Physics 2107
Moments of Inertia
Experiment 1
Read the following background/setup and ensure you are familiar with the
theory required for the experiment. Please also fill in the missing equations 5,
7 and 9.
Background/Setup
The moment of inertia, I, of a body is a measure of how hard it is to get it rotating
about some axis. The moment I is to rotation as mass m is to translation. The larger
the value of I, the more work must be done in order to get the object spinning. This is
analogous to the larger the mass, the more work must be done in order to get it
moving in a straight line. Alloy rim wheels on a bicycle have a lower moment of
inertia than steel rim wheels, thereby making them easier to set spinning and, as a
result, making it easier to accelerate the bicycle.
The moment of inertia of a body is always defined with respect to a particular axis of
rotation. This is frequently the symmetry axis of the body, but it can in fact be any
axis – even one that is outside the body. The moment of inertia of a body about a
particular axis is defined as:
(1)
I = ∑ mi ri 2 ,
i
Figure 1 A rotating disk is
composed of many particles, two
of which are shown.
where the sum is over all the body parts (of index i),
mi is the mass of part I and ri is the distance from
part i to the axis of rotation. This sum is easy to
perform if the object consists of discrete point
masses (Figure 1). If the body is a continuous object
of arbitrary shape, performing the sum requires
using integral calculus. In this practical we will
determine the moment of inertia of a number of
objects and compare the values obtained using two
different methods.
For a disk with an axis through its centre of symmetry (Figure 2) the moment of
inertia is given by:
1
(2)
I = mr 2 .
2
Axis of
rotation
Note that the thickness of the disk has no
influence on the value for I, which depends
only on the radius, r and the total mass, m.
r
mass, m
Figure 2 Disk with axis through its centre.
1.1
PY2107
Moments of Inertia
Experiment 1
__________________________________________________________________________________
In this experiment you will determine I for two different systems: (i) a disk and axle
rolling down an incline and (b) a ball-bearing oscillating on a concave spherical
surface. For both systems you will determine I in two ways. First, you will measure
the mass and radii of the systems under investigation and compute I from given
formulae. Then you will compute I by experimental investigation and using the
principle of conservation of energy. You will be expected to compare the values that
you obtain.
Part 1 Moment of Inertia of a Disk and Axle
In this experiment you must measure I for a disk mounted on an axle. The axle can
be thought of as a very thick disk and you use the same expression to compute I disk
and I axle . The total I of the disk + axle is the sum I disk + I axle .
R
=
radius r
+
maxle
mdisk
Method 1: Calculate the moment of inertia of the disk and axle about the axis of
symmetry using the equation:
1
1
(3)
I = I wheel + I axle = mdisk R 2 + maxle r 2 .
2
2
Measure the masses and radii of the disk and the axle and then compute I using the
expression above. Note that we are assuming that the disk is complete i.e. we are
ignoring the missing section through which the axle passes by assuming that the
difference is negligible since r << R.
mdisk =
R=
maxle =
r=
Ensure you include the maximum error in each of your measurements, e.g.
maxle = (300 ± 0.05) g . All data that you record should show the measured value and
its associated maximum error (and, of course, units).
Result using method 1 for axle + disk
I=
1.2
PY2107
Moments of Inertia
Experiment 1
__________________________________________________________________________________
Method 2: In the second method you will compute I by timing the wheel as it rolls
down inclined rails and using the principle of conservation of energy.
Consider a wheel consisting of disk and axle, rolling down an inclined set of rails
after starting from rest at the top. The wheel will move down the plane with constant
acceleration and its total energy will consist of the sum of the translational kinetic
energy, the rotational kinetic energy and the gravitational potential energy.
1
1
(4)
Energy = KEtrans + KE rot + PE = Mv 2 + Iω 2 + Mgh.
2
2
Here, M is the total mass of disk + axle, v is its translational speed, ω is its angular
velocity and h is the height of the centre of mass.
If the wheel starts from rest at position A and rolls downwards to position B then the
loss in potential energy must equal the gain in kinetic energy.
A
v
B
l
h
Zero PE height
α
Figure 3: Experimental arrangement
At point A, the energy is entirely potential since the wheel is at rest:
Energyinitial = PE =
(5)
At point B, the energy is entirely kinetic:
Energy final = KEtrans + KE rot =
1
1
Mv 2f + Iω 2f ,
2
2
(6)
where v f is the final translational speed and ω f is the final angular speed.
By assuming that friction is very small, we can assume that the total energy is
constant as the wheel rolls down the rails and so the initial energy is equal to the final
energy.
1.3
PY2107
Moments of Inertia
Experiment 1
__________________________________________________________________________________
(7)
=
For an axle or wheel that rolls without slipping, the angular velocity and the
translational speed are related by:
v
(8)
ω= .
r
Note that here, r is the radius of the axle, NOT the radius of the larger disk. Using
equations (7) and (8) one can find I in terms of M, r, g, v f and h.
I=
(9)
Since the body starts from rest and moves with a constant acceleration we can
determine v f in terms of the distance travelled l and the time taken t.
Newton’s equations of motion give that:
v f = vi + at ⇒ a =
vf
t
if vi = 0
1
1 vf 2 vft
2l
l = vi t + at 2 ⇒ l =
t =
⇒ vf =
2
2 t
2
t
(10)
Substituting this into the expression for I in (9) we get that:
⎡ gh ⎛ t 2 ⎞ ⎤
I = Mr 2 ⎢ ⎜⎜ 2 ⎟⎟ − 1⎥
⎣ 2 ⎝ l ⎠ ⎦
(11)
Looking at figure 3 and ensuring that the slope of the plane is kept constant at an
angle α with the horizontal we can rewrite expression (11) as:
2
⎡
⎤
2 g sin α ⎛ t ⎞
(12)
⎜⎜ ⎟⎟ − 1⎥
I = Mr ⎢
⎣ 2 ⎝ l ⎠ ⎦
since sin α = h l .
Keeping the slope of the rails constant, measure the time t it takes the wheel to move
through different distances l along the rails from rest using a stopwatch. The same
person should use the stopwatch and release the wheel and make a few trial runs to
determine the best procedure.
For each distance l take a number of measurements of t in order to determine the
average time and estimate the uncertainty in t. Plot a graph of t 2 versus l and find
the slope. Substitute this value into (12) and calculate the moment of inertia of the
disk plus axle. Refer to Appendix A for the error analysis.
Result using method 2 for axle + disk
I=
1.4
PY2107
Moments of Inertia
Experiment 1
__________________________________________________________________________________
How does the answer you obtained for method 1 compare with that obtained for
method 2? Mention possible sources of error that would account for this discrepancy.
1.5
PY2107
Moments of Inertia
Experiment 1
__________________________________________________________________________________
Part 2: Moment of Inertia of a Ball-Bearing
In this experiment you must measure I for a ball-bearing.
Method 1: Measure carefully the mass m and radius r of the ball-bearing and
determine its moment of inertia using the expression:
2
(13)
I = mr 2
5
Ensure you answer contains estimates on the error in I.
Result using method 1 for ball-bearing
I=
Method 2: In this method you will determine I for the ball-bearing using the
principle of conservation of energy and referring to Appendix B.
Consider a uniform sphere of mass m and radius r rolling back and forth without
slipping on a concave spherical surface of radius of curvature R, it will execute small
amplitude oscillations in a vertical plane. By showing these oscillations are simple
harmonic in nature it is possible to determine an expression for the period and, hence,
the moment of inertia for the sphere.
As with the previous method, we will ignore any frictional effects and assume that
energy must be conserved. If we consider the schematic of the problem shown in
figure 4, we can assume that when the oscillations reach maximum amplitude
(position A) the energy of the sphere is entirely potential. When it reaches the
equilibrium position (position B) the energy will be entirely kinetic.
O
R−r
x
r
A
y
B
1.6
PY2107
Moments of Inertia
Experiment 1
__________________________________________________________________________________
A
R−r
O
x
C
D
y
B
Figure 4: Experimental parameters
The centre-of-mass of the sphere moves in a vertical circle of radius R – r. Applying
the geometrical theorem AC x CB = CD2, we see that:
(2(R − r ) − y )y = x 2 → 2(R − r )y − y 2 = x 2 ⇒ 2(R − r )y = x 2 ,
y << R. (14)
Applying conservation of energy we have that:
E=
where v = rω .
1 2 1 2
mv + Iω + mgy = constant,
2
2
(15)
Differentiating equation (15) with respect to time we obtain:
mv
dv
dω
dy
dv
v dv
d ⎛ x 2 ⎞
⎟
+ Iω
+ mg
= mv + I 2
+ mg ⎜⎜
dt
dt
dt
dt
dt ⎝ 2(R − r ) ⎟⎠
r dt
= mv
dv
v dv
xv
+I 2
+ mg
=0
(R − r )
dt
r dt
I ⎞ dv ⎛
I ⎞ d 2 x
g
⎛
→ ⎜1 + 2 ⎟ = ⎜1 + 2 ⎟ 2 = −
x
R−r
⎝ mr ⎠ dt ⎝ mr ⎠ dt
(16)
(17)
g ⎛ mr 2 ⎞
⎜
⎟ . The motion is
R − r ⎜⎝ mr 2 + I ⎟⎠
therefore simple harmonic in nature. We can define the period of the motion as
T = 2π ω = 2π γ .
Equation (17) has the form x = −γ x , where γ = −
1.7
PY2107
Moments of Inertia
Experiment 1
__________________________________________________________________________________
(R − r )⎛⎜ m +
⎝
mg
⇒ T = 2π
I ⎞
⎟
r 2 ⎠
.
Hence the moment of inertia is given by:
⎤
gT 2
2 ⎡
I = mr ⎢ 2
− 1⎥ .
⎣ 4π (R − r ) ⎦
(18)
(19)
Time the period of, say, 10 oscillations of the sphere on the curved surface and,
thence, determine the average period for the oscillations, T. Make sure your answer
is in the SI unit of time, i.e. the second.
Measure the radius of the ball-bearing r and determine the radius of curvature of the
surface R using the spherometer as described in Appendix B. Note: please do this on
the concave spherical surface (rather than the convex one on page 1.10) – otherwise
you may get the wrong answer. Ensure you include the errors on each value and make
an estimate on the error in I determined using this method.
Result using method 2 for ball-bearing
I=
Compare the values obtained for the moment of inertia of a sphere using both
methods compare and comment on where additional sources of error may arise.
1.8
PY2107
Moments of Inertia
Experiment 1
__________________________________________________________________________________
Appendix A
Example of Error Analysis for the Disk + Axle Experiment
⎡ g sin α ⎛ t 2 ⎞ ⎤
⎜⎜ ⎟⎟ − 1⎥
I = Mr 2 ⎢
2
⎝ l ⎠ ⎦
⎣
= Mr 2 [XS − 1],
where X =
(
g sin α
t2
and S =
is the slope.
2
l
)
ln I = ln Mr 2 + ln( XS − 1)
= ln M + 2 ln r + ln( XS − 1)
ΔI ΔM
Δr Δ( XS )
=
+2
+
I
M
r
XS − 1
ΔM
Δr SΔX + XΔS
=
+2
+
M
r
XS − 1
Here X =
g sin α
g
, ⇒ ΔX = cos αΔα
2
2
NB: The error Δα must be expressed in radians.
1.9
PY2107
Moments of Inertia
Experiment 1
__________________________________________________________________________________
Appendix B
Determination of the radius of curvature of a convex surface using a
spherometer
Apparatus
Spherometer, flat surface, curved surface
Method
Step 1 The spherometer is first inspected to determine how
far (as measured on the vertical scale), the screw moves
when rotated through one complete revolution of the circular
scale. In general this will be 0.5 or 1 mm. The value of one
division on the circular scale is then known.
Step 2 The spherometer is next placed on a slab of glass and
the centre leg is adjusted until its point just touches he
surface of the glass. This is best determined by observing
the image of the leg in the glass surface when viewed at an
angle. In this position the zero on the two scales should
align. If not, the zero error must be determined by taking
the average of several settings.
Step 3 The centre leg is now screwed upwards and the spherometer is placed on the
curved surface so that the three legs are in contact with the surface. The centre leg is
again screwed downwards until it just touches the surface (best determined when
viewed optically as above) and the readings of the two scales are taken. This
procedure is repeated several times and the average of the readings is taken.
Step 4 Finally the spherometer is pressed onto a piece of paper and the average
distance, l, between the centre point and outer legs is determined.
Theory
On placing the spherometer on the curved
surface, the three outer legs stand on the
circumference of a circle of radius l, of
diameter AB. Let leg A, the spherometer
screw and the centre of curvature O define a
plane perpendicular to this circle. Let the
height through which the centre leg is raised
be h. If R is the radius of curvature of the
surface, from the properties of intersecting
chords, we have that: AC × CB = DC × CE
D
A
l
h
C
O
2
In other words: l = h(2 R − h ).
On rearranging we get that: R =
l 2 + h2
.
2h
1.10
R
B
PY2107
Moments of Inertia
Experiment 1
__________________________________________________________________________________
E
Results
#1
#2
#3
#4
Average Value
Zero error readings
∴ Average zero error =
Readings on curved
surface
Average =
∴h =
Readings of l
Average =
∴R =
cm
1.11