertia of the combination about the axis of the cylinder is given by: 
?? ?? 2
(
?? 2
+
4?? 15
)
(?? +
2?? 3
)
 
2.2 A uniform lamina is bounded by a parabolic arc of latus rectum ?? ?? and a 
double ordinate at a distance ?? from the vertex. If ?? =
?? ?? (?? +?? v?? ) , show that two of 
the principal axes at the end of a lotus rectum are tangent and normal there. 
(2010 : 12 Marks) 
Solution: 
Let the equation of parabola be 
?? 2
=4???? (1) 
Coordinates of the end ?? of the latus rectum ?? ?? '
 are (?? ,2?? ) (see fig.) Differentiating eqn. 
(1), we get 
? At ?? (?? ?? ,2?? ) , 
2?? ·?? '
=4?? ??? '
=
2?? ?? 
? Equation of the tangent ???? at ?? is 
????
????
=
2?? 2?? =1 
 or ?
?? -2?? ?=1·(?? -?? )
?? -?? -?? ?=0
 
and equation of normal ???? at ?? is 
Page 4


Edurev123 
2. Moment of Inertia 
2.1 The flat surface of a hemisphere of radius ' ?? ' is cemented to one flat surface 
of a cylinder of the same radius and of the same material. If the length of the 
cylinder be ?? and the total mass be ?? , show tiat the moment of inertia of the 
combination about the axis of the cylinder is given by : 
?? ?? ?? (
?? ?? +
?? ????
?? )
(?? +
?? ?? ?? )
 
(2009 : 12 Marks) 
Solution: 
The combined moment of inertia about the axis of cylinder will be given by adding their 
moment of inertia separately about this axis, i.e., 
?? ?? Tolal 
=?? ?? ?? +?? ?? ?? (1)
 
Now, let ?? be the density of material and ?? ?? ,?? ?? be the masses of hemisphere and 
cylinder respectively. So, 
?? ?=?? ?? +?? ?? ?? ?=?? ×
2
3
?? ?? 3
+?? ×?? ?? 2
?? 
 
Now, we know that moment of inertia of a spherical body (solid) or hemispherical body 
(solid) is given by 
2
5
?? ?? 2
 
So, 
?? ?? ?=
2
5
???? ?? 2
=
2
5
×?? ×
2
3
?? ?? 3
?? 2
 
Similarly, for calculating moment of inertia of cylinder (solid) can be calculated by 
assuming a disc of length/ width ???? 
Now 
???? ?? ?? =
???? ?? 2
2
 
(????? of a disc about its axis is 
?? ?? 2
2
 ) 
Now, 
 mass of disc =???? =?? ×?? ?? 2
???? 
So, 
???? ?? ?? ?=
???? ?? 2
·?? 2
????
2
 
 
From(1), 
?? ?? Total 
?=?? ?? ?? +?? ?? ?? ?=
4???? ?? 5
15
+
???? ?? 4
?? 2
 
[from (4) and (5)] 
From (3), put value of ?? in (6) 
?? ?? Total 
=
?? ?? 4
×?? ?? ?? 2
(?? +
2?? 3
)
(
?? 2
+
4?? 15
)=
?? ?? 2
(
?? 2
+
4?? 15
)
?? +
2?? 3
 
Hence, moment of inertia of the combination about the axis of the cylinder is given by: 
?? ?? 2
(
?? 2
+
4?? 15
)
(?? +
2?? 3
)
 
2.2 A uniform lamina is bounded by a parabolic arc of latus rectum ?? ?? and a 
double ordinate at a distance ?? from the vertex. If ?? =
?? ?? (?? +?? v?? ) , show that two of 
the principal axes at the end of a lotus rectum are tangent and normal there. 
(2010 : 12 Marks) 
Solution: 
Let the equation of parabola be 
?? 2
=4???? (1) 
Coordinates of the end ?? of the latus rectum ?? ?? '
 are (?? ,2?? ) (see fig.) Differentiating eqn. 
(1), we get 
? At ?? (?? ?? ,2?? ) , 
2?? ·?? '
=4?? ??? '
=
2?? ?? 
? Equation of the tangent ???? at ?? is 
????
????
=
2?? 2?? =1 
 or ?
?? -2?? ?=1·(?? -?? )
?? -?? -?? ?=0
 
and equation of normal ???? at ?? is 
 
 or ??? +?? -3?? =0 (3) 
Consider an element ???????? at the point ?? (?? ,?? ) of the lamina. 
???? = length of the perpendicular from ?? on the tangent ????
 given by (2) 
 
=
?? -?? -?? v1+1
=
?? -?? -?? v2
 
and 
???? = length of the perpendicular from ?? on the normal ???? given by (3) 
=
?? +?? -3?? v2
 
Product of inertia of the element about ???? and ???? 
=???? ·?? :???? =
(?? -?? -?? )
v2
(?? +?? -3?? )
v2
·?????????? 
If the tangent and normal at ?? are the principal axes, then product of inertia of the lamina 
about these will be zero. 
Hence, product of inertia of lamina about ???? and ???? is equal to zero, i.e., 
? ?
?? ?? =0
? ?
?? 0
? ?
2v????
-2v????
{?? 2
-4???? +(3?? 2
+2???? -?? 2
)}???????? =0 
Page 5


Edurev123 
2. Moment of Inertia 
2.1 The flat surface of a hemisphere of radius ' ?? ' is cemented to one flat surface 
of a cylinder of the same radius and of the same material. If the length of the 
cylinder be ?? and the total mass be ?? , show tiat the moment of inertia of the 
combination about the axis of the cylinder is given by : 
?? ?? ?? (
?? ?? +
?? ????
?? )
(?? +
?? ?? ?? )
 
(2009 : 12 Marks) 
Solution: 
The combined moment of inertia about the axis of cylinder will be given by adding their 
moment of inertia separately about this axis, i.e., 
?? ?? Tolal 
=?? ?? ?? +?? ?? ?? (1)
 
Now, let ?? be the density of material and ?? ?? ,?? ?? be the masses of hemisphere and 
cylinder respectively. So, 
?? ?=?? ?? +?? ?? ?? ?=?? ×
2
3
?? ?? 3
+?? ×?? ?? 2
?? 
 
Now, we know that moment of inertia of a spherical body (solid) or hemispherical body 
(solid) is given by 
2
5
?? ?? 2
 
So, 
?? ?? ?=
2
5
???? ?? 2
=
2
5
×?? ×
2
3
?? ?? 3
?? 2
 
Similarly, for calculating moment of inertia of cylinder (solid) can be calculated by 
assuming a disc of length/ width ???? 
Now 
???? ?? ?? =
???? ?? 2
2
 
(????? of a disc about its axis is 
?? ?? 2
2
 ) 
Now, 
 mass of disc =???? =?? ×?? ?? 2
???? 
So, 
???? ?? ?? ?=
???? ?? 2
·?? 2
????
2
 
 
From(1), 
?? ?? Total 
?=?? ?? ?? +?? ?? ?? ?=
4???? ?? 5
15
+
???? ?? 4
?? 2
 
[from (4) and (5)] 
From (3), put value of ?? in (6) 
?? ?? Total 
=
?? ?? 4
×?? ?? ?? 2
(?? +
2?? 3
)
(
?? 2
+
4?? 15
)=
?? ?? 2
(
?? 2
+
4?? 15
)
?? +
2?? 3
 
Hence, moment of inertia of the combination about the axis of the cylinder is given by: 
?? ?? 2
(
?? 2
+
4?? 15
)
(?? +
2?? 3
)
 
2.2 A uniform lamina is bounded by a parabolic arc of latus rectum ?? ?? and a 
double ordinate at a distance ?? from the vertex. If ?? =
?? ?? (?? +?? v?? ) , show that two of 
the principal axes at the end of a lotus rectum are tangent and normal there. 
(2010 : 12 Marks) 
Solution: 
Let the equation of parabola be 
?? 2
=4???? (1) 
Coordinates of the end ?? of the latus rectum ?? ?? '
 are (?? ,2?? ) (see fig.) Differentiating eqn. 
(1), we get 
? At ?? (?? ?? ,2?? ) , 
2?? ·?? '
=4?? ??? '
=
2?? ?? 
? Equation of the tangent ???? at ?? is 
????
????
=
2?? 2?? =1 
 or ?
?? -2?? ?=1·(?? -?? )
?? -?? -?? ?=0
 
and equation of normal ???? at ?? is 
 
 or ??? +?? -3?? =0 (3) 
Consider an element ???????? at the point ?? (?? ,?? ) of the lamina. 
???? = length of the perpendicular from ?? on the tangent ????
 given by (2) 
 
=
?? -?? -?? v1+1
=
?? -?? -?? v2
 
and 
???? = length of the perpendicular from ?? on the normal ???? given by (3) 
=
?? +?? -3?? v2
 
Product of inertia of the element about ???? and ???? 
=???? ·?? :???? =
(?? -?? -?? )
v2
(?? +?? -3?? )
v2
·?????????? 
If the tangent and normal at ?? are the principal axes, then product of inertia of the lamina 
about these will be zero. 
Hence, product of inertia of lamina about ???? and ???? is equal to zero, i.e., 
? ?
?? ?? =0
? ?
?? 0
? ?
2v????
-2v????
{?? 2
-4???? +(3?? 2
+2???? -?? 2
)}???????? =0 
???
?? 2
? ?
?? 0
?? ?
2v????
-2v????
?{?? 2
-4???? +(3?? 2
+2???? -?? 2
)}???????? =0
???? ?
?? 0
?[
1
3
?? 3
?2?? ?? 2
+(3?? 2
+2???? -?? 2
)·?? ]
-2v????
2v????
???? =0
???2? ?
?? 0
?{
8
3
???? v???? +2(3?? 2
+2???? -?? 2
)
v???? }???? =0
???? ?
?? 0
?(
8
3
?? 3/2
?? 3/2
+6?? 5/2
?? 1/2
+4?? 3/2
?? 3/2
-2?? 1/2
?? 5/2
)???? =0
???[
16
15
?? 3/2
?? 5/2
+4?? 5/2
?? 3/2
+
8
5
?? 3/2
?? 5/2
-
4
7
?? 1/2
?? 7/2
]=0
???
16
15
???? +4?? 2
+
8
5
???? -
4
7
?? 2
=0
????? 2
-
14
3
???? -7?? 2
=0
????? =
14?? 3
±
v
196
9
?? 2
+28?? 2
2
=
1
2
(
14
3
±
8
3
v7)?? ????? =
?? 3
(7+4v7) (-ve sign rejected as ?? is always positive) 
 
Hence, if ?? =
?? 3
(?? +4v7) , then principal axes at ?? are the tangents and normal there. 
2.3 Let ?? be the radius of the base of a right circular cone of height ?? and mass ?? . 
Find the moment of inertia of that right circular cone about a line through the 
vertex perpendicular to the axis. 
(2011 : 12 Marks) 
Solution: 
Let ?????? be the given cone whose semi-vertical angle is ?? and mass per unit volume is 
?? .