[ PHYS 1152, Sec. 45] Experiment 9 PDF

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Report for Experiment #9 Maxwell’s Wheel

Abstract: The goal of this experiment was to investigate the motion of Maxwell’s wheel unwinding due to the force of gravity, the relationship of acceleration and the moment of inertia, and measure the difference in results when additional weights are added to the wheel. Investigation 1 found a moment of inertia of 3.80E-3 ± 5.93E-4 kg*m2. In Investigation 2, the downward acceleration was calculated to be 0.0904± 0.00622 m/s2. With added weights in investigation 3, the acceleration decreased to 0.0606 ± 0.0814 m/s2, while the total moment of inertia increased to 0.00691± 5.93E-4 kg*m2.

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Introduction For this experiment, our materials included a Maxwell’s Wheel, two support rods, two brass knobs, a vernier caliper, a meter stick, two meters of string, a stopwatch, two clamps, a rod with the string clamps, and a digital scale. The Maxwell’s wheel hung vertically by the two strings on either side of the axle attached to the rod with string clamps across the two support rods. The wheel was rolled upwards from it’s hanging position 8 times, 6 times, and 4 times and released. Each of the separate positions were repeated four times to get a more accurate reading of the speed of the wheel as it comes down from different heights. After completing investigations 1 and 2, masses were attached to the wheel edge of the wheel to study the motion of wheels with extra mass. The goal of this experiment was to investigate the motion of Maxwell’s wheel unwinding due to the force of gravity, the relationship of acceleration and the moment of inertia, and measure the difference in results when additional weights are added to the wheel. In Investigation 1, I found the moment of inertia of the wheel. The wheel is constructed of a rod and a disk. They have the same equation, but different components of mass and radius, so I found the separate moments of inertia and added them to find the total inertia. In order to find this, I first had to calculate the radius and mass through the volume and density of the parts. I = 0.5*M*R2 In Investigation 2, I found the downward acceleration of the Maxwell’s wheel without additional weights. To find the downward acceleration, I found the time it took the wheel to fully unwind. I know the distance based on the circumference of the pivot point of the wheel. I created a graph of squared time vs. distance and the slope of the best-fit line of the graph represented half acceleration. Therefore, I multiplied this slope by two to find the acceleration. There are two equations that relate the downward linear and angular acceleration with the net force and torque on the wheel... The tension in the string and the weight of the wheel each create a torque on the axle, causing the device to spin and move. 𝐹 = 𝑀𝑎 and 𝜏 = 𝐼⍺ We related the moment of inertia to the downward acceleration by coupling the equations for linear and angular motion. The angular acceleration then cancels out leaving us with the equation: 𝑎=

𝑔 1+

𝐼 2

𝑀𝑟

In Investigation 3, two knobs were attached to opposite ends of the wheel. The acceleration of the wheel decreased and the total inertia increased. This occurred because it became harder for the wheel to spin with extra weight on the ends of the disk.

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Investigation 1 In this investigation, I calculated the moment of inertia of the Maxwell’s wheel, so no setup was required. A digital scale was used to calculate the mass of the wheel and a vernier caliper was used to measure diameters of the Maxwell’s wheel. As stated above, the wheel is constructed of two distinct shapes: a rod and a disk. They have the same equation for moment of inertia, but different components of mass and radius, so I found the separate moments of inertia and added them to find the total inertia. I = ½*M*R2 V = πR2 *d Figure 1.1 - Table for Disk Measurement and Error Disk R (m)

d (m)

V (m3)

δR (m)

δd (m)

δV/V (m3)

δV (m3)

0.108

0.00659

0.000241

0.0005

0.0005

0.0764

1.84E-5

Figure 1.2 - Table for Rod Measurement and Error Rod R (m)

d (m)

V (m3)

δR (m)

δd (m)

δV/V (m3)

δV (m3)

0.00643

0.199

2.58E-5

0.0005

0.0005

0.156

4.01E-6

The error in radius(R) and diameter(d) were found by taking ½ of the smallest increment of the ruler. The error in volume(δV) was calculated using a propagated error formula in Appendix A of the lab manual. [2] δV = V*

(2 *

δ𝑅 2 ) 𝑅

+ (

δ𝑑 2 ) 𝑑

The density, ρ, of the wheel was then calculated using the mass of the wheel, 0.7183 kg, and the volumes of the disk and axle. ρ=

𝑉

𝑚𝑎𝑠𝑠 +𝑉

𝑑𝑖𝑠𝑘

𝑟𝑜𝑑

=

0.7183 2.41𝐸−4+2.58𝐸−5

= 2692 kg/m3

The actual density of Aluminum, the material of Maxwell’s wheel, is 2700 kg/m 3. The equation for percent error,

|𝑎𝑐𝑡𝑢𝑎𝑙 − 𝑒𝑥𝑝| 𝑎𝑐𝑡𝑢𝑎𝑙

*100, yielded a 0.286% error.

In order to find the mass of each component, the disk and rod, of the Maxwell’s wheel, I rearranged the formula for density: m = ρ*V. mdisk = (2692)(2.42E-4) = 0.651 kg mrod = (2692)(2.58E-5) = 0.0695 kg

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To calculate the errors in the masses, the propagated error formula in Appendix A was again used. [2] δm = m*

(

δ𝑝 𝑝

2

2

) , where δρ = ) + ( δ𝑉 𝑉

(

δ𝑚 𝑚

2

) ) + ( δ𝑉 𝑉

2

First, I calculated δρ for the disk and the rod: δρdisk= 0.0764 kg/m3 and δρrod=0.156 kg/m3. These values were then used to calculate δm for the disk and rod: δmdisk= 0.0497 kg and δmrod= 0.0108 kg. Finally, the moment of inertia of the disk and the rod were calculated using the equation from above, I = ½*M*R2. Idisk= ½(0.651)(0.108)2 = 3.79E-3 kg*m2 Irod= ½(0.0695)(0.00643)2 = 2.87E-6 kg*m2 Iwheel = Idisk + Irod = 3.80E-3 kg*m2 The error in the moment of inertia, δI, was calculated using the propagated error formulas: 2

2

δI = I* (δ𝐼𝑑) + (δ𝐼𝑟) , where δId,r =

(

δ𝑚 2 ) 𝑚

+ (2 *

δ𝑅 2 ) 𝑅

Therefore we find our error in moment of inertia to be δI = 3.80E-3*

2

2

(0. 00929) + (0. 156) = 5.93E-4 kg*m2

In this investigation, we found the moment of inertia of the wheel to be 3.80E-3 ± 5.93E-4 kg*m2. The disk had holes drilled into it, which slightly affects these results of the data. The measured volume of the disk did not account for these holes. Investigation 2 The setup of Investigation 2 consisted of a Maxwell’s Wheel, two support rods, two brass knobs, two meters of string, a stopwatch, two clamps, and a rod with the string clamps. The Maxwell’s wheel hung vertically by the two strings on either side of the axle attached to the rod with string clamps across the two support rods. The wheel was rolled upwards from it’s hanging position 8 times, 6 times, and 4 times and released. Each of the separate positions were repeated four times to get a more accurate reading of the speed of the wheel as it comes down from different heights. We were given the diameter of the string, r’ = 0.3 ± 0.1 mm. The time, t, it took for the wheel to completely unwind from release was measured and recorded. The average time and the average time squared was recorded for all measurements.

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Figure 2.1 - Table of Measurements of Time For Varying Rotations # rotation

t1 (s)

t2 (s)

t3 (s)

t4 (s)

tavg (s)

δt (s)

t2 (s2)

δt2 (s2)

8

2.85

2.63

2.57

2.71

2.69

0.0606

7.34

0.0451

6

2.39

2.19

2.17

2.25

2.25

0.00493

5.06

0.00438

4

1.75

1.79

1.72

1.98

1.81

0.00683

3.06

0.00755

To find tavg, the sum of each trial was divided by four: tavg = of tavg, δt, was found using the equation δt =

σ 4

𝑡1 +𝑡2+𝑡3+𝑡4 4

. The uncertainty

. I found δt2 with the propagated error formula:

δ𝑡

δt2 = |2| 𝑡𝑎𝑣𝑔 . Then, I needed to find the radius of the rod and the string. R′ = Rrod + Rstring = 0.00643 + ½*0.0003 = 0.00658 m Figure 2.2 - Distance Dropped # rotation

R’ (m)

y (m)

δR’ (m)

δy (m)

8

0.00658

0.331

0.0005

0.0251

6

0.00658

0.248

0.0005

0.0189

4

0.00658

0.165

0.0005

0.0126

For each of the rotations, y was found using the equation: y = 2π𝑟'𝑛 , with n being the number of rotations. The uncertainty in y, δy, was found using the propagated error formula: 2

δy = |2𝜋𝑛|δr′, where δr′ =

2

(δ𝑟𝑟𝑜𝑑) + (δ𝑟 ) 𝑠𝑡𝑟

The uncertainty of the radius of the string and the rod were calculated by taking ½ of the smallest increment of the instrument used to measure them. Therefore δ𝑟𝑟𝑜𝑑 = 0.0005 m and δ𝑟𝑠𝑡𝑟= 5.0E-6 m.

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Figure 2.3 - Graph of y vs. t2

The slope of this graph, 0.0452 m/s2, was calculated using a line of best-fit straight and the error in the slope, 0.00622 m/s2, was determined using the IPL online straight-line fit calculator. To find the downward acceleration of the wheel, a, I used the formula: y = yo + vyot + ½ at2. The wheel starts at rest so the initial velocity is zero and the equation can be rewritten as a =

2𝑦 2

𝑡

.

I found the downward acceleration to be a = 0.0904 m/s2. Using the formula 𝑎 =

𝑔 1+

to find a new value for the moment of inertia, I calculated I

𝐼 2

𝑀𝑟

to be 3.35E-3 kg*m2. The moment of inertia calculated in Investigation 1 was 3.80E-3 kg*m2. This new value for the moment of inertia is not as precise as the one found in the first investigation. There is no account for the human error in the timing and release points during the experiment, therefore there is more error in it’s value. Investigation 3 The setup of investigation 3 was very similar to setup of investigations 2, with brass knobs added to opposing sides of the disk. The added weight changed the downward acceleration and moment of inertia of the disk. The investigation was conducted with the same procedure.

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The masses of the knobs were measured to be m1 = 0.1833 kg and m2 = 0.1836. The distances from the rod and the outer hole on the disk were measured to be r1 = 0.092 m and r2 = 0.092 m. The average of the two measurements was 0.092 m. To find the new moment of inertia, the moment of inertia of each knob, MR2, must be added to the previous moment of inertia. I = 3.80E-3 + (0.1833)(0.092)2 + (0.1836)(0.092)2 = 0.00691 kg*m2 Figure 3.1 - Table of Measurements of Time # rot

t1 (s)

t2 (s)

t3 (s)

t4 (s)

tavg (s)

δt (s)

t2 (s2)

δt2 (s2)

y (m)

δy (m)

8

3.16

3.21

3.19

3.19

3.19

2.13E-4

10.2

1.34E-4

.308

.0251

The downward acceleration, a, was calculated using the same equation as last time and I found a to be 0.0606 ± 0.0814 m/s2. I found the uncertainty of a using the propagated error formula: δa = a

(

δ𝑦 2 ) 𝑦

+ (

δ𝑡 𝑡

)

2

The acceleration found in this investigation, 0.0606 ±0.0814 m/s2, is less than the acceleration found in investigation 2, 0.0904± 0.00622 m/s2. This makes sense because the added mass should slow down the descent of the wheel. Conclusion The purpose of this experiment was to understand the law of conservation of energy and momentum. Investigations 1 and 2 found the moment of inertia of the wheel in two ways: equations and measurements of the motion of the wheel under the force of gravity. Investigation 3 measured the effect that additional masses had on the object’s moment of inertia. Investigation 1 found the moment of inertia of the wheel to be 3.80E-3 ± 5.93E-4 kg*m2. Investigation 2 found the moment of inertia to be 3.35E-3 kg*m2. The second value is within the uncertainty of the first investigation’s value of the moment of inertia. In investigation 3, with the added masses, the moment of inertia was calculated to be 0.00691± 5.93E-4 kg*m2. In Investigation 2, the downward acceleration of the wheel was calculated to be 0.0904 ± 0.00622 m/s2. With the added mass, the downward acceleration decreased in investigation 3 to be0.0606 ±0.0814 m/s2. There was a lot of room for error in this lab. There was human error in the winding of the wheel. In order for it to be completely accurate, the wheel must have been wound exactly 8 times and released at the same spot every time. Also, there was human error in the timing of the descent of the wheel. Another source of error, as discussed above is the holes in the wheel. These holes were not accounted for in the measurement of volume of the disk.

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Questions 1. The law of conservation of total energy states that mgh = ½ Iw2. Therefore, w =

2𝑚𝑔ℎ . 𝐼

Angular velocity without knobs is 23.6±8.6 rad/s for 8 rotations, 24.4±4.2 for 6, and 23.1±3.9 for 4. The angular velocities with knobs is 40.5±3.7 for 8, 23.2±3.4 for 6, and 23.7±2.5 for 4. 2. The equation is 𝑎 =

𝑔 1+

. The total changed moment of inertia was obtained to be

𝐼 2

𝑀𝑟

0.0117±0.0001 kg*m2. 3. a) The downward acceleration would be smaller because the mass decreases. Therefore, a larger radius will result in a larger downward acceleration. b) The acceleration with halved radius was calculated to be 0.259 m/s2. 4. The ratio of Idisk over Ihoop is ½. 5. If Maxwell’s wheel is to rotate twice as fast, the KEf = 4KEi.

Acknowledgements I would like to thank our TA, Alex Hyde, for all the help he provided with the lab. Thank you too, to Northeastern University for providing a virtual lab and all other necessary components for a successful experiment. References [1] Hyde, Batishchev, and Altunkaynak, Introductory Physics Laboratory, p. 1, Hayden-McNeil, 2020. [2] IPL Straight Line Fit Calculator, http://www.northeastern.edu/ipl/data-analysis/straight-line-fit/ [3]Virtual Data, https://web.northeastern.edu/ipl/maxwells-wheel/...