Carleton University
Laboratory Report
Course #: PHYS 1007-L9
Experiment #: 4
Atwood’s Machine Experiment
Nicholas Adeyeye
101010922
Date Performed: November 9th, 2016
Date Submitted: November 23th, 2016
Lab Period: L9
Partner: Abdullahi
Station #: 28
TA: Matt. E
Purpose
The purpose of the conducted experiment was to find the frictional torque and acceleration due to
gravity using the Atwood’s Machine.
Theory
In this experiment the acceleration due to gravity and the frictional torque were calculated using
Atwood’s machine.
The equation used to calculate the acceleration due to gravity was:
g = 2mhA
(1)
where g is the acceleration due to gravity, m is the slope of the graph of 1/t2 vs ∆m in 1/s2/g, h is
the distance travelled by the first mass in m, and A is the total mass plus the related inertia.
The equation to determine the error on the acceleration due to gravity calculated is as follows:
𝜎g=g×√(𝜎2m/m2)+(𝜎2h/h2) +(𝜎2A/A2)
(2)
where 𝜎g is the statistical error on the acceleration due to gravity, g is the calculated acceleration
due to gravity in m/s2, 𝜎m is the error on the slope, m is the slope in 1/s2/g, 𝜎A is the error on the
value determined for A, A is the total mass plus the related inertia, 𝜎h is the error on the
measurement of the distance travelled by the first mass and h is the distance travelled in metres.
The equation used to calculate the frictional torque was:
Γ = -2bhrA
(3)
where Γ is the frictional torque in N/m, b is the y-intercept of the graph of 1/t 2 vs ∆m in 1/s2, and h
is the distance travelled by first mass in metres, r is the radius of the pulley in metres.
The equation for the statistical error on the frictional torque is:
𝜎Γ= Γ×√(𝜎2b/b2) + (𝜎2h/h2) +(𝜎2A/A2) +(𝜎2r/r2)
(4)
where 𝜎Γ is the statistical error on the frictional torque, Γ is the frictional torque in N/m, 𝜎b is the
error on the y-intercept, b is the y-intercept of the graph of 1/t 2 vs ∆m in 1/s2, A is the total mass
plus the related inertia, 𝜎A is the error for the A value, h is the distance travelled by the first mass
in metres, 𝜎h is the error for h.
The equation used to determine the density using the Archimedes principle is:
M= m1 + m2 + 10mwasher
(5)
Where M is the total mass of the machine in g, m1 is the mass of the iron core and the screw in g,
m2 is the mass of the weight without the iron core and the screw in g, and 10mwasher is the mass of ten
washers in g.
The error on the total mass is calculated using the following equation:
𝜎M = √(𝜎2m1+𝜎2m2+100𝜎2mw)
(6)
where 𝜎M is the statistical error on the total mass, 𝜎m1 is the error on the first mass, 𝜎m2 is the error on
the second mass, 𝜎mw is the error on the mass of ten washers.
The equation for the A value is as follows:
A = M + I/r2
(7)
where A is the total mass plus the related inertia, M is the total mass of the machine, I is the
inertia, r is the radius in m, and the value for I/r2 of the apparatus is 80 ± 1 g.
The equation for the error on A is the following:
𝜎A = √ (𝜎2M+𝜎2I/r2)
(8)
where 𝜎A is the error for the A value, 𝜎M is the error on the mass, 𝜎I/r2 is the error on the related
inertia.
The reciprocal of the average time squared was calculated using the following:
1/(tav2)
(9)
where tav is the average time it takes for m1 to hit the switch at the top of the apparatus.
The error on the reciprocal of the average time squared was calculate by the following equation:
2
σ1/(tav2) = 𝑡 3 σtav
𝑎𝑣
(10)
The equation used to calculate Δm is as follows:
Δmn-x = (m2 + nmw) – (m1 + xmw)
(11)
where Δm is the change in mass between the two masses connected to the apparatus in g, n is the
number of washers on m2, x is the number of washers on m1, and mw is the mass of one washer in
g.
The equation for the error on Δm is as follows:
σΔm = √(𝜎2m1+𝜎2m2+100𝜎2mw)
(12)
where σΔm is the error on Δm, 𝜎m1 is the error on the first mass, 𝜎m2 is the error on the second mass,
𝜎mw is the error on the mass of ten washers.
The equation of the graph of 1/t2 vs ∆m for the data collected is:
1/tav2 = 0.007695x – 0.004906
(13)
The standard deviation of a collection of measurements is found using the inefficient statistic
equation:
xmax−xmin
√N
𝜎𝐼𝑆=
(14)
where 𝜎𝐼𝑆 is the approximation of the standard deviation, 𝑥𝑚𝑎𝑥 is the maximum value, 𝑥𝑚𝑖𝑛 is the
minimum value, and N is the number of measurements.
The averages were determined using the following equation:
𝑥 +𝑥 +𝑥 +𝑥
Average = 1 2 𝑁 3 𝑁
(15)
The standard deviation of the mean is calculated using the following equation:
𝜎mean
=
σ𝐼𝑆
√𝑁
(16)
The consistency test equation is as follows:
|x1−x2|
𝑡=
2
2)
(√σ1 +σ2
(17)
The combined weighted average equation is the following:
𝑥𝑤𝑒𝑖𝑔ℎ𝑡𝑒𝑑 = (𝜎22𝑥1 + 𝜎12𝑥2) / (𝜎12 + 𝜎22 )
(18)
The combined weighted error equation is the following:
𝜎𝑤𝑒𝑖𝑔ℎ𝑡𝑒𝑑 = 𝜎1 𝜎2 / √ 𝜎12 + 𝜎22
(19)
Apparatus
Figure 1: Atwood’s Machine Apparatus
Electronic Balance
Error: ±0.005 g
Range: 0 - __ g
Vernier Caliper
Range length: 0-15 cm
Range width: 0-10mm
Error: ±0.0025 mm
Observations
Table 1. Masses Used, Diameter and Radius of Pulley in Apparatus
Masses
Lengths
Readings Iron core
Other weight one washer Distance
+ screw
+ screw
(mass of
travelled
m₁(g)
m₂(g)
all/10)
by m₁
mw (g)
h(cm)
254.47
255.44
0.995
99.4
254.47
255.45
0.994
99.3
254.44
255.44
0.994
99.3
Average 254.46
255.44
0.994
99.33
RE
0.02
0.02
0.002
0.05
0.02
0.005
0.0005
0.05
𝜎𝐼𝑆
0.01
0.002
0.0003
0.02
𝜎𝑚𝑒𝑎𝑛
Error
RE
RE
RE
RE
chosen
Final
254.46 ±
255.44 ±
0.994 ±
99.33 ±
Value ± 0.02 g
0.02 g
0.002 g
0.05 cm
error
Radius
of the
pulley
r(cm)
6.18
6.045
6.065
6.097
0.025
0.07
0.04
RE
6.097 ±
0.025
cm
∆m
(g)
σΔm
(g)
Time
1 (s)
10.92
8.94
6.95
4.96
2.97
0.011
0.011
0.011
0.011
0.011
3.580
3.911
4.477
5.558
7.552
Table 2. Change in Mass, Time and Error Measurements
Time Time Time Time Reading σ (s) Average
2 (s) 3 (s) 4 (s) 5 (s)
Error
Time
(s)
(s)
3.567 3.547 3.554 3.526 0.002
0.024 3.555
3.919 3.889 3.893 3.897 0.002
0.013 3.902
4.450 4.467 4.487 4.477 0.002
0.017 4.472
5.428 5.472 5.594 5.488 0.002
0.074 5.508
7.550 7.515 7.691 7.852 0.002
0.151 7.632
Calculations
The calculation for Δm was done using Equation 11:
Δmn-x = (m2 + nmw) – (m1 + xmw)
Sample Calculation:
Δm9-1 = (255.44 + 9(0.994)) – (254.46 + 0.994)
Δm9-1 = (264.386g) – (255.454g)
Δm9-1 = 9.532g
The calculation for the average times were done using Equation 15:
𝑥 +𝑥 +𝑥 +𝑥
Average = 1 2 𝑁 3 𝑁
Sample Calculation:
3.580+3.567+3.547+3.554+3.526
Average time for Δm10-0 =
5
Average time for Δm10-0 = 3.555 s
The calculation for the error on the average time was done using Equation 14 & 16:
xmax−xmin
𝜎𝐼𝑆=
√N
3.580−3.526
𝜎𝐼𝑆=
𝜎𝐼𝑆=
√5
0.024
𝜎mean
=
σ𝐼𝑆
√𝑁
= 0.024
√5
𝜎mean = 0.011
𝜎mean
Therefore, the average time for the Δm10-0 data was 3.555 ± 0.011 s
σmean
(s)
1/Tav2
(1/s2)
0.011
0.006
0.007
0.033
0.067
0.079
0.066
0.050
0.033
0.017
Error on
1/Tav2
(1/s2)
0.000481
0.000202
0.000166
0.000397
0.000303
The calculation for the 1/t av2 was done using Equation 9:
Sample Calculation for Δm10-0 times:
1/tav2 = 1 / (3.555s)2
1/tav2 = 0.079 1/s2
The error on 1/tav2 was calculated using Equation 10:
σ1/(tav2) =
σ1/(tav2) =
2
3
𝑡𝑎𝑣
σtav
2
(3.555)3
(0.011 s)
σ1/(tav2) = 0.000
Therefore, the value for 1/tav2 for the Δm10-0 data was 0.079 ± 0.00 1/s2
The calculation for the total mass was done using Equation 5:
M= m1 + m2 + 10mwasher
M= 254.46 g + 255.44 g + 10(0.994 g)
M = 519.84 g
The calculation for 𝜎 for the total mass was done using Equation 6:
𝜎M = √(𝜎2m1+𝜎2m2+100𝜎2mw)
𝜎M = √[(0.01)2+(0.003)2+100(0.003)2]
𝜎M = 0.011
Therefore, the total mass is 519.84 ± 0.011 g
The calculation for the A value was done using Equation 7:
A = M + I/r2
A = 519.84 g + 80 g
A = 599.84 1/s2g
The calculation for the error on the A value was done using Equation 8:
𝜎A = √ (𝜎2M+𝜎2I/r2)
𝜎A = √ (0.011)2+(1)2)
𝜎A = 1.00006
Therefore, the value of A is 599.84 ± 1.000 1/ s2g
The calculation for the acceleration due to gravity was done using Equation 1:
g = 2mhA
The slope of the graph was: 0.007880 ± 1.797e-41/s2/g
g = 2(0.007880 1/s2/g) (0.9933 m) (599.84 1/s2g)
g = 9.39 m/s2
The error for the calculated acceleration due to gravity was determined using Equation 2:
𝜎g= (9.39 m/s2)×√(1.797e-4)2/(0.007880)2+(0.05)2/(0.9933)2 +(1.00006)2/(599.842)
𝜎g= 0.519
Therefore, the calculated value for the acceleration due to gravity was 9.39 ± 0.519 m/s2
The calculation for the frictional torque was done using Equation 3:
Γ = -2bhrA
The y-intercept of the graph was -0.005761 ± 0.001347 1/s2
Γ = -2(-0.005761 1/s2)(0.9933 m)(0.06091 m)(0.59984 1/s2kg)
Γ = 0.000418 Nm
The error for the frictional torque was determined using Equation 4:
𝜎Γ= Γ×√(𝜎2b/b2) + (𝜎2h/h2) +(𝜎2A/A2) +(𝜎2r/r2)
𝜎Γ= (0.000418 Nm) ×√(0.001347)2/(-0.005761)2 + (0.0005)2/(0.9933)2
+(1.00006)2/(0.59984)2) +(0.025)2/(0.0697)2
𝜎Γ= 0.0001
Therefore, the value of Γ was 0.000418 ± 0.0001 Nm
The calculation for the consistency test was done using Equation 17:
𝑡=
𝑡=
|x1−x2|
√(σ21 +σ22 )
|9.81−9.39|
√(0.01)2 +(0.519)2 )
𝑡= 0.81
Since 𝑡 < 2, the results are consistent.
The calculation for combining the accepted value of g with the calculated value was done using
Equation 18:
𝑥𝑤𝑒𝑖𝑔ℎ𝑡𝑒𝑑 = [(0.519)2(9.81) + (0.01)2(9.39)] / (0.519)2 + (0.01)2
𝑥𝑤𝑒𝑖𝑔ℎ𝑡𝑒𝑑 = 9.81 m/s2
The calculation for the error of the combined value was done using Equation 19:
𝜎𝑤𝑒𝑖𝑔ℎ𝑡𝑒𝑑 = (0.519) (0.01) / √ (0.519)2 + (0.01)2
𝜎𝑤𝑒𝑖𝑔ℎ𝑡𝑒𝑑 = 0.001
Therefore, the combined value for g is 9.81 ± 0.001 m/s2
Results
The goal of the conducted experiment was to calculate the value of acceleration due to gravity as
well as the frictional torque of the Atwood’s machine. The value of acceleration due to gravity
was calculated to be 9.39 ± 0.519 m/s2 which was combined with the value of 9.81 ± 0.01 m/s2 to
yield a value of 9.81 ± 0.001 the value for the frictional torque was calculated to be 4.18 x 10-4 ±
0.0001 Nm.
Discussion
The results of the conducted experiment produced a value of acceleration due to gravity equal to
9.39 ± 1.003 m/s2 and a value of frictional torque equal to 4.18 x 10-4 ± 0.0001 Nm. The accepted
value for acceleration due to gravity is 9.81 ± 0.01 m/s2, compared to this value the calculated
value has a much more significant error value, however the two values are consistent with each
other seeing as t = 0.81. Since the values are consistent, they are combined and the smaller error,
that of the accepted in this case, becomes the more important one. The experiment could be
improved if the apparatus was no as long, because if the string was shorter it would be easier to
control and thus easier to avoid a centripetal force that would cause error in collected data. The
friction of the pulley that exists contributes to the error of the calculation of the acceleration of
gravity because it opposes the motion that is being observed, slowing down the time and thus
causing error in data. The air resistance could also contribute to the error of the measurement of g
because the heavier mass that is descending, propelling the other mass upwards, would be
experiencing air resistance slowing it down, so the time measurements would be recorded as
being lower than they should be in the ideal system, increasing the error. The elasticity and the
weight of the string on the pulley is important because if the string has considerable weight and
low elasticity, the friction will be increased which means that the tension force would not be the
same throughout the string, which is an assumption made when the calculations were done,
therefore increasing risk of error. Also, if the string had very low elasticity, then two masses of
different weight would be able to balance on the pulley due to increased friction and the timing
for one mass to reach the top and the other to reach the bottom would be measured as being
slower than it really is, increasing the error of the results. Thus, another improvement for this
experiment would be to use the lightest and most elastic string possible to be able to sustain the
weight of the masses used.
Questions
1. How does this experiment compare to one where you would calculate 𝑔 by measuring the
time it takes an object to fall on the ground from a certain height?
Measuring the time it takes an object to fall to the ground would have to take into account
visual and physical reaction times when starting the clock when the object is let go and
stopping it when the object hits the ground; yielding less accurate results. This experiment was
done using an apparatus that hit stop immediately without anyone have to react in time to do
anything, so the results yielded by the experiment conducted would be more accurate.
2. Geologists are able to discover oil deposits by measuring variations on the gravitational
acceleration in the order of ~0.00005 m/s2. Is your measurement precise enough to find a
variation of such magnitude?
The measurement calculated is not precise enough to calculate gravitational acceleration in the
order of 0.00005 m/s2 as the calculated value only goes to two decimal places, which means a
variation of magnitude with anything more than two decimal places would not be found.
3. Under what circumstances the acceleration of the Atwood’s machine is equal to zero?
Explain.
The Atwood’s machine’s acceleration would equal to zero if there were equal masses placed
on both sides either at rest, or at constant velocity. In both cases the force of gravity is the
same on both sides, which means the tension in the string is equal to the force opposing
gravity; thus, the net force is zero. If the net force is zero, then there is no acceleration. In the
scenario with constant velocity, the upward velocity is the same in magnitude as the
downwards velocity but the directions are different, this means the net velocity is zero and the
acceleration is therefore also equal to zero.