Moment of Inertia - 3
Rotational Motion 3
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Moment of Inertia - 3
Rotational Motion 3
Let’s BEGIN !
Let’s BEGIN !
Radius of Gyration
Radius = k
Radius of Gyration
Radius of Gyration or Gyradius of a body about an axis of rotation is defined as the radial
distance to a point which would have a moment of inertia the same as the body's actual
distribution of mass, if the total mass of the body were concentrated.
It represents a length, hence, it has the unit, metre and dimensions, M0L1T0.
I
M
Consider a body
Suppose the same mass M is located at
a point at a distance k from the axis of
rotation then,Its MOI is given by I = Mk2
I
I
M
k
2
I = Mk
Find the radius of gyration of a disc about a ⊥r axis passing through its
centre.
R
A.
√2
2R
B.
√2 M,R
C. 3R
√2
D. 4R
√2
Solution:
MR2
I =
2
MK2 = MR2
2
K = R
√2
Find the radius of gyration of a body having MOI 20 kg m2 and mass 4 kg.
A. 5 m
B. √5 m I = 20 kg m2
C. 25 m
D. 2.5 m
4 kg
Solution:
I = MK2
20 = 4K2
K = √5m
Find the ratio of the radius of gyration of a hollow sphere and that of a
solid sphere of the same mass & the same radius
A. 2:3
B. 3:5
C. √5 : √3
D. √3 : √5
Solution:
Hollow sphere Solid sphere
2MR2 2MR2
I1 = I2 =
3 5
2MR2
2MR2 M(K2)2 =
M(K1)2 =
3 5
(K1)2
(K2)2
=
(2/3)R2
(2/5)R2
⇒
(K1)
(K2)
= √ 5
3
Problems on Moment of Inertia
2 rods of mass M and 2M having the same length L are joined as shown in
the figure. Find the MOI about an axis passing through their common
point and perpendicular to them.
A. ML2
B. 2ML2
C. 4ML2
D. 6ML2 2M, L
O M, L
Solution:
Steps to be followed
1. Take one object at a time , say rod of mass 2M
2. Find its MOI about the given axis
2ML2
I1 =
3
3. Find MOI of the other object about the given axis
ML2
I2 =
3
4. Add MOI of all objects
2ML2 ML2
Inet = I1 + I2 ⇒ + ⇒ ML2
3 3
2 rods of mass M & 2M having same length L are joined as shown in the
figure. Find MOI about the given perpendicular axis.
A. 1 ML2
2
B. 3 ML2 2M, L
4
C. 5 ML2
6
D. 7 ML2
8
L/2
M, L
2M, L
L/2
M, L
Solution:
Steps to be followed 2M, L
O L
1. Take one object at one time, say rod of mass 2M
L/2
√2
2ML2 M, L
2. Find its MOI about given axis I1 =
12 L/2
3. Find MOI of another object about given axis MOI of rod about ⊥lar
axis passing through
I2 = ICOM + Md2 ML2
its centre is
12
ML2
I2 = + Md2
12
2
ML2 L ML2 ML2 7ML2
I2 = + M ⇒ + ⇒
12 √2 12 2 12
4. Add MOI of all objects
2ML2 7ML2 9ML2 2M, L
Inet = I1 + I2 ⇒ + ⇒
12 12 12
O M, L
Disc of mass M & radius R and rod of mass M & length L = 2R are joined as
shown. Find MOI about the given perpendicular axis.
A. 9MR2 M, R
7
B. 19MR
2
6
C. 29MR
2
6
D. 37MR
2
6 L = 2R
M,L
Solution:
MOI of disc about ⊥lar
Steps to be followed
axis passing through centre is
1. Take one object at one time, say disc MR2
2. Find its MOI about given axis 2
MR2
I1 = M, R As axis is not at
2 COM, so use Parallel
Axis Theorem
3. Find MOI of rod
O
I2 = ICOM + Md2 R MOI of rod about ⊥lar
axis passing through
ML2 its center is
I2 = + Md2 R
12 ML2
C L = 2R 12
M
I2 = (2R)2 + M (2R)2 M
12
M
I2 = (2R)2 + M (2R)2
12
M 13MR2
⇒ 4R2 + M × 4R2 ⇒ M, R
12 3
O
4. Add MOI of all objects
MR2 13MR2
Inet = I1 + I2 ⇒ + L = 2R
2 3
29MR M
⇒ 2
6
Find moment of inertia of a spherical ball of mass m, radius r with its
centre attached at the end of a thin straight rod of mass M & length L
about axis passing through the other end of rod is perpendicular to it.
m,r
M,L
Solution:
Irad = ML2 (across y y’ or A OR)
m,r
Isphere = Icm + ML2 ( || axes theorem) M,L
Isphere = 2 mR2 + mL2
5
Itotal = Irod + Isphere
Itotal = ML + 2 mr2 + mL2
2
3 5
Find MI of a half disc of radius R2 from which a smaller half disc of
radius R1 is cut as shown in figure, about an axis of rotation passing
through centre & perpendicular to plane of disc.
M R2
R1 2M
R2
M R2
I
2I
Solution - I
2 I = Ibig disc - Ismall disc m2- m1 = 2M
2M
2I=
R2
2I=
2I
Solution - II
Mass of the differential half ring
R2
dm = M
x
R1
dx C M
MI of the elemental half ring
dI = dmx2
Find MI of the triangular lamina of mass M about the axis of rotation AB
as shown in figure.
A. 1/2 ML2
B. 1/4 ML2
C. 1/6 ML2
D. 1/8 ML2
L
M
L
L
M
L
Solution:
MI of the square (through centre ⊥ to plane)
1(4M)(l2 + b2)
Ic =
12 L
1(4M) (√2l)2 + √2l)2
= M
12
L
4
Ic = ML2
3
⊥ axes theorem
4I + 4I = Ic
1
I= ML2
6
Find Moment of Inertia of Triangular lamina.
A. 1/6 Mh2
B. 1/12 Mh2
C. 1/3 Mh2
D. 1/8 Mh2
h
h
Find moment of inertia of Equilateral triangular frame made from 3 rods.
A. 5/8 Ma2
B. 5/4 Ma2
C. 8/5 Ma2
D. 4/5 Ma2
a 3M
60°
a
60°
a
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INSTALL
11th / 9, 10
12th / Drop
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NAMOLIVE
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