UNESCO-NIGERIA TECHNICAL &
VOCATIONAL EDUCATION
REVITALISATION PROJECT-PHASE II
NATIONAL DIPLOMA IN
MECHANICAL ENGINEERING TECHNOLOGY
STRENGTH OF MATERIALS
COURSE CODE: MEC 222
YEAR 2- SEMESTER 2
PRACTICALS
Version 1: December 2008
1
COURSE: STRENGTH OF MATERIALS
COURSE CODE: MEC 222
CONTACT HRS: 4HRS/WK
COURSE SPECIFICATION: PRACTICAL CONTENT
TABLE OF CONTENT
Week 1
1. Experiment no.1 Principle of Moments {i}
Week 2
2. Experiment no.2 Principle of Moments {ii}
Week 3
3
Experiment no.3 Reaction at Beam supports
4.
Experiment no.4 Complete Tensile test on a Mild Steel Specimen
Week 4
Week 5
5. Experiment no.5 Complete Torsion test
Week 6
6
Experiment no.6 Torsion of shafts (i)
Week 7
7. Experiment no.7 torsion of shafts (ii)
Week 8
8. Experiment no.8 bending of beams (i)
Week 9
9. Experiment no.9 bending of beams (ii)
Week 10
10. Experiment no.10 Modulus of Rigidity
Week 11
11. Experiment no.11 Modulus of Elasticity
Week 12
12.Experiment 12 notched bar impact testing of materials
2
Week 13
13.Experiment 13 mechanical testing-impact & hardness testing
Week 14
14. Experiment 14 moment of inertia
Week 15
15.Experiment 15: Rotational Motion. Moment of Inertia
3
EXPERIMENT NO.1 PRINCIPLE OF MOMENTS {i}
Objective:
To illustrate the principle of moments.
Apparatus:
Metre rule with a hole drilled centrally through it, various weights, thread, a suitable
stand, pivot or fulcrum.
Method:
1. Set up metre rule as shown in fig.19, with several weights hanging on each side of
the fulcrum.
2. Adjust the weights until the rule is in equilibrium.
3. Note the weights W1,W2,W3,W4,etc.,and their corresponding distance
x1,x2,x3,x4,etc., from fulcrum.
(N.B.-The distance must be perpendicular to the force in each.)
4. Note which weighs are tending to turn the beam clockwise and which anticlockwise.
5. Multiply each force by its distance from the fulcrum, i.e. calculate the moment of each
force about the fulcrum.
4
6. Find the sum of the clockwise moment.
7. Find the sum of the anticlockwise moment.
Observations:
Anticlockwise Moments
Weight in N Distance
From
Fulcrum (m)
W1
x1
W2
x2
Clockwise Moments
Weight in N Distance
From
Fulcrum (m)
W3
X3
W4
x4
Moment
W1 x1
W2 x2
Sum
Moment
W3 X3
W4 X4
Sum
Conclusions:
1. What do you notice about the clockwise moments and the sum of the anticlockwise
Moments?
2. State the theorem you set out to illustrate.
5
EXPERIMENT NO.2 PRINCIPLE OF MOMENTS {2}
Objective:
To illustrate the principle of moments.
Apparatus:
Metre rule with a hole drilled centrally through it, various weights, thread, a suitable
stand, pivot or fulcrum.
Fig.2
Method:
1. Set up metre rule as shown in fig.2 with several weights hanging on each side of
the fulcrum.
2. Adjust the weights until the rule is in equilibrium.
3. Note the weights W1,W2,W3,W4,etc.,and their corresponding distance
x1,x2,x3,x4,etc., from fulcrum.
(N.B.-The distance must be perpendicular to the force in each.)
4. Note which weighs are tending to turn the beam clockwise and which anticlockwise.
6
5. Multiply each force by its distance from the fulcrum, i.e. calculate the moment of each
force about the fulcrum.
6. Find the sum of the clockwise moment.
7. Find the sum of the anticlockwise moment.
8. Repeat the experiment with the apparatus arranged as in fig. above
Observations:
Anticlockwise Moments
Weight in N Distance
From
Fulcrum (m)
W1
x1
W2
x2
Clockwise Moments
Weight in N Distance
From
Fulcrum (m)
W3
X3
W4
x4
Moment
W1 x1
W2 x2
Sum
Moment
W3 X3
W4 X4
Sum
Conclusions:
1. What do you notice about the clockwise moments and the sum of the anticlockwise?
moments?
2. State the theorem you set out to illustrate.
7
EXPERIMENT NO.3 REACTION AT BEAM
SUPPORTS
Object:
To measure the reactions on the supports of a loaded beam.
Apparatus: Wooden beam, two spring balances, various small weights.
8
Fig.22 Beam Forces
Method:
1. Suspend the beam from the two spring balances as shown in Fig.21.
2. Before placing any loads on the beam, note the reading on the spring balances.
(Let these be P1 and Q1 lb. Respectively.)
3. Place some weights on the beam.
4. Read the spring balances. (These values are P and Q lb.)
5. The differences (P-P1) and (Q-Q1) give the reactions on the supports due to
the added weights.
6. Calculate the reactions on the supports.
7. Repeat the experiment for several different loadings.
Conclusions:
1. Do the calculated values agree with the observed values?
2. What do you notice about the sum of the upward forces and the sum of the
downward forces?
9
EXPERIMENT NO.4 COMPLETE TENSILE TEST ON A MILD STEEL
SPECIMENT
Object:
To carry out a complete tensile test on a specimen of mild steel and determine the main
properties of the material.
Apparatus:
Tensile testing machine and extensometer, mild steel specimen, micrometer, rule and
engineer’s dividers. Make a neat line diagram of the testing machine and extensometer
used and write a short note on each. Describe how the specimen was prepared for testing,
dimensions of specimen, gauge length, etc.
Theory:
Graph 1: Load against extension (elastic range only)
stress
Young’s modulus=E =
= constant
strain
W .L
L
E=
= (slope of graph) ×
A
A .x
Where L=gauge length
and A=cross-sectional area
Load at yield po int
Yield stress =
A
= N/m2
10
Ultimate stress =
Ultimate Load
A
= N/m2
Figure 1 Tensile test of an AlMgSi alloy. The local necking and the cup and cone
fracture surfaces are typical for ductile metals.
11
Graph 3: Elongation against gauge length
After fracture the pieces are placed together again and the extension measured on as
many gauge lengths as possible over the fracture.
From graph (Fig.3), elongation on a B.S.S. gauge length of 5.65√A = %
reduction in area
% reduction in area at fracture =
× 100
A
=== %
Method:
1. Using the micrometer and rule, determine the dimensions of the specimen.
2. Adjust the load range of the machine to cover adequately the estimated load
required to cause fracture of the specimen.
3. Place the specimen in the machine and apply a small load just sufficient to grip it,
and then release the load back to almost zero.
4. Attach the extensometer to the specimen over a fixed gauge length (usually
50.8mm).
5. Apply load to the specimen in suitable increments and note the corresponding
extensometer readings. (Load should be applied at a steady rate.)
6. Plot the graph of load against extension and when the yield point has been
reached remove the extensometer.
7. With the dividers set to, say, 129.54mm increase load until this extension of
2.54mm on a 127mm gauge length has been reached. Note the load.
8. Repeat (7) with the dividers set to 132.08mm and continue this procedure until
fracture occurs.
9. Remove the specimen from thee machine.
12
10. Measure the diameter at fracture and the elongation lf the specimen over several
gauge lengths.
Observations:
Original diameter of specimen = m
Diameter of fracture
=m
Gauge length for elastic range = m
Gauge length for complete range = m
Elastic range:
Load W( N)
Complete range:
Length (m)
5.1
5.2
Elongation:
Original length (m)
0.5
1.5
2.5
e.t.c.
Extensometer reading
Extension (m)
Load ( N)
Extension (m)
Final length (m)
% Elongation
Calculations:
From Graph 1 (Fig.14)
Young’s modulus E= (slope of graph) ×L/A
= N/m2
Yield stress = Load at yield point/A
= N/m2
From Fig.3 B.S.S. gauge length=5.56√A = (m)
% elongation on this gauge length=
%
Also, % reduction in area at fracture =
%
Conclusions:
1.
2.
3.
4.
5.
State the values obtained for Young modulus, yield stress e.t.c.
How do these values compare with the generally accepted values for mild steel?
Give reasons for any differences in the experimental and reference book values.
Sketch the specimen before and after fracture and describe the type of failure.
Common on the graphs.
13
EXPERIMENT NO.5 COMPLETE TORSION TEST
Object:
To carry out a torsion test on a mild steel specimen.
Apparatus:
Torsion testing machine, torsion meter, micrometer. Make a neat line diagram of the
testing machine and torsion meter used and write a short note on each. Describe how the
specimen was prepared for testing, dimensions of the specimen, gauge length, etc.
14
Theory:
The general torsion equation is given by :
T
τ
Gθ
=
=
J
r
L
τ .r
Therefore T=
(Nm)……………………… (i)
J
Where τ = shear stress at outer surface
T .L
Also, G =
J.θ
1
=
(slope of graph) =N/m2……………….(ii)
Jθ
Method:
1.
2.
3.
4.
5.
6.
7.
8.
Measure the diameter of the specimen.
Attach the torsion meter to the specimen.
Grip the specimen in the jaws of the torsion machine.
Zero the torsion meter.
Place a load on the lever arm of the machine.
Apply a torque to the specimen until the lever arm is balance.
Note the angle of twist.
Repeat (5), (6) and (7) for a series of increasing loads until the limit of
proportionality is reach.(it will depend on type of torsion meter whether it is removed
at the limit of proportionality or whether remains until fracture occurs.)
9. Continue the test until fracture occurs.
Observations:
Diameter of specimen = d (m)
Gauge length
= L (m)
Load W (N)
Radius arm R (m)
Torque T=W.R
Angle of twist θ
Graphs
Plot the following graphs:
1. Torque against angle of twist up to the limit of proportionality.
2. Torque against angle of twist for the complete test.
15
Calculations:
Shear the stress at the limit of proportionality = N/m2
Modulus of rigidity G
= N/m2
Shear stress at fracture
= N/m2
Conclusions:
State the results obtained, i.e. modulus of rigidity shear stress at fracture, etc.
1. How do these values compare with the values taken from reference books?
2. Sketch and describe the fracture.
3. Comment on the shape of the graphs.
16
EXPERIMENT NO.6 TORSION OF SHAFTS (i)
Object:
To show how the angle of twist of a shaft varies with:
(a) the applied torque (length and diameter constant);
(b) the length of the shaft (torque and diameter constant);
(c) the diameter of the shaft (torque and length constant).
Apparatus:
Norwood or similar torsion testing apparatus, dial gauges. Weights.
Micrometer and several specimens each of a different diameter.
17
Theory:
Let W = weight on the bar
R= radius of torque arm
W
W
Torque on the shaft= (
X R )+(
X R )
2
2
T = W. R……….. (1)
To determine the angle of twist:
Let x = length of deflection arm
∆1= deflection of dial gauge 1
∆2= deflection of dial gauge 2
Hence, since ∆ is small, θ1= ∆1/x and θ2=∆2/x
Therefore angle of twist on length L = θ = (θ1- θ2)
Θ = I/x (∆1-∆2)……………… (2)
PART A: Length and diameter constant
Method:
1.Put a specimen in the apparatus and firmly fix the end remote from the weight.
2. Attach the torque arm to the other end of the specimen.
3.Attach the two deflection arms to the specimen and adjust the distance between
them to the gauge length, l. (Make sure the arms are horizontal.)
4.Measure the radius of the torque arm and length of the deflection arms.
5.Apply a load W to the torque arm.
6.Note the deflections on the dial gauges.
7.Calculate the angle of twist, θ.
8.Repeat (5), (6) and (7) for a series of increasing loads.
9.Check the deflections when unloading to ensure that the specimen has not been
loaded beyond the elastic limit.
10. Plot the torque against angle of twist graph.
18
Observations:
W (N)
R (m)
T (Nm)
∆1
∆2
Θ1
x
Θ2
Θ
Conclusions:
Comment on the shape of the graph and state the relationship between the torque and
angle of twist.
PART B:
Torque and diameter constant
Method:
Use same specimen as for part A.
1. Adjust the deflection arms until gauge length is 3m.
2. Place a load on the hanger that will give a reasonable angle of twist.
3. Calculate the angle of twist.
4. Remove the load and increase the gauge length by 3m.
5. Replace the same load and again calculate the angle of twist.
6. Repeat (4) and (5) for several different gauge lengths.
7. Plot a graph of length against angle of twist.
Observations:
Load = N
L(m)
∆1
∆2
Θ1
x
Θ2
θ
Conclusions:
Comment on the shape of the graph and state the relationship between the gauge length
and angle of twist.
PART C: Torque and length constant
Method:
1.Place a shaft in the apparatus.
2.Set the deflection arms at chosen gauge length.
3.Place a load on the hanger and calculate the angle of twist.
4.Repeat (1), (2) and (3) for several specimens of the same material, each with a
different diameter but using the same gauge length and load
19
EXPERIMENT NO.7 TORSION OF SHAFTS (ii)
Object:
To determine the modulus of rigidity of a material by means of a torsion test.
Apparatus:
As in previous experiment; specimens of different materials, e.g. mild steel, brass,
bronze, aluminium, etc.
20
Theory:
The general torsion equation is given by :
T
τ
Gθ
=
=
J
r
L
τ .r
(Nm)……………………… (i)
Therefore T=
J
Where τ = shear stress at outer surface
T .L
Also, G =
J.θ
1
(slope of graph) =N/m2……………….(ii)
=
Jθ
(Where J = πd4/32 m4 = polar moment of inertia of shaft)
If J and L are kept constant as in Part A of the previous experiment, then
T= Kθ, where K= a constant
From the graph of T against θ, as in part A of the previous experiment, the slope of the
Graph= K
G. J
Slope K =
L
Slope x L
G=
(N/m2)
J
Method:
As for Part A of previous experiment, for each material.
Observations
Diameter of the shaft = d(mm)
Gauge length of the shaft = 25.4mm
Conclusions:
1.State the value of the modulus of rigidity for each material.
2.How do these compare with the values stated in reference books?
3.Try to give 4 reasons for any difference between the experimental values and the
values from reference books.
21
EXPERIMENT NO.8 BENDING OF BEAMS (i)
Object:
To determine the value of Young’s modulus for the material of the beam.
Apparatus:
A beam about 1 m long, knife edge supports,
W
W
Dial gauge
Fig.9
Theory:
The beam is arranged on the knife edge supports so that it has equal overhanging ends.
This gives a constant bending moment between the knife edges.
Let M = bending moment = W.x (Nm)
I = second moment of area of the cross-section of the beam = BD3 for a rectangular
section
B = breadth of beam
D = depth of beam
∆ = deflection of the beam at mid-span
R = radius of curvature of deflected beam
L = distance between the knife edge
From the theory of simple bending,
M
E
=
I
R
M R
E=
L
Since E, M and I are constants, R will be constant and the beam will bend in a circular
arc between the supports.
∆
L/
L/2
( 2R∆)
Fig.9a
22
By property of intersection chords,
2(R-∆) ∆= (L/2)2
2R∆-∆2=L2/4
∆2 is negligible if ∆ is small
Therefore 2R∆=L2/4
R=L2/8∆
MR
WxL2
Hence E= /I=
/8I∆
xL2
= /8I × (slope of graph)
Method:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Measure the breadth and depth of the beam.
Place the beam on the knife edges so that there are equal overhanging ends.
Place the dial gauge on the top of the beam mid-way between the supports.
Adjust the dial gauge to zero.
Place equal loads on the overhanging ends.
Note the deflection on the clock gauge.
Repeat (5) and (6) for a series of increasing loads.
Plot the graph of load against deflection.
Calculate the value of Young’s modulus.
Observations:
Breadth of beam =
B = mm
Depth of beam = D = mm
Distance between knife edges = L = mm
Overhang = x = mm
Load W
Deflection
∆
Conclusion:
1. State the value found for Young’s modulus.
2. How does the value compare with that given in textbooks?
3. Comment on the shape of the graph.
23
EXPERIMENT NO.9 BENDING OF BEAMS (ii)
Object:
To show how the deflection of a beam varies with
(a) the load (span, breadth, and depth, constant);
(b) the span (load, breadth and depth constant);
(c) the breadth (load, span and depth constant);
(d) the depth (load, span and depth constant).
Apparatus:
Set of 6 beams of the same material each 76.2mm long, 24.6mm wide and the
depths varying in increments of 3.175mm from 9.144mm to 24.6mm. Knife edge
Supports, weights and vernier or dial gauge.
Beam Forces
PART A: Deflection against load (span, breadth and depth constant)
24
Method:
1.
2.
3.
4.
5.
Place the beam on the knife edge supports.
Measure the span of the beam.
Zero the dial gauge.
Place a central load on the beam and note the deflection.
Note the deflection for each of a series of increasing loads. (Be careful not to
exceed the elastic limit of the wood.)
6. Note the deflection readings when unloading and record the mean deflection for
each load.
7. Plot a graph of load against deflection.
Observations:
Load
Deflection
mm
(loading)
Deflection
mm
(Unloading)
Mean deflection
Conclusion:
1. Comment on the shape of the graph.
2. When do you learn from this regarding the formula for deflection?
PART B: Deflection against span (load, breadth and depth constant)
Method:
1. Use the maximum load employed in Part A and notes the deflection due to the
central load.
2. Decrease the span 76.2mm.
3. Make the load central.
4. Note the span and deflection.
5. Repeat (2), (3) & (4) for a series of spans, each decreasing by 76.2mm.
6. Plot the deflection against span graph.
7. Plot the deflection against span3 graph.
Observations:
Load = 1N
Span, in l
(span)3
Deflection,
mm
25
Conclusions:
1. Comment on the shape of the deflection against span graph.
2. Comment on the shape of the deflection against span3 graph.
3. What do you learn from these regarding the formula for deflection?
PART C: Deflection against depth (load, span and breadth constant)
Method:
1.
2.
3.
4.
5.
Place the 9.14mm deep beam on the knife edge supports.
Place a load on the beam at mid-span and note the deflection.
Repeat with each beam in turn, using the same load span and breadth each time.
Plot the deflection against depth graph.
Plot the deflection against I/depth3 graph.
Observations:
Load = 1N
Conclusion:
1. Comment on the shape of the deflection against depth graph.
2. Comment on the shape of the deflection against I/depth3
3. What do you learn from this graph regarding the formula for deflection?
PART
D: Deflection against breadth (load span and depth constant)
Method:
Repeat as in Part C but varying the breadth. Plot a graph of deflection against breadth, b.
Plot a graph of deflection against 1/b.
Observations:
Load= 1N
Conclusions:
1. Comment on the shape of the deflection against breadth graph.
2. Comment on the shape of the deflection I/breadth against graph.
3. What do you learn from this graph regarding the formula for deflection?
General Conclusion of Experiment:
From the conclusions to Parts A, B and C give a formula relating the deflection of a
beam with the load W, the span l, the breadth b and the depth d, and compare this with
the formula obtained from theory.
26
Experiment No.10 Modulus of Rigidity
Objective:
To determine the modulus of rigidity of rubber.
Apparatus:
Rubber block, vernier gauge, weights
Theory:
The Modulus of rigidity, G, is the ratio of the shear stress to shear strain:
27
Method:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Measure (in mm) the length L, depth l and thickness t of the block.
Apply a small load to the specimen as shown in Fig.
Adjust vernier to zero.
Add a small load and note the vernier reading.
Calculate the deformation.
Repeat for various loads.
Plot a graph of load (N) against extension (mm).
From the graph select a load W and it’s corresponding extension x.
Calculate G.
Conclusions:
1. What is the shape of the load- deformation graph?
2. What value of did you obtain for G?
3. Is your result within the range 70 – 220 N/m2.
28
Experiment No.11 Modulus of Elasticity
Objective:
To determine the Young’s modulus of Elasticity for Steel.
Apparatus:
Long steel wire secure to a rigid hook; vernier scale suspended near the base of wire,
with the sliding vernier scale attached to the wire; a micrometer; weights and hanger.
Method:
The apparatus is assemble as shown in Fig below:
Fig. Determining Modulus of Elasticity
Theory:
To calculate the Young’s Modulus for the steel wire, select a point A on the graph.
At A, load = N, Extension = x mm,
Load
Stress =
=
Area of cross − sec tion
Extension
Strain =
=
Original Lenght
29
Young’s Modulus E =
Stress
=
Strain
Method:
10. Measure (in mm) the length L,.
11. Apply a small load to the specimen as shown in Fig.
12. Adjust vernier to zero.
13. Add a small load and note the vernier reading.
14. Calculate the extension.
15. Repeat for various loads.
16. Plot a graph of load (N) against extension (mm).
17. From the graph select a load W and it’s corresponding extension x.
18. Calculate G.
Conclusions:
4. State the value for the Young’s Modulus
5. What is the shape of the load- extension graph?
6. Compare the value found of E With the one from Textbooks..
30
Experiment 12 notched bar impact testing of materials
Background
Materials sometimes display brittleness which precludes their use in a given
design. Brittleness is characterized by fracturing with low energy under impact. The
fracture energy is proportional to the area under the tensile stress-strain curve and is
called the toughness. A tough steel is generally ductile and requires 100 ft-lbs of
energy to cause failure. A brittle steel does not deform very much during failure and
requires less than 15 ft-lbs energy to cause failure.
Characterizing the toughness of a material is done in several ways. The most
common method is the notched-bar impact test for which two types of specimens
prevail, Charpy and Izod. By subjecting a specimen to an impact load, it will fail if the
load exceeds the breaking strength of the material. By using a swinging pendulum to
impart the load, the energy required to fracture the specimen can be calculated by
observing the height the pendulum swings after fracture, as shown in Figure 2-1.
This test has been used almost exclusively with body-centered-cubic (bcc)
crystalline materials. These materials show a transition from ductile to brittle behavior
with temperature (see Fig. 2-2). This means that at low temperature the fracture
energy is low. Very often bcc materials are ductile until they are heat treated.
31
The mechanical behavior of materials often exhibits variations even for
seemingly identical specimens and materials. The steel specimens for this experiment
are manufactured from a single ingot of steel and are machined to a single drawing.
Despite these precautions, the results from identical tests will not always be identical.
Variations in properties are often evaluated by means of statistics in order to
establish the average value and the possible variance in the results.
The fraction of shear in the surface failure of steels can be determined by looking
at the fresh failure surface under low-power magnification (approx 3X). A smooth
surface is characteristic of shear. A fine grained fracture surface is characteristic of
cleavage and brittleness. Often failures are mixed (part shear and part cleavage).
If no plastic deformation accompanies fracture, it is generally a brittle fracture,
i.e. cleavage. In the impact test the amount of plastic deformation is characterized by
lateral
expansion.
Lateral expansion is a thickening of the specimen during fracturing.
Looking at half the failed specimen, the lateral expansion is measured as shown below.
32
where lateral expansion = Dw = wf - wi
wf = final lateral dimension
wi = initial lateral dimension
Heat Treatment of Steels
Common steels, which are really solid solutions of carbon in iron, are bodycentered-cubic. However, the carbon has a low solubility in bcc iron and precipitates
as iron carbide when a steel is cooled from 1600oF (870oC).
The processes of
precipitation can be altered by adjusting the cooling rate. This changes the distribution
and size of the carbide which forms a laminar structure called pearlite during slow
cooling processes.
If a steel is quenched into water or oil from 1600oF (870oC) a metastable phase
called martensite forms, which is body-centered-tetragonal. This phase sets up large
internal stresses and prevents carbide from forming. The internal stresses produce a
high hardness and unfortunately, low toughness.
After cooling, to restore toughness, steels are tempered by reheating them to a
lower temperature around 800oF (426oC) and cooling. The tempering relieves the
internal stresses and also allows some iron carbide to form. It also restores ductility.
References
Deiter, Mechanical Metallurgy
ASM Handbook on Heat Treatment, Vol. 2
Flinn/Trojan, Engineering Materials and Their Applications, 2nd Ed.
Technical Approach
The experiment consists of investigating the ductile-to-brittle transition in 1018steel and 2024-aluminum as a function of temperature. The Charpy impact specimens
of each material are immersed in a bath for 10 minutes to reach thermal equilibrium at
five separate temperatures ranging from -110oF (-79oC) to 212oF (100oC).
The
specimens are quickly transferred to the Charpy testing machine, fractured, and the
impact energy is measured.
After failure the fracture surfaces are examined for
evidence of shear or cleavage failure and lateral expansion at the root of the notch.
33
Data for this experiment have been obtained from other groups during last
semester laboratory periods. This data will be supplied to you and should be combined
with your own data to produce a data base.
Apparatus
Thermocouple
Temperature Baths
Charpy Impact Machine
Tongs
Dial caliper
SAFETY INFORMATION
DURING THE COURSE OF THIS EXPERIMENT YOU WILL BE USING A
DEWAR OF ACETONE WHICH HAS BEEN COOLED USING DRY ICE TO THE
TEMPERATURE OF DRY ICE (–110oF or -79oC). CARE MUST BE TAKEN TO
AVOID EXPOSURE TO SKIN OR EYES. THE EXPOSURE MAY CAUSE BURNES
AND OTHER EFFECTS. STUDENTS ARE ADVISED TO TAKE THE FOLLOWING
PRECAUTIONS:
1.
Be careful to slowly insert specimens into the baths.
2.
Use tongs to insert and remove specimens.
3.
Do not drop specimens into bath. (Dewar will break).
4.
While transporting the Dewar, use the special cart set up for that purpose.
Procedure
1. Review safety considerations. Label each specimen using a hammer and punch.
2. Identify and measure the initial lateral dimensions of all specimens.
3. Conduct impact tests on five as received 1018-steel and five 2024-Al specimens
heated or cooled to dry ice, antifreeze bath with some dry ice, ice-pure water, room
temperature, and boiling water.
For consistent results, it is important that all
specimens be positioned identically in the anvil; use the special tongs provided to
correctly position the specimen. The notch on the specimen must face 180o opposite
to the hammer.
34
4. Measure lateral dimensions. Observe the nature of the fracture surface of steel
specimens. The fracture specimens should all be carefully examined and particular
attention should be paid to the type of fracture which is obtained in each particular
case. Try to relate the type of fracture to the energy absorbed by the metal being
fractured. A binocular microscope is available in order to study these fractures
carefully. Look for cleavage and shear and estimate the fraction area of each to
nearest 20%, i.e. 20% cleavage.
SAFETY INFORMATION
BEFORE USING THE IMPACT MACHINE, YOU SHOULD GET THE
INSTRUCTOR TO SHOW YOU EXACTLY HOW TO OPERATE IT.
ALWAYS
TAKE GREAT CARE IN HOW YOU HANDLE THE MACHINE IN ORDER TO
AVOID ANY ACCIDENT.
If the specimens are transferred rapidly to the machine, it can be assumed that the
temperatures at which they are broken are those of the baths in which they have been
held. Desired bath temperatures are:
dry ice/acetone bath
- 110oF- 79oC
- 40oF- 40oC
antifreeze/water mix with some dry ice
ice/pure water
32oF
room temperature
72oF
23oC
212oF
100oC
boiling water
0oC
Glossary of Terms
Understanding the following terms will aid in understanding this experiment:
Body-centered cubic. Common atomic arrangement for metals consisting of eight
atoms sitting on the corners of a cube and a ninth atom at the cubes center.
Brittle. Lacking in deformability.
Cleavage. Brittle fracture along particular crystallographic planes in the grains of the
material.
35
Ductile fracture. Fracture of a material with significant deformation required.
Ductility. The ability of a material to be permanently deformed without breaking
when a force is applied.
Face-centered cubic. Common atomic arrangement for metals consisting of eight
atoms sitting on the corners of a cube and six additional atoms sitting in the center of
each face of the cube.
Fracture. Failure or breakage of a material.
Impact energy. The energy required to fracture a standard specimen when the load is
suddenly applied.
Impact test. Measures the ability of a material to absorb a sudden application of a
load without breaking. The Charpy test is a commonly used impact test.
Lateral expansion. The lateral change in dimension of a Charpy impact specimen due
to fracture. The dimension measured is the width opposite the v-notch (see fig. 1-3).
Plastic deformation. Permanent deformation of the material when a load is applied,
then removed.
Shear. Deformation due to parallel crystallographic planes.
Shear lip. The surface formed by ductile fracture that is at a 45o angle to the direction
of the applied stress.
Toughness. A qualitative measure of the impact properties of a material. A material
that resists failure by impact is said to be tough (also given as the total area under the
stress-strain curve).
36
Transition temperature. The temperature below which a material behaves in a brittle
manner in an impact test.
Write-Up and Discussion
1. Prepare a memo report including the following:
2. Relate the type of fracture observed to the energy absorbed by the metal being
fractured. Report the fraction area of cleavage and shear to nearest 20%, i.e. 20%
cleavage.
3. Plot the results obtained graphically, relating notched bar impact energies to the
temperature of testing, and changes in width to the temperature of testing.
4. Try to relate the type of fractures observed (e.g. shear or cleavage) to the test
temperatures.
5. Compare the variation of impact energy with temperature observed for 1018-steel
and 2024-Al.
6. Using the data provided by your instructor, calculate the average value of the
fracture energy at each temperature. Construct a plot of average fracture energy versus
temperature for steel and for aluminum. Calculate the standard deviation for n data
points. The Standard Deviation is given by:
SD =
( xi − x) 2
(n − 1)
where: xi = individual fracture energy value
x=
xi
n
n = number of values (data points)
7. Plot the upper and lower values of the standard deviation versus temperature on the
same chart as the average value of the fracture energy. Data from other groups will be
supplied to make these calculations.
37
EXPERIMENT 13 MECHANICAL
TESTING-IMPACT
& HARDNESS
Experimen11Experiment:
Mechanical TestingImpact & Hardness
Testing TESTING
38
39
40
41
42
43
44
45
46
Experiment: Mechanical Testing- Impact & Hardness Testing
47
Experiment 14 MOMENT OF INERTIA
Moment of Inertia
PRELAB EXERCISES
1. Consider an ideal system of two masses attached to a massless platform at the same
distance from the center. If that distance is doubled, how does the moment of inertia of
the system about the center change?
2. Imagine a solid disk, made of uniform material, of radius R and thickness L. What is
the ratio of L/R, if the moment of inertia of this disk about the axis passing through the
center and perpendicular to the plane of the disk is the same as the moment of inertia
about the axis passing through the center and parallel to the disk’s plane?
3. Is it possible to have a torque act on an object while the net force applied to it is zero?
If no, why not, if yes, give an example.
48
I.
OBJECTIVE
There are three objectives for this experiment.
1. To study Newton's Second Law in rotational form.
2. To elucidate the analogies between quantities in translational motion and quantities in
rotational motion (
,
,
) and to use these analogies to develop
expressions for angular quantities such as rotational kinetic energy.
3. To illustrate the dependence of the moment of inertia on the shape of the object as
well as on the choice of the axis of rotation.
II.
INTRODUCTION
The motion of any rigid body can be analyzed by separately considering the translation of
its center of mass and the rotation about its center of mass. If all points of an object
move along parallel straight lines, then the translational description alone suffices. If the
center of mass of a rotating object is fixed, then the rotational description alone suffices.
In many other cases, such as a flight of a football after a free kick, both descriptions are
needed simultaneously to describe the complicated motion.
The translational version of Newton’s Second Law for a rigid body can be written as
,
where is the linear momentum and all vector quantities refer to the center of mass of
the body. Using our “dictionary” (see objective 2 above), this can be translated into
rotational language
,
where L is the angular momentum and
is the angular acceleration.
For simplicity, we have written this rotational law of motion in its scalar version. All
rotational quantities therein must be calculated about the same axis of rotation, which we
will take as an axis passing through the center of mass of the body.
The moment of inertia I depends not only on the mass of the object, but also on how its
mass is distributed. The greater the portion of mass located far away from the rotation
axis, the greater the moment of inertia about that axis. Thus, for a point mass m rotating
a distance R from an axis (or a thin hoop of radius R) the moment of inertia about that
axis (or hoop’s axis of symmetry) is I=mR2, but a uniform disk of the same mass and
radius has the moment of inertia of only half this value! In turn, the same uniform disk
49
rotating about an axis passing through the center and in the plane of the disk has the
moment of inertia I=mR2/4+mL2/12 where L is the disk’s thickness. Unless L is large,
this is even smaller than the moment of inertia of a disk rotating horizontally. Can you
explain why? Why does the moment of inertia of a disk depend on L for one
configuration of the disk and not for the other?
Starting with these expressions, you will be able to determine moments of inertia of more
complicated arrangements by summing the moments of inertia of all constituent parts
(again, all theses moments must refer to the same axis of rotation!). In this way, you will
be able to determine the moment of inertia of the rotating part of the apparatus in this
experiment.
Now consider the system as shown in the photograph on page 39. If the pulley is ideal,
then the tensions in the horizontal and vertical parts of the string are equal. The tension,
T, is the force which produces a torque on the rotating shaft given by τ=rT, where r is the
radius of the spool on which the string winds. This torque is calculated for the axis
passing along the shaft through its center, as are all subsequent rotational quantities.
From Newton’s Second Law (in translational form) for the falling mass, we obtain
mg-T=ma, where m is the mass of the hanger and weights and a its acceleration.
In practice, as you will observe, the acceleration of the falling mass a is much less than
the acceleration due to gravity g, a<<g, and the quantity ma is small compared to other
terms in this equation. Then
and the torque can be approximated by τapp=mrg.
More accurately, by writing Newton’s Second Law in rotational form for a rotating
assembly with the moment of inertia I, and by using
which relates the angular acceleration of the rotating shaft to the linear acceleration of the
falling hanger, it can be shown that
.
Under what condition does this expression for torque reduce to the approximate
expression derived above? The accuracy of the approximate expression for the torque,
τapp, can be estimated directly by calculating the relative difference from the exact
expression
In this experiment we will measure the linear acceleration of the mass hanger directly
using a photogate attached to the pulley. Once the mass is released, it will fall down and
50
then start moving up again, with the apparatus reversing its direction of spin, it will then
start falling again, etc. We will observe multiple such oscillations. Note that while the
acceleration predicted above is applicable to both up and down motion, friction in the
system will result in
. Here, as in M4, the average of these two values should
agree with the prediction for an ideal system, and the frictional torque is given by their
difference
.
III.
EQUIPMENT
The equipment for this experiment consists of a PASCO rotational base and several
different shapes that can be attached to it (a disk, a ring, an aluminum platform, and a pair
of block masses). The masses (and associated errors) of these shapes are given below. A
mass set and pulley provide the angular acceleration in the system, and a photogate
connected to the DataStudio interface allow the motion of the system to be carefully
analyzed. A ruler and calipers are also available to measure the appropriate dimensions
of the system.
51
As show in the picture above, the aluminum platform and block masses can be mounted
on the rotational base. The position of each block on the platform can be varied. In
effect, this allows one to change the moment of inertia about the axis passing along and
through the center of the shaft of the rotating assembly.
System masses
plastic disk: m= 1.43 0.05 kg
metal ring: m= 1.43 0.05 kg
mass block (each): m= 0.27 0.01 kg
aluminum platform: m= 0.58 0.02 kg
IV.
PROCEDURE
Part I – Horizontal Disk
1. Open the DataStudio workbook M7 Moment of Inertia.
2. Measure and record the radius r of the middle groove of the vertical shaft with the
vernier calipers. Also, measure and record the relevant dimensions of the disk.
3. With no added masses on the hanger, wind the string evenly and tightly around the
middle groove of the shaft. Make sure the string does not overlap.
4. Press Start. Release the hanger and observe the resulting plot. Allow the hanger to
“bounce” and the string to wind itself up onto the groove (in the opposite direction!).
After five of these cycles, press the Stop button.
5. From the first oscillation on plot 1, use the Fit function to determine the magnitude of
the downward acceleration, ad, of the hanger. Pay special attention to the points used in
your data fit. Now, fit the data for the upward acceleration, au, during the first cycle.
Note that that the minus sign in the slope should be discarded to obtain the actual
acceleration.
To obtain the error associated with your measurements, repeat the fits for the remaining
oscillations on your graph, both up and down. Then perform the usual statistical analysis.
6. Notice that the two accelerations are not equal. As in experiment M4, friction breaks
the symmetry of the motion. Using the same arguments outlined in the previous
experiment, the ideal acceleration should be the average of the magnitudes of au and ad.
It is this average that you will use to determine the moment of inertia of the system.
7. Add 50 grams to the hanger, rewind the string onto the shaft, and repeat steps 4 - 6.
Does the direction of the winding matter? How did the change in mass affect your plot?
Does this agree with your expectations? Continue to add mass to the hanger and repeat
until you have five different au and ad pairs and their averages.
52
For each of these runs, we will consider the relative error in au and ad to be the same as in
the first run. Is this a good approximation? Can it be verified?
8. Using the masses and their corresponding accelerations, calculate the approximate
torque (τapp = mrg) applied by the masses, with the associated error, and the angular
acceleration (α=a/r) of the disk. You can assume the error in hanging mass is negligible.
Input these values into the table on page 2 of the DataStudio workbook. By examining
plot 2, determine the moment of inertia, I, of the horizontal disk and the associated
experimental error. Record this value. Select landscape printing under page setup in the
file menu, and print your workbook.
Part II – Other Moments of Inertia
9. Select one of the mass configurations from those used in Part I. You will use this
mass for Part II of the experiment.
Note: you will not need to print any data from Part II; however, you should make
detailed notes and sketches.
10. Measure and record the relevant dimension(s) of the ring. Place the ring
concentrically on top of the disk. Predict the effect on the moment of inertia of the
system. Using your chosen mass, repeat steps 4 - 6. How have the plot and the
acceleration changed from Part I? Calculate the torque produced by the mass and the
resulting angular acceleration. Using the relationship τ=Iα, calculate the moment of
inertia of the system and the associated error. How does it compare with I for the
horizontal disk?
11. Remove the ring and the disk from the vertical shaft. With the assistance of your
instructor, mount the disk vertically on the shaft (using one of the holes along the
perimeter). How should the moment of inertia change from what was observed in Part
I? Make sure the disk is properly secured before proceeding. As in step 10, determine I
for the vertical disk. Does it agree with your expectations?
12. With the assistance of your instructor, remove the disk. Mount the aluminum
platform on the vertical shaft. Make sure it is properly secured with the thumbscrew.
How do you think the moment of inertia of the platform will compare with the disk?
Determine I for the platform using the procedure in step 10.
13. Measure and record the mass of the black blocks. For the purposes of this
experiment, we will treat these as point masses. Is this a good approximation? Attach
the blocks symmetrically to the aluminum bar. Record the distance from the center of the
rotating shaft to the center of one of the blocks. How will the blocks affect the moment
of inertia of the system? Measure I for the bar with attached blocks.
14. Repeat step 13 for a different (but still symmetric) position of the two blocks. How
does the moment of inertia change as you move the blocks?
Caveats:
53
1. Depending on the moment of inertia and the torque applied, the system can spin quite
rapidly. Exercise caution while the apparatus is spinning.
2. When you wrap the string around the shaft, do it in such a way that the string
connecting the shaft to the pulley is as close to horizontal as possible at all times while
the string unwinds. Why is this important?
3. Your results will depend on the value of r, the radius of the spool on which the string
wraps. If you wrap the string so that it overlaps itself creating layers you will change r,
thus changing values derived from it.
4. As you release the mass hanger, make sure that it falls straight down. Any side-toside oscillations can affect your data and are not accounted for in our analysis.
5. As the direction of the hanger’s motion changes, the data will usually appear “noisy”.
Try to avoid using these transitional data points in your fits. Can you conjecture why
there is this “noise” effect?
V.
REPORT
1. Include all measured and predicted moments of inertia and their comparisons. In each
case, note whether prediction agrees with the measurement within experimental error
(you do not need to predict the moment of inertia of the platform by itself).
2. Based on your results, can you confirm that the moment of inertia is an additive
quantity, i.e., the moment of inertia of any object is a sum of the moments of all its
constituent parts (all measure with respect to the same fixed axis)?
3. Does your data show that the mechanical energy of the system decreases as the
oscillations continue? Where does the energy go? Is the conservation of mechanical
energy theorem violated for this system? Is this rate of decrease the same for all
configurations? Use you plots to qualitatively answer these questions.
4. In calculating the torque, it was assumed that
. Estimate the percent error
introduced by this assumption. Hint: Calculate the tension using
and then
the torque for the 50 and 500 gram runs using
and
, and compare these
values. Alternatively, use the ratio of actual (translational) acceleration to gravitational
acceleration g.
5. Comment on all your results and the applicability of the approximation used.
54
6. Answer all the questions included in the introduction and procedure sections.
55
Experiment 15: Rotational Motion. Moment of Inertia
Aleksandr Shpiler
Section E2
Abstract
This experiment was conducted to figure out and analyze the effects of force on
circular rotating objects about their centers by analyzing their basic components such as
moment of inertia and angular momentum.
In part A we will measure the moment of inertia of a solid disk about an axis of
rotation through the center of the disk and perpendicular to the face and in part B we will
test the conservation of angular momentum for a collision process in the absence of a net
external torque.
Data
Preliminary Data (raw)
Diameter of the disk
Diameter of the spool
Inner diameter of the ring
Outer diameter of the ring
Mass of the disk
Mass of the ring
Values
10.76
38.24
107.4
127.74
1444
1439
Uncertainty
0.05
0.81
0.005
0.005
0.5
0.5
cm
mm
mm
mm
g
g
***uncertainties are incorrect because we forgot to use ∆tfi =
cm
mm
mm
mm
g
g
2
σ 2fi + ∆t ins
***all the calculations in this experiment will be based on the original uncertainty!!!
Preliminary Data (converted to SI)
Values
Diameter of the disk
0.1076
Diameter of the spool
0.3824
Inner diameter of the ring
0.1074
Outer diameter of the ring
0.12774
Mass of the disk
1.444
Mass of the ring
1.439
m
m
m
m
kg
kg
56
Original
Uncertainty
0.0005
0.00081
0.000005
0.000005
0.0005
0.0005
m
m
m
m
kg
kg
Correct Uncertainty
0.305255
0.00081
0.002743
0.165045
Part A (General raw data)
Part A
xG1
xG3
Delta m
25.7
69.8
0.5
cm
cm
g
xG2
Delta x
Delta t
50
0.05
0.001
cm
cm
s
0.5
0.0005
0.001
m
m
s
12.68
0.005
mm
mm
General data converted into SI units.
Part A
xG1
xG3
Delta m
0.257
0.698
0.0005
m
m
kg
xG2
Delta x
Delta t
Part A Obtained Data
Trial
1
2
3
4
5
6
7
8
Suspended
mass, g
50
100
150
200
250
300
350
400
tG2tG1, s
2.488
1.928
1.362
1.256
1.158
1.041
0.983
0.883
tG3tG1, s
4.047
3.074
2.28
2.082
1.907
1.721
1.614
1.463
Part A Obtained Data in SI units
Trial
1
2
3
4
5
6
7
8
Suspended
mass, kg
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
tG2tG1, s
2.488
1.928
1.362
1.256
1.158
1.041
0.983
0.883
tG3tG1, s
4.047
3.074
2.28
2.082
1.907
1.721
1.614
1.463
Part B (General raw data)
Part B
Rp
Delta Rp
Delta t
48.05
0.005
0.001
mm
mm
s
w
Delta w
57
Part B (General data converted to SI)
Part B
Rp
Delta Rp
Delta t
0.04805
0.000005
0.001
m
m
s
w
Delta w
0.01268
0.000005
m
m
Obtained Data in SI units
Trial
1
2
3
tb1, s
0.012
0.018
0.008
tb2, s
0.016
0.027
0.011
Analysis
Experimental values of the moment of inertia and friction torque.
x13 x12
−
t13 t12
we calculate the linear acceleration, a, for each
1) Using the equation a = 2
t13 − t12
mass used.
x12 = xG2 – xG1 and x13 = xG3 – xG1
x12 = 0.5 – 0.257 = 0.243
x13 = 0.698-0.257=0.441
t12 = tG2-tG1 and t13 = tG3-tG1
t12 = 2.488
t13 = 4.047
x13 x12
0.441 0.243
−
−
t13 t12
m
a=2
= 2 4.047 2.488 = 0.0145 2
4.047 − 2.488
t13 − t12
s
*** rest of the values for linear acceleration are in the table on the next page
2) Using equation τ = mr ( g − a) we calculate the applied torque, and equation α =
we calculate the corresponding angular acceleration,α again for each mass.
τ = mr ( g − a )
τ = 0.05 * 0.01912 * (9.81 − 0.0145) = 0.00937 Nm
58
a
r
α=
α=
a
r
0.0145
rad
= 0.7582 2
0.01912
s
*** to save time and space without writing out every single equation I will use a table for
the rest of the trials on the next page
3)
a (acceleration)
Trails
1
2
3
4
5
6
7
8
τ (torque)
α (angular acc.)
x13 x12
−
t13 t12 m
a=2
t13 − t12 s 2
τ = mr ( g − a ) Nm
0.05
0.014497
0.009365
0.758237
0.1
0.030408
0.009349
1.590391
0.15
0.032695
0.009347
1.709984
0.2
0.044417
0.009336
2.323065
0.25
0.057166
0.009324
2.989861
0.3
0.067109
0.009314
3.509872
0.35
0.08251
0.009299
4.315352
0.4
0.090473
0.009292
4.73186
Mass kg
59
α=
a rad
r s2
4) Plot of the applied torque,τ (y-axis) versus the angular acceleration, α (x-axis).
The slope of the trend line is the best estimate value for the moment of inertia of the disk,
Id, and the y-axis intercept gives the frictional torque τf.
alpha vs torque
torque Nm
0.00937
0.00936
0.00935
0.00934
0.00933
0.00932
y = -2E-05x + 0.0094
0.00931
0.0093
0.00929
0.00928
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
alpha rad/sec^2
Since the applied torque is linear we use linear function with intercept. .
From the graph we obtain that the slope is = -2E^-05 so I d = -2E-05 kg * m 2
From the graph we also obtain that the y-intercept is 0.0094 so the τ f = 0.0094 Nm
¿¿Is the graph supposed to look like that??
The frictional torque makes sense, I think, but shouldn’t the graph increase?
I think there’s an error somewhere but I couldn’t find it, of course I could be wrong.
5) Here we found the uncertainties for the slope and the intercept using the LINEST
function. This is a measure of ∆Id and ∆τf, the uncertainties in the measured moment of
inertia for the disk and in the frictional torque.
60
-1.82787E-05
2.83012E-18
0.00937836
8.58979E-18
From here we can see that
I d =-1.82787E-05 kg * m 2
∆I d = 0.00937836 kg * m 2
τ f =2.83012E-18Nm
∆τ f = 8.58979E-18 Nm
Theoretical value of the moment of inertia
0
7) Now we calculate the theoretical value for the moment of inertia of the disk, I d , using
1
I d0 = MR 2
2
equation
1
0.1076 2
I d0 = * 1.444 * (
) = 0.0388 kg * m 2
2
2
1
8) Based on expression I d0 = MR 2 and using the uncertainty propagation rule, and
2
known values of R, M, ∆R, and ∆M, we derive the formulas for the moment of inertia
uncertainty components due to uncertainty in the mass (∆I d )M and radius (∆I d )R .
0
0
We take derivatives with respect to M and then with respect to R and get the following

1  ∂MR 2
∂MR 2
∆I d0 = 
∆M +
∆R  =
2  ∂M
∂R

=
1 2
R × 0.0005 + (2MR × 0.0005)
2
[(
)
]
***using the appropriate uncertainty of the Diameter of the disk divided by 2.
As well as the uncertainty of the mass
∆I d0 = 0.5 [( 0.01912 2 * 0.0005) + (2* 1.444 * 0.01912 * 0.0005)] = 1.3896E-05 kg * m 2 .
or 1.39*10-5 kg * m 2
61
0
0
9) Now we calculate (∆I d )M , (∆I d )R and the total uncertainty
( ) + (∆I )
∆I d0 = ∆I d0
0
d R
M
.
The experimental values for the moment of inertia are
I d =1.82787E-05 kg * m 2
∆I d = 0.00937836 kg * m 2
I d min = -0.0094 kg * m 2
I d max = 0.00936 kg * m 2
The theoretical values for the moment of inertia are :
I d0 = 0.0388 kg * m 2
∆ I d0 = 1.39*10-5 kg * m 2
I d min = 0.038786104 kg * m 2
4)
∆v =
I d max = 0.038813896 kg * m 2
uncertainties in the velocity
∂v
∂v
w
1
∆tb +
∆w = 2 ∆tb + w
∂tb
∂w
tb
tb
∆ V=1/tb1i* ∆ W1+W1/tb1i^2* ∆ tb1i
deltaTB1I = Squareroot(uncert^2+.001^2)
dv1i
dv1f
dv2f
0.019729
0.004283469
0.022041
∆v1i =
0.024
1
m
0.00141 +
0.001 = 0.019729
2
s
0.0739
0.0739
∆v1f =
0.024
1
m
0.005686 +
0.001 = 0.00428347
2
0.33
s
0.33
∆v2f =
0.024
1
m
0.001125 +
0.001 = 0.022041
2
0.0644
s
0.0644
5) Initial and Final momentum
Pi = m1v1i
Pf = m1v1 f + m2 v 2 f
0.09915
0.098527
62
Pi = m1v1i = 0.3053 kg * 0.324763
Pf = m1v1 f + m2 v 2 f = 0.3053 kg * 0.072727
m
kg × m
= 0.09915
s
s
m
m
kg × m
+ 0.2048 kg * 0.372671 = 0.098527
s
s
s
6) Uncertainty in the momentums
partial derivative of p
delta Pi=V1i*dM1+M1*dV1i
Delta Pf=V1f*dM1+M1*dVf1+V2f*dM2+M2*dV2f
delta Pi =
delta Pf =
0.006039642
0.005843968
∆pi =
(+)
0.10519
0.104371
Pi +/- dPi
Pf +/- dPf
(-)
0.093111
0.092683
∂pi
∂p
∆m1 + i ∆v1i = v1i × ∆m1 + m1 × ∆v1i
∂m1
∂v1i
∆pi = 0.324763 × 0.00005 + 0.3053 × 0.019729 = 0.006039642
∆p f =
∂p f
∂m1
∆m1 +
∂p f
∂v1 f
∆v1 f +
∂p f
∂m 2
∆m 2 +
∂p f
∂v 2 f
kg × m
s
∆v 2 f = v1 f ∆m1 + m1∆v1 f + v 2 f ∆m2 + m2 ∆v2 f
∆p f = 0.072727 * 0.00005 + 0.3053 * 0.004283469 + 0.372671 * 0.00005 + 0.2048 * 0.022041
= 0.005843968
kg × m
s
The results clearly indicate that momentum (p) was conserved during these collisions.
P
0.006039642
∆ Pi / Pi = ∆ i =
= 0.06091
0.09915
Pi
Pf
0.005843968
∆ Pf / Pf = ∆
=
= 0.0593134
Pf
0.098527
10) Kinetic Energy
Ki=1/2M1V1i^2
0.0161
Ki =
Kf =
m1v12f
2
+
m 2 v 22 f
2
Kf=1/2M1V1f^2+1/2M2V2f^2
0.015029
m1v12i 0.3053 × 0.324763 2
kg × m 2
=
= 0.0161
2
2
s2
0.3053 × 0.072727 2 0.2048 × 0.3726712
kg × m 2
=
+
= 0.015029
2
2
s2
63
11) Uncertainty in Kinetic Energy
partial derivative of k
delta Ki=dM1(V1i^2)/2+dV1iM1V1i
delta Kf=dM1(V1f^2)/2+dV1fM1V1f+dM2(V2f^2)/2+dV2fM2V2f
(+)
delta Ki= 0.001958817
Ki+/- dKi= 0.018059
delta Kf= 0.001780932
Kf+/- dKf=
0.01681
(-)
0.014319
0.013248
2
v
∂K i
∂K i
∆K i =
∆m1 +
∆v1i = 1i × ∆m1 + m1v1i × ∆v1i
∂m1
∂v1i
2
kg × m 2
∆K i = (0.324763 * 0.00005)*0.5 + (0.019729 * 0.3053 * 0.324763) = 0.001958817
s2
2
2
∂K f
2
∂K f
v1 f
v2 f
∂Kf
∂Kf
∆K f =
∆m1 +
∆v1 f +
∆m 2 +
∆v 2 f =
× ∆m1 + m1v1 f × ∆v1 f +
× ∆m2 + m2 v2 f × ∆v2 f
∂m1
∂v1 f
∂m2
∂v2 f
2
2
∆K f = (0.00005* 0.0727272 )* 0.5 + (0.004283469*0.3053*0.072727) + (0.00005*0.3726712)*0.5 +
(0.022041*0.2048*0.372671) = 0.001780932
kg × m 2
s2
The results clearly indicate that kinetic energy was conserved during these collisions.
Fractional Uncertainties
K
0.001958817
∆ Ki / Ki = ∆ i =
= 0.1217
0.0161
Ki
Kf
0.001780932
∆K f /K f = ∆
=
= 0.1185
Kf
0.015029
64
Part B
1)
∆w =
tb1i
xb1+ xb2 =
0.0005+
0.0005 =
0.001 m
0.0728s
σ tb1i
0.001033s
deltaTB1I
∆t b1i = 0.001033 2 + 0.0012
0.001438s
tb(1+2)f
0.1274s
σ tb1f
0.002633s
deltaTB1F
∆t b1 f = 0.002633 2 + 0.0012
0.002817s
2)
v1i =
v1 =
w1
t b1i
v1 f =
0.024
m
= 0.32967
0.0728
s
v2 =
w1
tb 2 f
0.024
m
= 1.8838
0.1274
s
Uncertainties in the velocity
∂v
∂v
w
1
∆tb +
∆w = 2 ∆tb + w
∂tb
∂w
tb
tb
∆v =
∆ V=1/tb1i* ∆ W1+W1/tb1i^2* ∆ tb1i
deltaTB1I = Squareroot(uncert^2+.001^2)
dv1i
dv1f
0.02025
0.01201
∆v1 =
m
0.024
1
0.001 = 0.02025
0.001438 +
2
0.0728
0.0728
s
∆v2 =
m
0.024
1
0.002817 +
0.001 = 0.01201
2
0.1274
0.1274
s
Initial and Final momentums
Pi = m1v1i
Pf = (m1 + m2 )v 2
0.10065
0.9609
65
Pi = m1v1 = (0.3053) 0.32967
m
kg × m
= 0.10065
s
s
Pf = (m1 + m2 )v 2 = (0.3053 + 0.2048) 1.8838
m
kg × m
= 0.9609
s
s
6) Uncertainty in the momentums
0.006198
0.006
delta Pi =
delta Pf =
Pi +/- dPi
Pf +/- dPf
(+)
0.106848
0.9669
(-)
0.094452
0.9549
∆pi = v1 × ∆m1 + m1 × ∆v1 = 0.32967 × 0.00005 + 0.3053 × 0.020246 = 0.006198
kg × m
s
∆p f = v 2 ∆m1 + v 2 ∆m 2 + (m1 + m 2 )∆v 2 = 0.32967 × 0.00005 + 0.188383 × 0.00005 + 0.5101 × 0.012014 =0.006
The results clearly indicate that momentum (p) was conserved during these collisions.
P
0.006198
∆ Pi / Pi = ∆ i =
= 0.06158
0.10065
Pi
Pf
0.006
∆ Pf / Pf = ∆
=
= 0.006244
Pf
0.9609
) Kinetic Energy
Ki
Kf
0.01659
0. 00905
Ki =
m1v12i 0.3053 × 0.32967 2
kg × m 2
=
= 0.01659
2
2
s2
(m1 + m 2 )v 22 0.5101 × 0.188383 2
kg × m 2
Kf =
=
= 0.00905
.
2
2
s2
66
kg × m
s
11) Uncertainty in Kinetic Energy
0.002
0.00116
delta Ki=
delta Kf=
Ki+/- dKi=
Kf+/- dKf=
(+)
0.01859
0.01021
(-)
0.01459
0.00789
2
v1
kg × m 2
2
∆K i =
× ∆m1 + m1v1 × ∆v1i = 0.32967 x 0.5x0.00005 + 0.32967 x 0.3053 x 0.02025 = 0.002
2
s2
2
2
v
v
∆K f = 2 × ∆m1 + 2 × ∆m2 + (m1 + m2 )v 2 × ∆v 2 =0.1883832x 0.5x0.00005+0.1883832x 0.5x0.00005 + 0.5101 x
2
2
kg × m 2
0.188383 x 0.012014= 0.00116
s2
The results clearly indicate that kinetic energy was not conserved during these collisions.
Fractional Uncertainties
K
0.002
∆ Ki / Ki = ∆ i =
= 0.12055
0.01659
Ki
Kf
0.00116
∆K f /K f = ∆
=
= 0.1282
Kf
0.00905
Results
Part A
Velocity
m
m
± 0.019729
s
s
m
m
v1f = 0.072727 ± 0.004283469
s
s
m
m
v2f = 0.372671 ± 0.022041
s
s
v1i = 0.324763
Momentum
kg × m
kg × m
± 0.006039642
.
s
s
kg × m
kg × m
Pf = 0.098527
± 0.005843968
s
s
Pi = 0.09915
67
∆ Pi / Pi = 0.06091
∆ Pf / Pf = 0.0593134
Kinetic Energy
kg × m 2
kg × m 2
K i = 0.0161
± 0.001958817
s2
s2
kg × m 2
kg × m 2
K f = 0.015029
±
0.001780932
s2
s2
Fractional Uncertainties:
∆ K i / K i = 0.1217
∆ K f / K f = 0.1185
Part B
Velocity
m
m
± 0.02025
s
s
m
m
v2 = 1.8838 ± 0.01201
s
s
v1 = 0.32967
Momentum
g × cm
kg × m
± 0.006198
.
s
s
g × cm
kg × m
Pf = 0.9609
± 0.006
s
s
Pi = 0.10065
Fractional Uncertainties:
∆ Pi / Pi = 0.06158
∆ Pf / Pf = 0.006244
Kinetic Energy:
68
kg × m 2
kg × m 2
±
0.002
s2
s2
kg × m 2
kg × m 2
K f = 0.00905
±
0.00116
s2
s2
K i = 0.01659
Fractional Uncertainties:
∆ K i / K i = 0.12055
∆ K f / K f = 0.1282
Conclusion
For part A the theoretical and experimental values for the moment of inertia are in
0
0
agreement because the he range of possible values for the theoretical value I d ± ∆I d
overlap the range of possible values for the experimental result I d ± ∆ I d .
For part B the Momentum was conserved there is an intersection between the two
values and therefore the momentum is conserved. The kinetic energy obtained showed
that there is no intersection between the two answer sets and therefore kinetic energy was
not conserved. This proves that the collisions performed in part B are non-elastic and
contrariwise.
69