Moment of inertia of a body
OBJECT: i) To determine the moment of inertia of a body about an axis passing through its center of
gravity and perpendicular to its plane by dynamical method (Inertia Table)
ii) To determine the rigidity modulus of the material of the suspension wire.
Apparatus used:
Inertia Table, irregular body, regular body, stop watch, sprit level, physical balance,weight box, Vernier
caliper, Screw gauge
Theory:
The time period of the torsional oscillation of a body of moment of inertia I about the axis of suspension is
𝑇 = 2𝜋√𝐼⁄𝑐
Where ‘c’ is the moment of couple per unit angle of twist of the suspension wire: 𝑐 =
𝜂𝜋𝑟 4
…… (1)
2𝐿
Here r is the radius and L is the length of the suspension wire and η is the rigidity modulus of the
material of the wire. Then
𝐼
𝑇0 = 2𝜋√ 0
𝑐
……….. (2)
If T1 be the period of oscillation about the same suspension wire of the cradle and a body of known moment
of inertia I1 held on the cradle with its axis passing through the center of gravity coinciding with that of
the cradle, then
𝐼 +𝐼1
𝑇1 = 2𝜋√ 0
𝑐
…………… (3)
Let T2 be the period of oscillation about the same suspension wire of the cradle and a body of unknown
moment of inertia I2 placed with its axis of oscillation passing through the center of gravity coinciding with
that of the cradle. Then
𝐼 +𝐼2
𝑇2 = 2𝜋√ 0
𝑐
………….. (4)
From equations (2) and (3)
𝑇2 − 𝑇2
𝐼1 = 𝐼0 1 2 0 ……………… (5)
𝑇0
Similarly, from equations (2) and (4)
𝑇 2 −𝑇 2
𝐼2 = 𝐼0 2 2 0 ……………… (6)
𝑇0
From equations (5) and (6) we can obtain
𝑇 2 −𝑇02
𝐼2 = 𝐼1 22
𝑇1 −𝑇02
……………….. (7)
Comparing the equations (5), (6), (7) moment of inertia of different regular bodies can be compared
with the theoretical values and also moment of inertia of an irregular body can be obtained.
(Refer to Annexure- I for moment of inertia of different regular bodies)
Eliminating I0 between (2) and (3) gives
𝑐=
4𝜋2 𝐼1
𝑇12 −𝑇02
……………… (8)
Substituting for c from Eq. (1) into Eq. (8), we get
𝜂= 4
8𝜋𝐿𝐼1
𝑟 (𝑇12 −𝑇02 )
………………. (9)
Similarly, from Eqs. (1), (2) and (4) we can get
𝜂= 4
8𝜋𝐿𝐼2
𝑟 (𝑇22 −𝑇02 )
……………. (10)
So, the rigidity modulus η can be found either from Eq. (9) or Eq. (10).
In SI units, the lengths are measured in meter, the mass in kg and the time period in second. Then I
is obtained in kg.m2 and η in N/m2
Figure:
Procedure:
(1) Using sprit level, balance the plane of inertia table by rotating screws S1 and S2 so that its base
becomes horizontal.
(2) Balance the inertia table by placing the balancing rings in grooves such that pointer in base of
inertia disc just lies above the pointer on the base of base of inertia table.
(3) Give the slight twist to the table so that it begins to execute torsional in horizontal plane. Now note
the time for 20 oscillations with lamp, scale, telescope arrangement or by eye. Repeat it three
times. This will give you value of T0=t0/20.
(4) Now place the regular body on the table and do again the step 3. This will give you value of
T1=t1/20.
(5) After it, place the irregular body on the table and do again the step 3. This will give you value of
T2=t2/20.
(6) If given regular body is in shape of disc, then find its mass with help of physical balance and also
find its diameter with help of Vernier calipers.
(7) Measure the diameter of the suspension wire with help screw gauge. And length of the wire using
measuring tape.
Observations:
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛 𝑜𝑛 𝑚𝑎𝑖𝑛 𝑠𝑐𝑎𝑙𝑒
(1) Least count Vernier caliper = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛𝑠 𝑜𝑛 𝑣𝑒𝑟𝑛𝑖𝑒𝑟 𝑠𝑐𝑎𝑙𝑒 =……… cm
𝑝𝑖𝑡𝑐ℎ
(2) Least count of screw gauge = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑔 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑠𝑐𝑎𝑙𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛 =……….cm
(3) Least count of the measuring tape = ……….cm
(4) Least count of stop watch = ……… sec
(5) Table for calculation of the moment of inertia (I1) of the cylinder.
Mass of the cylinder(M1) = …….. kg
No. of
Obs.
Diameter of the cylinder(2R)
MSR(A)
VSR(B)
TOTAL(A+B) Mean
Radius
Zero
error
Corrected (R) in
value(2R) cm
Moment
of inertia
𝟏
I1=𝟐 𝑴𝟏 𝑹𝟐
(6) Calculation of moment of inertia of rectangular bar/ regular body
Mass of the body- (M2 )………..kg
parameter No of
obs.
Reading of the
Vernier scale
MSR(A)
Length
Breadth
VSR(B)
Total
reading
=A+B
mean
Zero
error
Corrected I2
value
(7) Determination of Time period of oscillation
Oscillating Body
No of obs.
Time for 20 oscillations (t)
Mean (t)
Time period(T)
T= t/20 (secs)
Cradle/ inertia
table
1
T0
2
3
Cradle and
cylinder
1
T1
2
3
Cradle and Bar
1
T2
2
3
Cradle and
irregular body
1
T2’
2
3
(8) Determination of the radius (r) of the suspension wire
No of
obs
Diameter of wire (2r)
ICSR
(I)
No of
FCSR
complete
(F)
rotation(n)
Pitch(p) CSR
(I-F) ×L.C
PSR
Total
n×p
CSR+PSR
Mean
Radius
(r)
1
2
3
4
5
(9) Length of the Suspension wire, L= …………….cm = ……… m
(take 3 observations for length and then take the mean value.)
Calculations:
Moment of inertia of irregular body –
Comparison of moment of inertia of regular body using time period and also theoretical formula-
Rigidity modulus of the suspension of the wire –
8𝜋𝐿𝐼
8𝜋𝐿𝐼
1
2
𝜂 = 𝑟 4 (𝑇 2 −𝑇
2 ) and 𝜂 = 𝑟 4 (𝑇 2 −𝑇 2 ) ( take the mean of the results)
1
0
2
0
Error calculation-
Results: The moment of inertia of irregular body= ............. kg-m2(
Precautions:
1. The base and inertia table should always be horizontal.
2. The table should execute torsional vibrations only.
3. The amplitude of twist should be small so that the wire is not twisted beyond the elastic limit.
4. There should be no kink in wire.
5. Periodic time should be noted carefully.
Note: If irregular body is approximately in shape of thick hollow cylinder then you can check your result with an
approximate calculation of I2 with following formula.
1
I 2 = m R2 + R2 ,
1
2
2
(
)
Here m=mass of irregular body, R1&R2 : external and internal radius of thick hollow cylinder