physics 2107 moments of inertia experiment 1 · 2018-08-30 · physics 2107 moments of inertia...

1.1

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: ∑=

iii rmI 2 , (1)

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:

2

21mrI = . (2)

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.

Figure 2 Disk with axis through its centre.

Figure 1 A rotating disk is composed of many particles, two of which are shown.

r

mass, m

Axis of rotation

PY2107 Moments of Inertia Experiment 1 __________________________________________________________________________________

1.2

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 diskI and axleI . The total I of the disk + axle is the sum axledisk II + .

Method 1: Calculate the moment of inertia of the disk and axle about the axis of symmetry using the equation:

22

21

21 rmRmIII axlediskaxlewheel +=+= . (3)

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 .Rr << =diskm =R =axlem =r Ensure you include the maximum error in each of your measurements, e.g.

( )g05.0300±=axlem . 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 =

= +

mdisk

maxle

R radius r


1.3

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.

.21

21 22 MghIMvPEKEKEEnergy rottrans ++=++= ω (4)

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.

At point A, the energy is entirely potential since the wheel is at rest: == PEEnergyinitial (5) At point B, the energy is entirely kinetic:

22

21

21

ffrottransfinal IMvKEKEEnergy ω+=+= , (6)

where fv is the final translational speed and fω 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.

A

Zero PE height

v

B

h

α

l

Figure 3: Experimental arrangement


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= (7) For an axle or wheel that rolls without slipping, the angular velocity and the translational speed are related by:

rv

=ω . (8)

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, fv and h. I = (9) Since the body starts from rest and moves with a constant acceleration we can determine fv in terms of the distance travelled l and the time taken t. Newton’s equations of motion give that:

tlv

tvt

tv

lattvl

vtv

aatvv

fff

i

if

if

222

121

0 if

22 =⇒==⇒+=

==⇒+= (10)

Substituting this into the expression for I in (9) we get that:

⎥⎦

⎤⎢⎣

⎡−⎟⎟⎠

⎞⎜⎜⎝

⎛= 1

2 2

22

ltghMrI (11)

Looking at figure 3 and ensuring that the slope of the plane is kept constant at an angleα with the horizontal we can rewrite expression (11) as:

⎥⎦

⎤⎢⎣

⎡−⎟⎟⎠

⎞⎜⎜⎝

⎛= 1

2sin 2

2

ltgMrI α (12)

since lh=αsin . 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 2t 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 =


1.5

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.


1.6

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

52mrI = (13)

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

y A

B

r


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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: ( )( ) ( ) ( ) .,222 2222 RyxyrRxyyrRxyyrR <<=−⇒=−−→=−− (14) Applying conservation of energy we have that:

constant,21

21 22 =++= mgyImvE ω (15)

where ωrv = . Differentiating equation (15) with respect to time we obtain:

( )

( )0

2

2

2

2

=−

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛

−++=++

rRxvmg

dtdv

rvI

dtdvmv

rRx

dtdmg

dtdv

rvI

dtdvmv

dtdymg

dtdI

dtdvmv ω

ω

(16)

xrR

gdtxd

mrI

dtdv

mrI

−−=⎟

⎠

⎞⎜⎝

⎛ +=⎟⎠

⎞⎜⎝

⎛ +→ 2

2

22 11 (17)

Equation (17) has the form xx γ−= , where ⎟⎟⎠

⎞⎜⎜⎝

⎛

+−−=

Imrmr

rRg

2

2

γ . The motion is

therefore simple harmonic in nature. We can define the period of the motion as .22 γπωπ ==T

O

x

y D C

A

B

R − r

Figure 4: Experimental parameters


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( )

mgrImrR

T⎟⎠

⎞⎜⎝

⎛ +−=⇒

2

2π . (18)

Hence the moment of inertia is given by:

( ) ⎥

⎦

⎤⎢⎣

⎡−

−= 1

4 2

22

rRgTmrI

π. (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.


1.9

Appendix A

Example of Error Analysis for the Disk + Axle Experiment

[ ],1

12sin

2

22

−=

⎥⎦

⎤⎢⎣

⎡−⎟⎟⎠

⎞⎜⎜⎝

⎛=

XSMr

ltgMrI α

where 2sinαgX = and

ltS2

= is the slope.

( ) ( )

( )1lnln2ln1lnlnln 2

−++=

−+=

XSrMXSMrI

( )

12

12

−

Δ+Δ+

Δ+

Δ=

−

Δ+

Δ+

Δ=

Δ

XSSXXS

rr

MM

XSXS

rr

MM

II

Here 2sinαgX = , ααΔ=Δ⇒ cos

2gX

NB: The error Δα must be expressed in radians.


1.10

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: CEDCCBAC ×=×

In other words: ( )hRhl −= 22 .

On rearranging we get that: .2

22

hhlR +

=

A B

D

C

O

R

h l


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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

E

physics 2107 moments of inertia experiment 1 · 2018-08-30 · physics 2107 moments of inertia...

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