Lab 3 (1D03)

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1 D3: Moment of Inertia NAME: Amanda Lykos STUDENT No.: 400297082 LAB/TUTORIAL SECTION: T7 LECTURE SECTION: C01 COMPLETE ALL SECTIONS OF THIS TEMPLATE AND SUBMIT AS ONE SINGLE DOCUMENT. SOME INSTRUCTIONS HAVE BEEN INCLUDED HERE. SEE LAB MANUAL FOR FULL DIRECTIONS. The following shows the value of all the questions in this lab: Laboratory 1 Grading Scheme Part 1 Part 1 Q1 Total Points /2 /2 Part 2 Part 2 Table (1) Calc Total Points /1 /2 /3 Part 3 Part 3 Table (1) Total Points /1 /1 Part 4 Part 4 Q2 Q3 Q4 Q5 Table (1) Plots (2) Total Points /1 /1 /1 /1 /2 /4 /10 Part 5 Part 5 Q6 Total Points /1 /4 TOTAL /20

2 Introduction In a translational system, "mass" is that characteristic of a body which determines how effective a given force will be in producing a translational motion. Mass might be considered a measure of the quantity of matter in a body. In a rotational system, "moment of inertia" is the analogue of mass. That is, the moment of inertia is that characteristic of a body which determines how effective a given torque will be in producing a rotational motion. The moment of inertia of an object depends not only on the mass of the object, but also on how the mass is distributed with respect to the axis of rotation. The kinetic energy of a rotating body likewise depends on the mass distribution. Those parts of the body farthest from the axis of rotation have the largest speed and hence the largest kinetic energy. In fact it is the moment of inertia introduced above that, along with the angular velocity, determines the kinetic energy of a rotating body. A simple system, whose moment of inertia can be easily calculated, is one with two equal point masses equidistant from the axis of rotation. The system you will investigate is slightly more complicated. It consists of a rotating shaft and cross bar on which two disks of total mass m are located at a distance R from the axis of rotation. It is designed so that the distance R can be easily changed and, hence, the moment of inertia, I , varied. By measuring both the rotational kinetic energy and the angular velocity of this system, I can be determined in a dynamical manner. The moment of inertia, I , consists of two parts: there is the moment of inertia of the shaft and crossbar, called I 0 , and the moment of inertia of mass m at distance R , called I m . Thus, ࠵? = ࠵? ! + ࠵? " (1) We want to find I m , by itself, for comparison with its theoretical value, so we use ࠵? " = ࠵? − 0 (2) where both quantities on the right side of this equation will be obtained from measurements of the motion of the system. The theoretical value for the moment of inertia of a total point mass m situated at a distance R from the axis of rotation is This equation will be tested graphically in two ways using the experimentally determined values of I m and the measured values of R . Don’t forget to read the Background Theory on page D3.6. I m = mR 2 (3)

3 Procedure: In this lab you will notice there are several videos for some of the experiments (ex. Videos 03 – 05). Use all of these videos as “multiple trials” in order to find an average value and experimental uncertainty (unless you are told not to calculate an uncertainty). For example, videos 03, 04, and 05 all measure the same thing – use each of these videos to make your measurements, and use all three values to calculate an average value and uncertainty. The figure below is a schematic of the experimental apparatus. A weight is hung from a cord which is stretched over a pulley, and connected to a rotating shaft. By rotating the shaft, lengths of cord can be wound onto the shaft, lifting the weight on the other end. Take a moment to understand the apparatus. Consider watching a few videos (Video 03 Determination and Video 08) before you begin to orient yourself with the experiment. Figure 1 rotating shaft (radius r ) R cross arm pulley driving weight M g ball bearing housing cord R m 1 m 2

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4 Part 1 (Video 01 - 05): In order to relate the translational motion of the driving weight to the rotational motion of the rotating shaft, we need to know r , the radius of the shaft. (This part uses Videos 01 and 02) ∙ Set up the apparatus as shown in Figure 1, without m 1 and m 2 attached to the crossbar. ∙ Be certain that the crossbar is centred on the rotating shaft and that the entire apparatus is level. ∙ Tie one end of the cord (about 125 cm long) to the screw-eye on the shaft and pass the other end of the cord over the pulley and attach to it a 100 g driving mass, M , remembering that the pan by itself weighs 50 g. Wind the cord evenly on the shaft (shown in Video 03- 05) ∙ Slowly allow the shaft to rotate at least 15 times, measuring the distance the driving mass falls. Use the bottom edge of the pan as a marker against the metre stick. ∙ The radius r can now be simply calculated. Question 1: How will the thickness of the string affect your measurements and subsequent calculations? Distance for 15 turns: 62.7cm±0.05cm Radius: 0.00796m ± 0.0005m The thickness of the string would only create a marginal difference if increased or decreased. If it were to have an affect it would cause a greater mg force therefore returning the amount of rotations.

5 Part 2 (Video 08 – 10): Determine the moment of inertia of the shaft and crossbar. ∙ Release the driving mass and measure the time, t 0 , it takes to fall a measured distance, h (try about 75 cm). ∙ Repeat this trial 4 times to ensure that an accurate value of t 0 has been obtained. Enter your values in Table 1. Table 1 Trial h 75± 0.05(cm) t 0 ± 0.25 (sec) 1 75cm 4.88sec 2 75cm 4.94sec 3 75cm 4.85sec Average time t 0 = 4.89±0.25 sec The moment of inertia can now be determined using Equations 4, 5 and 6 in the Background Theory at the end of the experiment. ∙ Use the average value of t 0 obtained from Table 1. ∙ Us ࠵? ࠵? = #$ % ! to calculate the final velocity of the driving mass M , and then ࠵? = & ’ for the final angular velocity of the shaft. ∙ Calculate the kinetic energy of the rotating system from ࠵?࠵? ’(% = ࠵?࠵?ℎ − )& " # ∙ Finally, use ࠵?࠵? ’(% = * ! + " # to obtain a value for I 0 , the moment of inertia of the shaft plus crossbar. Ignore error on M and m . Driving mass M = 100 g Calculation of I 0 I 0 = 0.969 ±1.01x 10^ − 4࠵?࠵? ࠵?^2 *Go to calculations page*

6 Part 3 (Video 07, 11 – 20): This part of the procedure is to determine the moment of inertia of a mass m placed on the crossbar at several distances, R , from the axis. ∙ Enter the data for this part in Table 2 . ∙ Using wing nuts, mount a 100 g mass on each side of the crossbar (Figure 1) at a distance of 15 cm from the shaft (measure from the centre of the shaft to the centre of the masses). You now have m = m 1 + m 2 = 100 g + 100 g = 200 g set at R = 15 cm. Actually m will be greater than 200 g since the mass of the wing nuts should be included. ∙ Release the driving mass and measure the time, t , it takes to fall the measured distance used in part 2. ∙ Use the method of analysis given in part 2 to determine the moment of inertia, I , of the rotating system. ∙ Calculate the moment of inertia of the two masses (plus wing nuts), I m . This is given by I m = I - I 0 from Equation (2). Value of I 0 from part 2: I 0 = 0.969 ±1.01x 10^ − 4࠵?࠵? ࠵?^2 Table 2 Do not calculate errors for these quantities R (cm) t (seconds) I (Nm) I m (Nm or kgm^2) 2.5 5.52 9.06E-4 2.74E-4 5.0 6.59 1.29E-3 6.58E-4 7.5 8.36 2.08E-3 1.45E-3 10.0 10.09 3.03E-3 2.41E-3 12.5 12.29 4.51E-3 3.88E-3 15 14.45 6.23E-3 5.60E-3

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7 It should be realized at this stage that this value of I m is the experimental value for the moment of inertia of mass m (the two 100 g masses plus the wing nuts). It is the experimental value because it is obtained from measured values of time and displacement, which effectively let you determine the rotational kinetic energy of the rotating system. The value of the mass, m , will be used only when the experimental value of I m is compared with the theoretical value for the moment of inertia from Equation (3). This comparison will be made in the graphical analysis. Remember the meaning of the various quantities: ∙ M is the "driving mass" hanging on the string ∙ m is the moveable mass added to the crossbar; you are trying to find the moment of inertia due to this mass m ∙ r is the radius of the shaft; this value is to be used in your calculations of I o and I . ∙ R is the distance between mass m and the axis of rotation. This parameter is not used in any calculation, but is an independent experimental variable. ∙ Repeat the above procedure in order to determine I m for the following values of R : R = 2.5 cm, 5.0 cm, 7.5 cm, 10.0 cm and 12.5 cm. Note that the mass m = m 1 + m 2 = 200 g (plus mass of 4 wing nuts) is kept the same for all these values of R .

8 Part 4: Graphical analysis. Log/Log Plot: ∙ Transfer R and I m to Table 3 Table 3 R (m) R 2 (m^2) log R I m (kgm^2) log I m 0.025 0.000625 -1.60 2.74E-4 -3.56 0.05 0.0025 -1.30 6.61E-4 -3.18 0.075 0.005625 -1.12 1.45E-3 -2.84 0.10 0.01 -1.0 2.41E-3 -2.62 0.12 0.0144 -0.92 3.88E-3 -2.41 0.15 0.0225 -0.82 5.60E-3 -2.25 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 max line of best fit y=1.87x-0.75 Log Im vs Log R Log R Log Im Min line of best fit

9 Plot a graph of log I m versus log R . Determine the slope and intercept (on the vertical axis) of the best-fit line drawn through the data points. Use the scatter of the data points to draw in the extreme-slope lines and hence estimate an uncertainty for the slope of the best-fit line. Question 2 : Do the slope and intercept of your first plot agree with the theoretical predictions? Yes, my slope and intercepts of the plot agree with the theoretical values calculated. There is a slight variance within the slope which could have been caused by human error such as perception and reaction time. I m /R 2 Plot: Plot a graph of I m versus R 2 and determine the slope and intercept (on the vertical axis) of the best-fit line drawn through the data points. Use the scatter of the data points to draw in the extreme-slope lines and hence estimate an uncertainty for the slope of the best-fit line . 0.00E+00 1.00E-03 2.00E-03 3.00E-03 4.00E-03 5.00E-03 6.00E-03 0 0.005 0.01 0.015 0.02 0.025 max line of best fit min line of best fit y=0.267x-7.5E-6 Im vs R^2 R^2(m^2) Im (kgm^2)

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10 Question 3: Do the slope and intercept of your second plot agree with the theoretical predictions? Yes, my slope and intercepts of the plot agree with the theoretical values calculated. The slope is very close to the predicted one but is slightly different because we are using an actual mass instead of assuming point masses. Question 4: What is the significance of any possible intercept in the I m versus R 2 plot? Note that in the general theoretical expression ࠵? = ∑ ࠵? ࠵? ࠵? ࠵? ࠵? ࠵? it is true that if all the masses m i are located at R i = 0 for all i, then I = 0. However, there is a finite size to the masses m 1 and m 2 in your experiment... The significant of possible intercepts in the Im versus R^@ plot is that my intercept was not exactly 0 while the theoretical expression assumes it is 0 because it follows point mass theory. Question 5: Much of the uncertainty in I m comes from I 0 . How does this uncertainty affect the scatter of the data points? This uncertainty affects the scatter plot points due to human error. Some examples of human error that could be found in this lab are perception issues reading the numbers and timing things from a video instead of actually doing it. Another example is reaction time from the person completing the experiment. These errors would affect the Im reading the most and therefore would alter the y axis for both scatter plots. The uncertainty of Im is heavily impacted by the value of Io. An increase of Io would decrease the Im and a decrease would make the Im greater. This would alter the slope of the line of best fit for the scatter plot.

11 Part 5: The analysis used above ignores the effects of friction and air resistance in this experiment. Question 6: What simple observation could be made which would indicate that there is a loss of mechanical energy due to these effects? Make this observation for the last value of R you used. Comment on your result but do not attempt any calculations. The loss of mechanical energy can be observed when the string is fully unwound as it begins to wind itself back up, not all the way to the starting position. It only winds up part way because it loses mechanical energy to friction. This can be observed when looking the results of my last R value which was 0.15, there is an obvious decrease in Im value.

12 Calculations : Part 1 ࠵? = ℎ # ࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? = 0.75 15(2࠵?) = 7.96࠵?10 ./ ࠵? Part 2 ࠵? = 2ℎ ࠵? = 2(0.75) 4.89 = 7.96࠵?10^ − 3 ࠵? = ࠵? ࠵? = 0.30675 0.00796 = 38.536 ࠵?࠵? = ࠵?࠵?࠵? − P ࠵?࠵? 2 Q = 0.1(9.81)(0.75) − (0.1)(0.30675) 2 = 0.72 ࠵?࠵? = ࠵?࠵? # ࠵? = 0.72 38.536 # = 9.65࠵?10 .0 ࠵?࠵?࠵? # Part 3 ࠵? = S2࠵?࠵?ℎ − T 4࠵?ℎ # ࠵? # UV ࠵? # ( 4ℎ # ࠵? # ) = ࠵?࠵? # S ࠵?࠵? # 2ℎ − 1V = 0.1(0.00676) # S 9.81(4.89 # ) (2)(0.75) − 1V = 7.1࠵?10 .0 ࠵?࠵? = ࠵? − ࠵?࠵? = 9.65࠵?10 .0 − 7.1࠵?10 .0 = 2.55࠵?10 .0 Part 4 LogIm vs LogR ࠵? = −0.75 − (−3) 0 − (−1.2) = 1.875 ࠵? = −3 − (1.875)(−1.2) = −0.75 Im vs R^2

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13 ࠵? = 0.006 − 0.004 0.0225 − 0.015 = 0.267 ࠵? = ࠵? − ࠵?࠵? = 0.006 − 0.267(0.0225) = −7.5࠵?10 .1 Uncertainty ࠵?ℎ = ±0.0005࠵? ࠵?࠵? = ±0.25࠵? ࠵?࠵? = 0.30675 ((0.0005/0.75) + (0.25/4.89)) = ±0.01589m/s ࠵?࠵? = 38.538 ((0.01589/0.30675) + (0.0005/0.00796)) = ±4.42rad/s ࠵?࠵?࠵? = 0.72 ((0.0005/0.75) + (0.01589/0.30675)) = ± 0.037777J ࠵?࠵? = 0.000965 ((0.037777/0.72) + (0.01589/0.30675)) = ±1.01x 10^ − 4 ࠵?࠵? ࠵?^2

14 Background Theory In this experiment the gravitational potential energy of a falling mass is converted into the translational kinetic energy of this mass and the rotational kinetic energy of the crossbar system. The conservation of energy principle requires the initial total energy (potential energy (PE) plus kinetic energy (KE)) to be equal to the total final energy (PE plus KE), provided frictional forces are negligible. Initial conditions (just before the system is released): The system is at rest, so KE = 0 If the final (lowest) position of the driving mass M is taken to be the zero point for potential energies, then the initial PE is Mgh , where h is the initial height of M above the final height. Hence, the initial energy = Mgh Final conditions: The driving mass has a translational KE, ࠵?࠵? %’234 = )& " # where v is its final velocity. The rotating assembly has a rotational KE, ࠵?࠵? ’(% = *+ " # where ω is the final angular velocity. The driving mass has PE = 0. Hence, the final energy = )& " # + *+ " # From the conservation of energy: ࠵?࠵?ℎ = )& " # + *+ " # (4) If v and ω are known, then I can be determined from this expression. In order to calculate v from t , the time it takes for M to fall through h , we use the constant acceleration formulae ࠵? = ࠵? ! + ࠵?࠵? and ࠵? # = ࠵? ! # + 2࠵?ℎ with ࠵? ! = 0 (5) to obtain ࠵? = #$ % (6) Finally, as the string which is wound around the shaft has the same speed as the driving mass M , ࠵? = & ’ (7) where r is the radius of the shaft.