Carleton University
Laboratory Report
Course: PHYS 1003
Experiment#: 4
Atwood’s Machine
Bolaji Benedict Lawanson
(101170303)
Date Performed: 15-11-2019
Date Submitted: 22-11-2019
Lab Period: Friday AM, A4
Partner: Jacob Van Groningen
Station #: 19
TA: Brandon
Purpose
The goal of this experiment is to determine the acceleration due to gravity. This
will be done using a device called Atwood’s Machine. The set-up involves using a
timer to measure the time interval between the motion of two unequal weights
held by a light inextensible string passing over a pulley.
Theory
If the two masses are represented by 𝑚1 𝑎𝑛𝑑 𝑚2 and 𝑚2 > 𝑚1 , applying the
second law of motion to these two weights will give the following equations:
𝑚2 𝑔 − 𝑇2 = 𝑚2 𝑎
(1)
𝑇1 − 𝑚1 𝑔 = 𝑚1 𝑎
(2)
Where 𝑇1 is the tension of the string pulling 𝑚1 and 𝑇2 is the tension of the string
pulling 𝑚2 . When dealing with rotating systems, the friction is in the form of a
torque,𝜏 that opposes the motion of the two masses. This torque acts
counterclockwise and the relationship between,𝜏, 𝑇2 and 𝑇1 is given by
𝜏 = 𝑇2′ 𝑟 − 𝑇1′ 𝑟
(3)
Where 𝑇2′ and 𝑇1′ and the reactions to tensions 𝑇1 and 𝑇2 and 𝑟 is the radius of the
pulley. Torque, 𝜏 makes the pulley to rotate with angular acceleration,𝛼 and the
equation for the rotation of the pulley is given by
𝐼𝛼 = 𝑇2 𝑟 − 𝑇1 𝑟 − 𝛤 (4)
Where 𝐼 represents the rotational inertia and 𝛤 is the torque due to the friction in
the axel. The angle between the string and the pulley is assumed to be 𝜃 = 90𝜊 so
that 𝑠𝑖𝑛𝜃 = 1. Assuming there is no slipping of the string over the pulley,
𝛼=
𝑎
𝑟
(5)
Substituting equation (5) into (4), we get
𝐼𝑎 𝛤
+ = 𝑇2 − 𝑇1 (6)
𝑟2 𝑟
If equations (1) and (2) are added together, we have
𝑚2 𝑔 − 𝑚1 𝑔 − (𝑇2 − 𝑇1 ) = 𝑎(𝑚2 + 𝑚1 ) (7)
Combining the two previous equations, we can derive a formula for acceleration,𝑎
which is
𝛤
𝛥𝑚𝑔 −
𝑟
𝑎=
𝐼
𝑀+ 2
𝑟
(8)
Where 𝛥𝑚 is the difference between the masses and 𝑀 is the sum of the two
masses. Since the masses move along vertical lines, 𝑎 can be determined by
measuring the time,𝑡 it takes for mass,𝑚1 to rise to a height,ℎ starting at rest using
the equation below:
𝑎𝑡 2
ℎ=
2
(9)
Combining the previous two equations, we have
1
𝑔
𝛤
=
𝛥𝑚
−
(10)
𝐼
𝑡 2 2ℎ(𝑀 + 𝐼 )
2ℎ𝑟 (𝑀 + 2 )
𝑟2
𝑟
From this equation, a straight-line graph of
1
𝑡2
against 𝛥𝑚 can be generated with
sufficient plot points. The slope and y-intercept of the graph will then be used to
calculate 𝑔 and 𝛤.
Apparatus
Instrument
Clock
Vernier Caliper
Meter Stick
Weighing machine
Masses
Pulley
Solenoid
Reading error,𝜎𝑅𝐸
0.02s
0.02mm
0.05cm
0.02g
N/A
N/A
N/A
Observation
Table 1: Measurements of height, radius and masses of the iron cores and
washers collected during the experiment. An average of three trials was carried
out.
Trial #
Units
1
2
3
Average
Reading Error
𝜎𝑅𝐸
Standard
Deviation,𝜎𝑆𝐷
Std Dev
mean,𝜎𝑚𝑒𝑎𝑛
Final
Measurement
(𝑎𝑣𝑒𝑟𝑎𝑔𝑒 ±
𝑒𝑟𝑟𝑜𝑟)𝑢𝑛𝑖𝑡𝑠
Masses
Weight with
the iron
core+ screw
𝑚1
g
254.62
254.55
254.49
254.55
0.02
Other
weight+
screw
𝑚2
g
255.41
255.34
255.36
255.37
0.02
0.075
0.043
(254.550
± 0.043)
Derivation of Equation (10):
(8)
𝑎𝑡 2
2ℎ
ℎ=
;𝑎= 2
2
𝑡
(9.1)
Radius of
the pulley
(diameter/2)
𝑟
mm
61.86
60.45
63.19
61.83
0.02
0.040
0
0.058
1.6
0.023
0
0.033
0.92
(255.370 ± (1.00 ± 0.02)𝑔 (104.53
(61.83
± 0.05)𝑐𝑚 ± 0.92)𝑚𝑚
0.023)g
Calculations
𝛤
𝛥𝑚𝑔 −
𝑟
𝑎=
𝐼
𝑀+ 2
𝑟
Mass of one
washer (mass
of all/10)
𝑚𝑤
g
1.00
1.00
1.00
1.00
0.02
Lengths
Distance
travelled
by 𝑚1
ℎ
cm
104.6
104.5
104.5
104.5
0.05
Equate (8) and (9.1)
𝛤
𝛥𝑚𝑔 −
𝑟 = 2ℎ
𝐼
𝑡2
𝑀+ 2
𝑟
Cross-multiply
𝑡 2𝛤
2ℎ𝐼
𝑡 𝛥𝑚𝑔 −
= 2ℎ𝑀 + 2
𝑟
𝑟
2
Multiply each term by 𝑟 2
𝑡 2 𝑟 2 𝛥𝑚𝑔 − 𝑡 2 𝛤𝑟 = 2𝑀ℎ𝑟 2 + 2ℎ𝐼
2𝑀ℎ𝑟 2 + 2ℎ𝐼
𝑡 = 2
𝑟 𝛥𝑚𝑔 − 𝛤𝑟
2
1
𝑟 2 𝛥𝑚𝑔 − 𝛤𝑟
=
𝑡 2 2𝑀ℎ𝑟 2 + 2ℎ𝐼
1
𝑟 2 𝛥𝑚𝑔
𝛤𝑟
=
−
𝑡 2 2𝑀ℎ𝑟 2 + 2ℎ𝐼 2𝑀ℎ𝑟 2 + 2ℎ𝐼
Divide each term on the Right-Hand Side by 𝑟 2
𝑟 2 𝛥𝑚𝑔
𝛤𝑟
1
2
𝑟
𝑟2
=
−
2𝑀ℎ𝑟 2 2ℎ𝐼 2𝑀ℎ𝑟 2 2ℎ𝐼
𝑡2
+ 2
+ 2
𝑟2
𝑟
𝑟2
𝑟
1
𝑔
𝛤
=
𝛥𝑚
−
𝐼
𝑡 2 2ℎ(𝑀 + 𝐼 )
2ℎ𝑟(𝑀 + 2 )
2
𝑟
𝑟
Calculation for (𝑨 ± 𝝈𝑨 ):
𝐴=𝑀+
𝐼
𝑟2
𝐴 = (519.92 + 80)𝑔
𝐴 = 600𝑔
To calculate 𝜎𝐴 :
2 + 𝜎 2 + 100𝜎 2
𝜎𝑀 = √𝜎𝑚
𝑚2
𝑚𝑤
1
𝜎𝑀 = (√0.0432 + 0.0232 + 100(0.02)2 ) 𝑔
𝜎𝑀 = 0.21𝑔
𝜎𝐴 = √𝜎𝑀2 + 𝜎 2𝐼
2
𝑟
𝜎𝐴 = (√0.212 + 12 ) 𝑔
𝜎𝐴 = 1.0𝑔
Calculation for (𝒈 ± 𝝈𝒈 ):
𝑆𝑙𝑜𝑝𝑒, 𝑆 =
𝑔
2ℎ𝐴
𝑔 = 2𝑆ℎ𝐴
𝑔 = (2(7.591 × 1.045 × 0.6))𝑚/𝑠 2
𝑔 = 9.52𝑚/𝑠 2
𝜎𝑔 = √(
𝜕𝑔 2 2
𝜕𝑔
𝜕𝑔
) 𝜎𝑆 + ( )2 𝜎ℎ2 + ( )2 𝜎𝐴2
𝜕𝑆
𝜕ℎ
𝜕𝐴
For the partial derivatives:
𝜕(2𝑆ℎ𝐴)
= 2ℎ𝐴 = 2(1.045 × 0.6) = 1.254
𝜕𝑆
𝜕(2𝑆ℎ𝐴)
= 2𝑆𝐴 = 2(7.591 × 0.6) = 9.109
𝜕ℎ
𝜕(2𝑆ℎ𝐴)
= 2𝑆ℎ = 2(7.591 × 1.045) = 15.865
𝜕𝐴
𝜎𝑔 = (√1.2542 × 0.12082 + 9.1092 × 0.00052 + 15.8652 × 0.0012 ) 𝑚/𝑠 2
𝜎𝑔 = 0.15𝑚/𝑠 2
Consistency test between experimental value of acceleration due to
gravity,(𝒈 ± 𝝈𝒈 ) and theoretical value, (𝒈𝒕𝒉 ± 𝝈𝒈𝒕𝒉 ):
𝛥
|𝑔𝑡ℎ − 𝑔|
=
𝜎𝛥
√𝜎𝑔2𝑡ℎ + 𝜎𝑔2
|9.81 − 9.52|𝑚/𝑠 2
𝛥
=
𝜎𝛥 (√0 + 0.152 )𝑚/𝑠 2
𝛥
= 1.93(𝐶𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡)
𝜎𝛥
Calculation for the Frictional Torque, (𝜞 ± 𝝈𝜞 ):
Unit Analysis:
𝑏 = −0.006535𝑠 −2
ℎ = 61.83𝑚𝑚 =
61.83
𝑚 = 0.06183𝑚
1000
104.5
𝑚 = 1.045𝑚
100
600
𝐴 = 600𝑔 =
𝑘𝑔 = 0.6𝑘𝑔
1000
𝑟 = 104.5𝑐𝑚 =
Unit for 𝛤 is 𝑠 −2 × 𝑚 × 𝑚 × 𝑘𝑔 = 𝑘𝑔𝑚2 𝑠 −2
𝑏=−
𝛤
2ℎ𝑟𝐴
𝛤 = −2𝑏ℎ𝑟𝐴
𝛤 = (−2(−0.006535 × 0.06183 × 1.045 × 0.6))𝑘𝑔𝑚2 𝑠 −2
𝛤 = 0.000507𝑘𝑔𝑚2 𝑠 −2
𝜎𝛤 = √(
𝜕𝛤 2 2
𝜕𝛤
𝜕𝛤
𝜕𝛤
) 𝜎𝑏 + ( )2 𝜎ℎ2 + ( )2 𝜎𝑟2 + ( )2 𝜎𝐴2
𝜕𝑏
𝜕ℎ
𝜕𝑟
𝜕𝐴
For Partial Derivatives:
𝜕(−2𝑏ℎ𝑟𝐴)
= −2ℎ𝑟𝐴 = −2(0.06183 × 1.045 × 0.6) = −0.0775
𝜕𝑏
𝜕(−2𝑏ℎ𝑟𝐴)
= −2𝑏𝑟𝐴 = −2(−0.006535 × 1.045 × 0.6) = 0.00819
𝜕ℎ
𝜕(−2𝑏ℎ𝑟𝐴)
= −2𝑏ℎ𝐴 = −2(−0.006535 × 0.06183 × 0.6) = 0.000484
𝜕𝑟
𝜕(−2𝑏ℎ𝑟𝐴)
= −2𝑏ℎ𝑟 = −2(−0.006535 × 0.06183 × 1.045) = 0.000844
𝜕𝐴
=
(√(−0.0775)2 (0.000891)2
𝜎𝛤
2
+ (0.00819) (0.05)2
+ (0.000484)2 (0.92)2 + (0.000844)2 (0.001)2 )𝑘𝑔𝑚2 𝑠 −2
= 0.00061 𝑘𝑔𝑚2 𝑠 −2
Sample Calculations for 𝜟𝒎 𝒂𝒏𝒅 (𝟏/(𝒕𝒂𝒗 )𝟐 ± 𝝈(𝟏/(𝒕𝒂𝒗)𝟐 ):
𝛥𝑚 = 𝑀 − 2𝑚1 − 2𝑁𝑚𝑤
𝛥𝑚 = (509.92 − 2(254.55) − 2(0)(1))𝑔
𝛥𝑚 = 10.82𝑔
𝑡𝑎𝑣 =
𝑡𝑎𝑣 = (
𝛴𝑡
𝑛
3.65 + 3.64 + 3.64 + 3.70 + 3.65
)𝑠
5
𝑡𝑎𝑣 = 3.656𝑠
1
1
=
(
)𝑠 −2
2
2
𝑡𝑎𝑣
3.656
1
= 0.07481𝑠 −2
2
𝑡𝑎𝑣
1
2 )
𝑡𝑎𝑣
√
2
= (
)2 𝜎1/𝑡
𝑎𝑣
𝜕𝑡𝑎𝑣
𝜕(
2
𝜎1/𝑡𝑎𝑣
For the partial derivatives:
1
2 )
−2
−2
𝑡𝑎𝑣
−2
= 𝑡𝑎𝑣
= 3 =
= −0.040927
𝜕𝑡𝑎𝑣
𝑡𝑎𝑣 3.6563
𝜕(
𝜎
1
= (√(−0.040927)2 (0.012)2 ) 𝑠 −2
2
𝑡𝑎𝑣
𝜎
1
= 0.00049𝑠 −2
2
𝑡𝑎𝑣
Results
The value for (𝐴 ± 𝜎𝐴 ) is (600.0 ± 1.0)𝑔
The final value for 𝑔 is (9.52 ± 0.15)𝑚/𝑠 2
The consistency test between 𝑔 𝑎𝑛𝑑 𝑔𝑡ℎ gave a value of 1.93 which shows
consistency
The final value for frictional torque,𝛤 is (0.00051 ± 0.00061)𝑘𝑔𝑚2 𝑠 −2
2
−2
2 ) gave (0.07481 ± 0.00049)𝑠
The sample calculation for (1/𝑡𝑎𝑣
± 𝜎1/𝑡𝑎𝑣
Discussion
The goal of this experiment was to determine an experimental value for
the acceleration due to gravity. The experiment was carried out
successfully with little to no setbacks. If I was to increase the precision
of my measurements, I would increase the number of trials in order to
get more precise values for the averages. It would be trickier to
calculate 𝑔 from the time it takes for an object to fall. This is because
compared to the Atwood’s machine, the motion of the object begins
much faster and I will not be able to start the timer at the exact instant
the object began to fall. This would less accurate values for the time
which would in turn, affect the values for the acceleration due to
gravity. It was established in the theory section that the direction of the
torque was counterclockwise. By convention, counterclockwise
directions are taken to be positive whilst clockwise directions are taken
to be negative. This goes in line with my value for 𝛤 as my calculations
gave a positive sign for the torque.
Since this experiment involves motion over surface, friction would be a
factor. Friction in this experiment was in the form of a torque and it
opposed the motion of the strings passing over the pulley. This would
result in a longer time interval which would affect the value for the
acceleration due to gravity. Air resistance is also a force that would
oppose the vertical movement of the masses. However, because of
how large the masses were, I do not believe that air resistance would
cause a significant error.
During the experiment, we measured the distance between the strings
to determine the diameter of the pulley. This is method was better
than directly taking measurements of the pulley because if we took
measurements directly from the pulley, we may have to consider
factors like outside diameter and inside diameter of the pulley making
the diameter more difficult to measure.
Although the Atwood’s Machine is mainly used in physics labs to
demonstrate principles of machines, there are still some real-life
structures that follow the same concept. Some examples would be
elevators or wells which resemble the Atwood’s machine structurewise and also make use of the same physics principles.
A source of systematic error would be the presence of frictional forces
in the pulley system. Also, one of the masses was placed on a solenoid.
An error could arise due to some traces of magnetic attraction between
the mass and solenoid which could result in a longer time interval.
Another source of error could be due to the fact that the mass of the
pulley was not taken into consideration. A possible source of
improvement to this experiment would be to take more trials for the
measurements. It might also help to clean the washers before
transferring them to another mass since traces of dust would be
removed