Lab# 22: Physical Pendulum Lab

Physical Pendulum 

Samuel
Ellis
Mia
11-28-2016

Purpose:

Derived expressions for the period of various physical pendulum. Verify your predicted period by experiment.

Theory:

Procedure:

1. derive expressions for the moment of inertia of the ring about this point and the period of this ring behaving as a physical pendulum
2. Make physical measurement of the relevant quantity for the ring
3. Measure the actual period of the ring, acting as a physical pendulum
4. compare your experimental result with your theoretical result
5. Cut our or find objects of the appropriate shapes. Measure the appropriate dimensions of each one, and their respective masses
6. Attach a thin piece of making tape to the bottom of each object
7. Use the same setup as above to measure the small-amplitude period of oscillation for each one.
8. Compare the actual and theoretical values for the periods using the actual measured dimensions for the various object.

Measured Data:

Ring_inner = 20.59 cm
Ring_outer = 21.8 cm
Height_1 = 12 cm
Based_1 = 13 cm
Height_2 = 13 cm
Based_2 = 13.4 cm

Graph:

 (Above, is the oscillating ring graph. The period slowly decreasing due to frictional force acting between rod and and the ring.) 

 (Above, is upright triangle graph. This graph has lesser decreased because we used paper clip instead of metal ring. So the frictional force is lesser.)

(Above, is upside down triangle. The period is faster than the upright triangle. Because it has less mass on the bottom)

Calculation:

I_pt = I_cm + m*r^2

        = m*r^2 + m*r^2

        = 2*m*r^2

I = I*𝝰

m*g*sin𝛉*r = 2*m*r^2*𝝰

𝝰 = g/2*r * sin𝛉

w^2 = g/(2*r)

w = sqrt(g/(2*r))

Based on above equation, I got 0.928 for my calculated period for the ring. 

Conclusion:

In this experiment, we derived a equation to determined the period of oscillating ring and compared it with actual period of the ring. As usual, we have made some assumption. First, the angle we pushed can not be larger than 10 degree. Because if it is less than 10 degree, the amount of error is so small, we can just ignore it. Secondly, we assumed there's no frictional force acting between two materials. And lastly, we assumed the rod for holding the object is parallel to the surface. 

      

Lab#21: Mass-Spring Oscillations

Mass-Spring Oscillation

Samuel
Ellis
Mia
11-23-2016

Purpose:

To determine the oscillation of the spring with different weight and compare it to see if simple harmonic equation is accurate.

Theory:

Based on simple harmonic equation, we know that spring and pendulum have same oscillation. And we know that oscillation is based on how much the spring stretched. When releasing spring and weight dropped, the up and down motion is determined by the spring constant. Since we have different spring with different spring constant, we can use oscillation equation to predict it period. 

Procedure:

1. Mount clap on the table, with a rod horizontal respect to the table. 

2. Hang a spring on the rod.

3. Hang 5 different weigh on it and record it period and mass.

4. Find spring constant of the spring.

Measured Data:

Data for Spring 4

m = 215 g

t_1 = 6.66 sec, t_2 = 6.63 sec, t_3 = 6.62   t_average = 6.64 sec

m = 315 g

t_1 = 8.05 sec, t_2 = 8.00 sec, t_3 = 8.08 sec  t_average = 8.05

m = 4.15 g

t_1 = 9.23 sec, t_2 = 9.21 sec, t_3 = 9.21 sec  t_average = 9.21 sec

m = 65 g

t_1 = 3.63 sec, t_2 = 3.62 sec, t_3 = 3.65 sec  t_average = 3.63 sec

Spring for All Group

Calculation:

Predicted value:

T = 2*pi*sqrt(m/k)

T = 2*3.14*sqrt(0.115/19.4)

T = 0.48 sec

Measured value:

T = time/how many times

T = 5.22/10

T = 0.522 sec

Analysis:

1. We calculate our spring by using spring potential energy equation, 0.5*k*x^2. k is spring constant and x is how much it stretch. 

2. 

3.  

5. Our spring constant is right. But if k is off by 5% or more, my first option will go with period. Usually we measure period for at least 3 times and those time should somewhere similar to each other. We then average those number to get most accurate data. However, when averaging 3 different number, there's will be uncertainty, and that's where those error come from.

6. According to our formula, period(T) = 2*pi*sqrt(m/k). In this case, m is mass & k is spring constant. k = (2*pi*m)/T. since 2*pi is constant, we can simply ignore it when determining variables. Since k = m/T(when ignoring 2*pi), when T is decreasing, it will make m divide by smaller number, which will have bigger product. In this case, it will be k. 

7. Use the same equation from above, k = m/T. We know that if we increase the mass(m), then k will become bigger number. Since m is increasing, bigger number divide by smaller number result in bigger resultant value. Therefore, when increasing period, mass is also increase.

Conclusion:

In this experiment, we used simple harmonic equation to find the period and spring constant of the spring. We used different weight and make it oscillating so we make sure our spring constant is the same. Thus, we used the same equation to calculate it period. However, the number we got are little bit different from the actual measured data. 0.48(calculated value) verses 0.522(measured value). And that difference might due to the rounding error of the measured period or the human error when stopping the timer. 

Lab#20: Conservation of Linear and Angular Momentum

Conservation of Linear and Angular Momentum

Samuel
Ellis
Mia
11-21-2016

Purpose:

To understand how conservation of momentum can translate to both linear and angular direction.

Theory: 

When object is rotating, it will have a tangential and linear direction. When the tension is release, the object will move toward tangential direction. And since we know that momentum is conserved, therefore, we can use tangential momentum and change that to angular momentum. In addition, we can use kinematic equation to find how far or how long object travel. Because velocity is the same throughout flying period.  

Procedure:

1. Use the aluminum to disk, and mount the ball-catcher on top of the small torque pulley using a gray-capped thumbscrew.

2. Using the method of Experiment 6, determine the moment of inertia of the disk and ball catcher. Also measure of the ball. 

3. Place the ramp on the edge of a table. Determine the horizontal velocity of the ball as it roll off the end of the ramp

4. Mark a starting point on the ramp

5. Release the ball from this staring point and note where the ball strikes the floor(a floor of carbon paper over a sheet of white paper will leave a mark for measuring)

6. Measure the distance L and h. The horizontal velocity of the ball is then equal to L/t where t = 2*h/g.

7. If you have time, you may want to calibrate the ramp by determining the horizontal velocity for a variety of starting points. If you plan on doing the optional part of the analysis for this experiment, you will also need to record the starting height of the ball. 

8. Roll the ball down the ramp so that it is caught by the ball-catcher. Quickly remove the ramp, so it doesn't disturb the rotational of the ball-catcher, and record the reading of the digital display. Record the horizontal velocity of the ball based on your earlier measurement, and also record the radius.

9. Repeat your measurement.

10. Using procedure outlined above, make the measurement necessary to investigate the relationship between the angular momentum imparted to the system.

Measured Data:

Mass_ball = 28.9 grams

Diameter_ball = 19 mm

h = 95 cm

L = 59.5 cm

d = 16 cm

hang mass = 24.6 g

diameter_pulley = 47.9 mm

w = 2.269 rad/s

r_ball to axis = 7.6 cm

Graph:

(Above, is the graph of ball-catcher rotating on top of disk. The consistent slope is cause by the rotating disk. Since it's angular velocity is the similar, so the slope looks almost identical.)

(Above, is the angular acceleration graph of it going up and down.)

Calculated Data:

Find hanging time in air
Δy = -0.5*g*t^2
0.95 = -4.9*t^2
t = sqrt(0.95/4.9)
t = 0.440

Δx = v*t
0.59*t = v*(0.440)
v = 1.35 m/s

Find w
1.
(0.076)(0.02898)(1.35) = (1.05*10^-3)+0.0289*0.076^2+ 2/5 *0.0289*0.009^2
w = (0.002965/0.001217)
w = 2.43 rad/s

2.
0.042*0.0289*1.35 = 0.00105 + 0.0289*0.042^2 + 2/5 *0.0289 * 0.009^2
w = (0.001638/0.001102)
w = 1.48 rad/s

Conclusion:

In this experiment, we used already prepared equipment to conduct this experiment. We used LoggerPro to calculated the angular velocity of the rotating disk with ball-catcher. We then measured the height, distance and length of the ramp. With all the measured data, we are able to use kinematic equation to find the air time during the fall. And according to conservation of momentum, all energy conserved. So for our calculated data for angular momentum, we got 2.43 and 1.48. And for theoretical we got 2.43 and 1.49. Which is almost identical and it also prove momentum is conserved. 

     

Lab#19: Conservation of Energy/Conservation of Angular Momentum

Conservation of Energy/Conservation of Angular Momentum

Samuel
Ellis
Mia
11-21-2016

Purpose:

To understand how momentum is conserve when energy is transferring.

Theory: 

Based on conservation of momentum, we know that we can transfer energy into momentum. Therefore, energy equal momentum. Since the stick swing down was cause by gravitational potential energy and inertia. In addition, when stick is vertically straight, that's when GPE equal 0. And at that moment, that's when stick collides with clay, and turn that into momentum.

Procedure:

1. Set up ring stand on the edge of the table

2. Insert ruler with hole (at the very end)into ring stand 

3. Make sure ruler swing without any external friction

4. Use tape taped around the ruler(inside out) and clay(inside out)

5. Place clay at the place that ruler can hit it

6. Use camera in your phone to videotape the whole movement

7. Upload video into computer

8. Dot every position of movement of the ruler and clay

9. And don't forget to set origin and distance ratio

Measurement:

Mass of clay: 36.7 grams

Mass of ruler: 92.3 grams

Graph:

(Above is the video capture software we used to capture the position of the whole movement. We set the intersection of the yellow line as our origin. And doted the position of ruler every second. And plot into the graph)

(Above is the graph of the center of the ruler verses end of the ruler. Since we set our origin at center of the ruler, therefore the blue dot(center of ruler) touch 0 when its vertically straight. And red dot(end of ruler) will go way under 0 because it is further than center of ruler)

Calculation:

Y-_center mass

y_cm = ((0.0923(0.9/2) + 0.0367(0.9))/(0.0923+0.0367)

          = 0.578

Energy = moment of inertia

mgh = 0.5*m*v^2 + 0.5*I*W^2

mgh = (0.5*m*r^2 * w^2) + (1/12 *m + 0.16 m)*w^2

0.9*g = 0.5 * 0.45^2 * w^2 + 0.5 (73/300)*w^2

17.64 = (0.45^2 + 73/30)*w^2

w = 6.29 rad/sec

Omega(w) after

I*w = (I_stick + I_clay)*w_after

(1/12 *M + 0.16*M)(6.29) = (1/12*M + 0.16*M + m*r^2)*w_after

0.14127 = 0.0521 * w_after

w_after = 2.7 rad/sec

Find Rising Distance

0.5*I*w^2 = m*g*Δh

0.5(1/12 *m + 0.16 * m +m*L^2)*w^2 = (0.0923 + 0.0367)*9.8*h

h = 0.15 m

Find angle

Δh = y_cm(1-cos𝛉)

0.0756 = 0.578(1-cos𝛉)

𝛉 = 30 degree

Conclusion:

In this experiment, we used conservation of momentum and inertia of the stick to determine how height and the angel of the stick after collides with clay on the floor. As usual, we make some assumptions to make this experiment easier. First, we assumed there's no frictional force acting between stick on the rod and clay on the floor. Also, we assume there's no human error when plotting dots on the video capture software. Lastly, we assume when we dropped the stick, we held it perfectly parallel respect to the horizontal floor. For our calculation, we make momentum equal to torque force. Since the momentum is based on gravitational force acting on the stick, we simply set it equal to torque force. In addition, since stick is rotating at 10 cm mark. We have to used axis parallel theorem to add the inertia to the total torque force. Therefore, we got the answer that is reasonable and close to our prediction.     

Lab#18: Moment of Inertia and Frictional Torque

Moment of Inertia and Frictional Torque

Samuel
Kyle
John
11-16-2016

Purpose:

To determine moment of inertia of combined mass and calculate frictional force acting against the rotating mass. 

Theory:

With given mass and the shape of an object, we can find the moment of inertia when object is rotating. Since, object is decreasing its angular velocity consistently. Therefore, we can calculate the fractional force acting on it, even more on coefficient of friction. In addition, with known constant force apply to the disk tangentially, we can calculate/time the velocity of the object with already known distance.

Procedure:

1. Make appropriate measurement of the rotating part of the apparatus and determine its moment of inertia.
2. Spin the apparatus. Use video capture to determine its angular deceleration as it slow down. Calculate the frictional torque acting on the apparatus.
3. You are going to be connecting this apparatus to a 500-gram dynamic cart. The cart will roll down an inclined track for a distance of 1 meter. Calculate how long it should take for the cart to travel 1 meter from rest. Assume for your preliminary calculations that the track is angled at 40 degree.
4. Set up the apparatus. Determine your actual angel measurement. Calculate what the time for the cart to travel 1 meter should be with this actual angle.
5. Run three trial where you measure the time for the cart to travel 1 meter. Be sure that your instructor witnesses at least one of the trial. If your average time is more than 4% off from what you calculated, figure out what went wrong and repeat the steps again until your predictions and your calculations match.

Measurement & Equation:

Big_diameter (diameter of big disk)= 20.05 cm  = 0.2005 m
Dept_ Big (thickness of big disk)= 14.6 mm = 0.0146 m
Small_diameter (small rod diameter)= 30.9 mm = 0.0309 m
Dept_small (dept of small rod)= 5.1 cm = 0.05 m
Mass (mass of both big disk and small rod)= 4615 g = 4.615 kg

w(omega) = sqrt(v_x^2 + v_y^2) / (0.2005/2) (For the purple graph below)
I(inertia) * ⍺(alpha) = 𝓣_friction
V(volume) = 𝜋*r^2 * h

Graph:

(Above is the graph we used to find the torque due to friction and angular deceleration. We first used camera to videotape the rotating disk. Then upload it to LoggerPro using air drop. Next, use video capture to apply dot in every 5 frames. At the end, we got this up and down curve signifying its position and where's it actually is. And the bottom graph shows how much does disk decelerate, we used from 1.3 second and beyond because it become more stable after that time)

Calculation:

Part 1. Find angular deceleration and frictional torque

Before finding angular deceleration, we first need to know how big of portion that each shape is and it's inertia.

We first find volume of each shape.

Disk = 𝜋*r^2 * h = 3.14 * 0.1002^2 * 0.0145 = 4.6*110^-4

Cylinder = 𝜋*r^2 * h = 3.14*0.01545^2*0.051 = 7.64*10^-4

and we got disk is 85.7% and cylinder is 7.1%(each cylinder)

We used the percentage multiply with weight of total mass to find it inertia.

I_cylinder = 1/2*2 * m*r^2 = 0.655*0.01593^2

I_disk = 1/2 * 3.955*.10^2

I(inertia) * ⍺(alpha) = 𝓣

𝓣 = 0.0062

and deceleration is -0.094

Part 2. Find time for traveling for 1 meter with 40 degree angle

Time (Stop watch)

1. 7.16 s

2. 7.22 s

3. 7.43 s

Average time: 7.27 s

Calculation

mgsin𝚹-I a/r -𝓣/r = m*a

(m+I/r^2)*a = m*g*sin𝚹 - 𝓣/r

a = (m*g*sin𝚹 - 𝓣/r) / (m+I/r^2)

a = 0.039571

then we used kinematic equation 

Δx = v*t + 1/2* a*t^2

Δx = 1/2 * a* t^2

1 = 1/2 * (0.039571)*t^2

t = 7.1 second

Conclusion:

In this experiment, we first have to find out inertia for each rotating shape and how big of portion does each shape weight. Then we can find inertia for the total shape when rotating. We know that torque = I * alpha, so we can find frictional torque(because only force acting on it is friction beside gravity) and we can find alpha using camera and LoggerPro. Since we know the inertia and frictional torque and we also know the tension of a cart pulling on a string in a inclined plane. We can use that to find how long does it take to travel certain distance. In this experiment, we make it travel for 1 meter. Furthermore, our calculated time is 7.1 seconds; stop watch time is 7.27 second. In this case, the number is almost identical. The difference might due to static and kinetic friction between wheels and track or human error when stopping stopwatch. At the end, our calculation is pretty decent and close to our prediction.    

Lab#17: Finding Moment of Inertia of Triangle

Finding Moment of Inertia of Uniform Triangle

Samuel
Kyle
John
11-16-2016

Purpose:

To determine the moment of inertia of a right triangle thin plate around its center of mass, for two perpendicular orientation of the triangle.

Theory:

The parallel axis theorem states that I_parallel axis = I_around cm + M(d_parallel axis displacement)^2. Because the limits of integration are simpler if we calculate the moment of inertia around a vertical end of the triangle, you can calculate that moment of inertia and then get I_cm from I_around cm = I_around one vertical end of the triangle - M(parallel axis displacement)^2

Procedure:

1. Use same experimental setup as Rotational Acceleration lab 

2. Mount the triangle on a holder and disk

3. Use string wrapped around a pulley on top of and attached to the disk 

4. The tension in the string exerts a torque on pulley-disk combination.

5. Measured angular acceleration of the system to determine moment of inertia of the system.

6. Mount triangle onto the disk-pulley-holder system and measure α 

7. Determine the I of the new system

Data:

m = 453.6 grams = 0.4536 kg

thickness = 7.7 mm = 0.0077 m

Base = 98.0 mm = 0.098 m

Height = 146.3 mm = 0.1463 m

(Above data is on rotating disk WITHOUT triangle. Since rotating with less moment arm the, it has more force acting on the rotating object. And that's how the slope is getting more steep than the next 2 graphs)   

( Above data is disk rotating with short triangle. Short triangle is with longer bass and shorter height. Since the moment arm is longer than the other 2, therefore the rotating speed will be slower than other 2.) 

(Above is the graph of higher triangle. In this graph, the slope is between no triangle and short triangle. Since, it's higher on the axis of rotation, so it won't affect on the speed of rotation. Therefore, the angular velocity will be faster than shorter triangle.)

Calculation:

Theoretical value : 1/18* M*B^2
Experimental Value : I(around cm) = I(around one vertical end of the triangle) - M(d_parallel axis displacement)^2

Short base
Theoretical value = 1/18 * M*B^2 = (0.4536*0.098^2)/18 = 2.4*10^-4 = 0.000242
Experimental value = m*g*r*(1/alpha(short triangle) - 1/alpha(no triangle)) = 0.025*0.027*9.8*(0.489-0.453) = 0.000238

In this case
m(mass of pulley) = 0.025 kg
r(radius of pulley) = 0.027 m 
alpha = (1.925+1.772)/2 = 1.8485 rad/s^2
B(base of triangle) = 0.1265 m
M(mass of triangle) = 0.4536 kg

The difference between theoretical value (0.000238) and experimental value(0.000242) is 0.000004. Which is about 1% error. The experimental and theoretical value we had calculated is almost identical.

Long Base

Theoretical value = 1/18 * M*B^2 = (0.4536*0.1463^2)/18 = 0.000539 = 5.39*10^-4
Experimental value = m*g*r*(1/alpha(long triangle) - 1/alpha(no triangle)) = 0.025*9.8*0.027*(0.544-0.453) = 0.000601 = 6.01*10^-4 

In this case
m(mass of pulley) = 0.025 kg
r(radius of pulley) = 0.027 m 
alpha = (2.16+1.92)/2 = 2.04 rad/s^2
B(base of triangle) = 0.1463 m
M(mass of triangle) = 0.4536 kg

The value difference in longer base triangle is 5.39*10^-4 and 6.01*10^-4. The error percentage is about 10% difference. We do not know what cause this big different in calculation. The value should have come out quite similar just like short triangle. 

Conclusion: 

In this experiment, we compare the inertia of no triangle, short triangle and long triangle to see the different it make when rotating on the same disk. However, to perform this experiment, we have made a quite few assumption. First, we assume there's no outside environment force acting in the system(no air resistance, constant gravity...). Secondly, there no mechanical error or resisting force(string is rotating not slipping, table is perfectly balance, there's no frictional force on disk or pulley). We then compare our theoretical value with experimental value, the resultant value are quite similar for shorter triangle, but bigger different for longer triangle. I assume some of our measurement are a bit off, and when all the calculation added together, the round off error accumulate into big different. The difference for longer triangle is about 10% error. But at the end, our prediction on rotating inertia of a triangle is close to what it actually turn out. We have used parallel axis theorem to find how much center of mass has shifted and add it with inertia of triangle. Furthermore, the resultant upward alpha(angular acceleration) is similar downward alpha. Which is a good sign that our angular acceleration are constant.  

Lab#16: Angular Acceleration

Angular Acceleration

Samuel
Ellis
Mia
11-2-2016

Purpose:

To use our knowledge about torque and apply it on the rotating object to measure the angular acceleration and its inertia and how the radius affect torque.

Theory:

There is a direct relationship between radius and apply force. We substitute different radius and forces to find their downward and upward speed, and compare it with radius, mass and acceleration. According to our formula, the final answer should come out exactly or somewhere similar.  

Procedure:

1. Measure each of the following to at least three significant figure

    - diameter and mass of top steel disk

    - diameter and mass of bottom steel disk

    - diameter and mass of top aluminum disk

    - diameter and mass smaller torque pulley

    - diameter and mass of larger torque pulley 

    - mass of hanging mass supplied with apparatus

2. Plug the power supply into the Pasco rotational sensor. If there is a cable with the yellow paint or tape, connect only that cable to the Lab Pro at Dig/Sonic 1, so the computer is reading the top disk. If there cables are the same, connect them both. You will need to discern which is measuring the top disk and ignore the other sensor. 

3. Set up the computer. Open LoggerPro. There is no defined sensor for this rotational apparatus so we will need to create something that works with this equipment. Choose Rotary Motion. There are 200 marks on your top disk, so you need to set up equation in the Sensor setting 200 counts per rotation. When you collect data, you can see graphs of angular position, angular velocity and angular acceleration vs. time. The graph of angular acceleration vs. time is useless due to poor timing resolution of the sensors. 

4. Make sure the hose clamp on the bottom is open so that the bottom disk will rotate independently of the top disk when the drop pin is in place. 

5. Turn on the compressed air so that the disks can rotate separately. You will not so much air that you pop the hose form the air source, but enough to keep things smooth. Set the disks spinning freely to test the equipment. 

6. With the string wrapped around the torque pulley and the hanging mass its highest point, start the measurements and release the mass. Use the graphs of angular velocity to measure the angular acceleration as the mass moves down and up. 

(Measure hanging string does not touch the edge of table and make sure clean disk with alcohol before started experiment) 

Data:

hanging mass- measure by electronic scale

α(down, up) - measured by LoggerPro 

α_average - measured by me

EXPT 1 -- hanging mass only - 24.9 grams, small torque pulley, Top steel, α(down, up)= 1.098,1.205(rad/sec^2), α_average= 1.151

EXPT 2-- 2 x hanging mass - 49.9 grans, small torque pulley, Top steel, α(down, up) = 2.206, 2.388(rad/sec^2), α_average = 2.297

EXPT 3 -- 3 x hanging mass - 74.9 grams, small torque pulley, Top steel, α(down, up) = 3.315,3.557(rad/sec^2), α_average = 6.872

EXPT 4 -- hanging mass only - 24.9 grams, large torque pulley, Top steel, α(down, up) = 2.125,2.343(rad/sec^2), α_average = 2.233

EXPT 5 -- hanging mass only - 24.9 grams, large torque pulley, Top aluminum, α(down,up) = 5.927,6.638(rad/sec^2), α_average = 6.2825

EXPT 6 -- hanging mass only - 24.9 grams, large torque pulley, Top steel + bottom steel, α(down, up) = 1.069,1.777(rad/sec^2), α_average = 2.246

Measured Data:

Top steel disk: 126.3 cm, 1353.3 grams

Bottom Steel disk: 126.3 cm, 1345.8 grams

Top aluminum disk : 126.3 cm, 465.8 grams

Small torque pulley: 27.8 cm, 10.0 grams

Large torque pulley: 53.2 cm, 36.3 grams

Graph and Calculated Data:

(Above, is the Angle vs Time(top graph) and Velocity vs Time(bottom graph). For Angle vs Time, you can see that as time increase, the angle of the disk is also increase. In addition, instead of increasing linearly, the line is more a curve like line. The reason behind that is because the velocity is changing constantly, since the disk is constantly changing its rotation and hanging mass gives acceleration to the disk. Therefore, it create a curve like graph. For Velocity vs Time graph, the disk is changing its direction, that's how it get those up and down line.)

Analysis:

According to the result from our data, there should have some kind of relationship from different hanging mass, rotating mass and radius. 

EXPTS. 1, 2, and 3: Effect of changing the hanging mass

The difference between 1, 2, and 3 is that the hanging mass is increase by 20 grams. The resultant angular acceleration is quite different. 

EXPT 1 = 1.152 rad/s^2, mass = 24.9 grams

EXPT 2 = 2.297 rad/s^2, mass = 49.9 grams

EXPT 3 = 3.436 rad/s^2, mass = 74.9 grams

Result, angular acceleration increase as hanging mass increase.

EXPT 1 and 4: Effect of changing the radius and which the hanging mass exerts a torque

With same amount of hanging mass and disk, we used different torque pulley. 

EXPT 1 (small torque pulley)= 1.152 rad/s^2, mass = 24.9 grams

EXPT 4 (large torque pulley) = 2.233 rad/s^2, mass = 24.9 grams

EXPT 4, 5, 6: Effect of changing the rotating mass

With same amount of hanging mass and same length of torque pulley, we changed the rotating mass(same radius but different mass)

Steel disk: 1353.3 grams

Aluminum disk: 465.8 grams

EXPT 4 = Top steel, 2.233 rad/s^2

EXPT 5 = Top aluminum, 6.2825 rad/s^2

EXPT 6 = Top steel + bottom steel, 1.123 rad/s^2

Part 2 

We compare theoretical value to experimental value ( it should come out close or similar value) by using I_disk = (mgr)/alpha - mr^2. When using this equation, we assumed there's no friction between two disk and string on pulley. 

Experimental value = (mgr)/alpha - mr^2

Theoretical value = 1/2* mr^2

EXPT 1

Experimental value = 0.00294 --> ((0.0139*9.8*0.0249)/1.1515) - 0.0139^2 * 0.0249

Theoretical value = 0.00269 -->  1/2 * 1.353*(0.1263/2)^2

EXPT 2

Experimental value = 0.00295 --> (0.0139*9.8*0.0499)/2.297 - 00139^2 * 0.0499

Theoretical value = 0.00269 -->  1/2 * 1.353*(0.1263/2)^2

EXPT 3
Experimental value = 0.00295 --> (0.0139*9.8*0.0749)/3.436 - 0.0139^2 *0.0749
Theoretical value = 0.00269 -->  1/2 * 1.353*(0.1263/2)^2

EXPT 4
Experimental value = 0.00288 --> (0.0266*9.8*0.0249)/2.233 - 0.0266^2 *0.0249
Theoretical value = 0.00269 --> 1/2 * 1.353*(0.1263/2)^2

EXPT 5
Experimental value = 0.0010 --> (0.0266*9.8*0.0249)/6.2825 - 0.0266^2 *0.0249
Theoretical value = 0.0011-->1/2* 0.4658* (0.1363/2)^2

Conclusion:

As we observed and analyzed the relationship between pulley radius, hanging mass and disk to angular acceleration, we found out that they all related to each other. When we increase hanging mass but pulley and disk remain the same, our angular acceleration increase. In addition, when we change small pulley to bigger pulley, the angular acceleration is also increase. But when we combined to steel disk with steel disk, the resultant acceleration become really. The reason behind that is because steel disk is heavier than regular disk that we performed previous experiment. When heavier object spin with each other, they will have more frictional force acting between them. When frictional force increase, the angular acceleration decrease, that's why our resultant alpha is the smallest out of 6 experiment.