Laboratory 4
Rotational Motion. Moment of Inertia
Michael Akopov
PH1004 Section E2
Abstract
In this experiment we analyze the rotating effects of force. Like linear motion, rotation motion also consists of basic components such as moment of inertia and angular momentum. The objective of this lab was to explore and analyze the rotational motion of circular objects about their centers. We will determine the moment of inertial of a ring, and a solid disk.
In part A of the experiment we had to analyze the rotation if a disc when a force is applied to it. In part B we had to check if there was a conservation of angular momentum when we drop the ring on the rotating disc. Both experiments achieved satisfactory results.
Data and Analysis
Variables
Symbol / Description / Measurement toolxd / Diameter of disk / Ruler
xs / Diameter of spool / Ruler
xir / Diameter inner ring / Ruler
xor / Diameter outer ring / Ruler
Dxd / Diameter of disk measurement uncertainty / Ruler
Dxs / Diameter of spool measurement uncertainty / Ruler
Dxir / Diameter inner ring measurement uncertainty / Ruler
Dxor / Diameter outer ring measurement uncertainty measurement uncertainty measurement uncertainty measurement uncertainty measurement uncertainty measurement uncertainty measurement uncertainty measurement uncertainty measurement uncertainty measurement uncertainty measurement uncertainty measurement uncertainty measurement uncertainty / Ruler
xG1 / Position of photogate G1 / Ruler
xG2 / Position of photogate G2 / Ruler
xG3 / Position of photogate G3 / Ruler
md / Mass of disk / Digital Scale
mr / Mass of ring / Digital Scale
mh / Mass of hanging object / Digital Scale
Dm / Mass measurement uncertainty / Digital Scale
tG1 / Time flag passes photogate G1 / Computer Clock
tG2 / Time flag passes photogate G2 / Computer Clock
tG3 / Time flag passes photogate G3 / Computer Clock
tb1 / Time during which notch is passing the photogate / Computer Clock
tb2 / Time during which notch is passing the photogate / Computer Clock
Dt / Time measurement uncertainty / Given (0.001s)
Rp / Distance from center of disk to center of post / Ruler
DRp / Distance measurement uncertainty / Ruler
w / Width of the post / Ruler
Dw / Width measurement uncertainty / Ruler
Part A
Diameter of the disk / 22.7 cm / 0.05 cmDiameter of the spool / 3.80 cm / 0.05 cm
Inner diameter of the ring / 10.7 cm / 0.05 cm
Outer diameter of the ring / 12.7 cm / 0.05 cm
xG1 / 10.0 cm / 0.05 cm
xG2 / 25.0 cm / 0.05 cm
xG3 / 35.0 cm / 0.05 cm
Mass of the disk / 1432 g / 0.05 g
Mass of the ring / 1411 g / 0.05 g
Trial / Suspended mass / tG2-tG1 / tG3-tG1
1 / 100 g / 0.761 s / 1.399 s
2 / 150 g / 0.622 s / 1.144 s
3 / 200 g / 0.537 s / 0.994 s
4 / 250 g / 0.483 s / 0.890 s
5 / 300 g / 0.441 s / 0.815 s
6 / 350 g / 0.402 s / 0.745 s
7 / 400 g / 0.376 s / 0.698 s
Rp / 105 mm
DRp / 0.5 mm
w / 13 mm
Dw / 0.5 mm
Dt / 0.001 s
Part B
Trial / tb1 / tb21 / 0.028 s / 0.049 s
2 / 0.010 s / 0.017 s
3 / 0.004 s / 0.007 s
Analysis
Part A
1. The acceleration in every trial was calculated using the following formula:
It yielded the following results for Trial 1:
a == 0.0289
2. The torque in every trial was calculated using the following formula:
It yielded the following results for Trial 1:
τ = = 0.0186 Nm.
The angular acceleration for every trial was calculated using the following formula:
α = a/r
It yielded the following results for Trial 1:
α = = 1.519
Table 1.2: results of calculations for the rest of the trials
Trial / mh / A / t / a1 / 100 g / 0.028856 m/s2 / 0.018565 Nm / 1.518726 rad/s2
2 / 150 g / 0.043345 m/s2 / 0.027806 Nm / 2.281316 rad/s2
3 / 200 g / 0.060876 m/s2 / 0.037009 Nm / 3.204026 rad/s2
4 / 250 g / 0.072875 m/s2 / 0.046204 Nm / 3.835527 rad/s2
5 / 300 g / 0.089272 m/s2 / 0.055351 Nm / 4.698507 rad/s2
6 / 350 g / 0.109516 m/s2 / 0.064442 Nm / 5.763980 rad/s2
7 / 400 g / 0.126615 m/s2 / 0.073518 Nm / 6.663939 rad/s2
3. Chart 1: torque vs. angular acceleration
5. The slope of the graph is given by:
m = 0.0107
This yields the moment of inertia of the disc.
= 0.0107
6. The intercept of the line is:
b = 0.0034
This yields the frictional torque:
= 0.0034 Nm.
7. Using the LINEST function in excel we retrieve the following data:
=0.010692 =0.000302
=0.003411 Nm = 0.001314 Nm
8. The theoretical value for moment of inertia was calculated using the following formula:
=
It yielded the following result:
= = 0.009224
9. The theoretical uncertainty was calculated using the error propagation rule:
=
= 0.5 [(0.11352 0.0005) + (2 1.432 0.1135 0.0025)] = 0.0004096 .
10. Comparing the experimental and the theoretical moments of Inertia:
Experimental:
=0.010692 =0.000302
min = 0.00768 max = 0.01100
The fractional uncertainty for was calculated by:
∆/= 0.028
Theoretical:
= 0.009224 ∆= 0. 0004096
min = 0.008814 max = 0.009634
The fractional uncertainty for was calculated by:
∆/= 0.044
The two answers are in experimental agreement because the experimental answer set and the theoretical answer set intersect in the range of values of:
= 0.008814and= 0.009634
Part B
1. The moment of inertia of the ring s given by the formula
This yielded the following result:
= 0.0048643 kgm2
2. The uncertainty of the inner radius of the ring was given by
This yielded the result:
= 0.001m
The uncertainty of outer radius of the ring was given by
This yielded the result:
= 0.001m
3. The uncertainty of the moment of inertia of the ring is given by:
This yielded the following result:
4. The total moment of inertia of the disk and ring was given by
Using the above formula the following result was obtained:
= 0.014089kgm2
5. The uncertainty of the moment of inertial of the disk and ring is given by:
Using the above the following result is obtained:
= 0.0004956kgm2
6. The angular velocity before collision was given by
This yields the following result for the first trial:
= 4.422 rad/s
The angular velocity before collision was given by
This yields the following result for the first trial:
= 2.527rad/s
Trial / w1 / w21 / 4.421769 rad / s / 2.526725 rad / s
2 / 12.38095 rad / s / 7.282913 rad / s
3 / 30.95238 rad / s / 17.68707 rad / s
7. The uncertainty of angular velocity before collision was given by
This yielded the following:
= -0.00891 rad/s
The uncertainty of angular velocity after collision was given by
This yielded the following:
= -0.03358 rad/s
Trial / Dw1 / Dw21 / -0.00891 rad / s / 0.033584 rad / s
2 / -0.821 rad / s / -0.18298 rad / s
3 / -6.69501 rad / s / -1.93068 rad / s
8. The angular momentum of the system before collision was given by
The above calculation yielded the following:
= 0.040786 kgm2/s
The angular momentum of the system after collision was given by
The above calculation yielded the following:
= 0.035599 kgm2/s
Trial / L1 / L21 / 0.040786 kgm2/s / 0.035599 kgm2/s
2 / 0.114202 kgm2/s / 0.102609 kgm2/s
3 / 0.285505 kgm2/s / 0.249193 kgm2/s
9. The uncertainty of angular momentum of the system before the collision was given by:
This yielded the following result:
= 0.001729kgm2/s
The uncertainty of angular momentum of the system before the collision was given by:
This yielded the following result:
= 0.001725kgm2
Trial / DL1 / DL21 / 0.001729 kgm2/s / 0.001725 kgm2/s
2 / -0.00250 kgm2/s / 0.001031 kgm2/s
3 / -0.04908 kgm2/s / -0.01844 kgm2/s
Results:
Part A:
Trial / mh / A / t / a1 / 100 g / 0.0289 m/s2 / 0.0186 Nm / 1.52 rad/s2
2 / 150 g / 0.0433 m/s2 / 0.0278 Nm / 2.28 rad/s2
3 / 200 g / 0.0609 m/s2 / 0.0370 Nm / 3.20 rad/s2
4 / 250 g / 0.0729 m/s2 / 0.0462 Nm / 3.84 rad/s2
5 / 300 g / 0.0893 m/s2 / 0.0554 Nm / 4.70 rad/s2
6 / 350 g / 0.1095 m/s2 / 0.0644 Nm / 5.76 rad/s2
7 / 400 g / 0.1266 m/s2 / 0.0735 Nm / 6.66 rad/s2
=0.01070.000302
∆/= 0.028
Part B:
Trial / w1 / w2 / Dw1 / Dw21 / 4.421769 rad / s / 2.53 rad / s / -0.00891 rad / s / 0.0336 rad / s
2 / 12.4 rad / s / 7.28 rad / s / -0.821 rad / s / -0.183 rad / s
3 / 31.0 rad / s / 17.7 rad / s / -6.70 rad / s / -1.93 rad / s
Trial 1:
L1 = 0.0408 0.00173 kgm2/s D L1/L1 = 0.0424
L2 = 0.0356 0.00173 kgm2/s D L2/L2 = 0.0485
Trial 2:
L1 = 0.114 0.00250 kgm2/s D L1/L1 = 0.0219
L2 = 0.103 0.00103 kgm2/s D L2/L2 = 0.0100
Trial 3:
L1 = 0.286 0.0491 kgm2/s D L1/L1 = 0.172
L2 = 0.249 0.0184 kgm2/s D L2/L2 = 0.0740
Conclusion:
For part A the two answers are in experimental agreement because the experimental answer set and the theoretical answer set intersect in the range of values of:
= 0.008814and= 0.009634
For part C, The answers for trial 1 are not in experimental agreement because the sets of two answers L1 = 0.0408 0.00173 kgm2/s and L2 = 0.0356 0.00173 kgm2/s do not intersect.
For trial 2, the answers are not in experimental agreement because the sets of two answers
L1 =0.114 0.00250 kgm2/s and L2 = 0.103 0.00103 kgm2/s do not intersect. For trial 3, the answers L1 = 0.286 0.0491 kgm2/s and L2 = 0.249 0.0184 kgm2/s have an intersection, which means that the answers are in experimental agreement.