18-May 2015 Parallel Axis Theorm

Purpose: To determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle, and the "holder."

Theory: The Parallel axis theorem states that the moment of inertia (I) around a parallel axis (Ipa) is equal to the moment of inertia around the center of mass plus the mass of the object multiplied by the displacement from the parallel axis (d) squared. By determining the moment of inertia of the disk, pulley, and "holder" we can subtract that from the moment of inertia calculated when we have the triangle included to fine the moment of inertia of just the triangle. If we compare the two calculations they should be fairly close with minimal difference error

Procedure: Have the apparatus spin without the triangle, but with the holder, to discover the moment of inertia of the system (use a 25 gram hanging mass). Average three angular accelerations (both up and down) and use the formula I= (Mass(hanging)*gravity*radius/avg angular acceleration)-Mass*radius^2 to determine the moment of inertia. Run the experiment twice more, with the triangle included, alternating the base and height

Ex: Triangle Max Height (Left) Triangle Max Length (Right)

Calculate this moment of inertia, and then subtract the previous calculated moment of inertia to find the moment of inertia of just the triangle. Finally calculate the moment of inertia of the triangle with the parallel axis theorem (Ipa=Icm+md^2), and compare it with your previous calculation.

Apparatus

Graph of angular momentum.

Results of Angular acceleration

	1 rad/s^2	2 rad/s^2	3 rad/s^2	Average
No Triangle up	-3.421	-3.416	-3.475
No Triangle down	3.059	3.038	3.081	3.248333
Triangle Max height up	-2.803	-2.824	-2.803
Triangle Max height down	2.514	2.488	2.514	2.657667
Triangle Max Length up	-2.156	-2.137	-2.175
Triangle Max Length down	1.914	1.912	1.92	2.035667

*Moment of inertia no triangle, 0.000939 kg*m

Sample calculation for Moment of inertia Triangle

*Moment of inertia of Triangle Max height, 0.00021 kg*m

*Moment of inertia of Triangle Max length, 0.000562 kg*m

Dimensions of the triangle

*Since the center of mass of a right triangle is always 1/3 of the base from the 90 degree angle, there is no need to calculate center of mass.

Sample calculation for Moment of inertia Triangle Using parallel axis theorem

*Triangle max height 0.000005056 kg*m, error 97%

*Triangle max length 0.000001137 kg*m, error 99%

Conclusion 

*Because of the large error, I believe I must have measured something wrong.

*Perhaps I was mistaken in my assumption that the distance from the center of mass was 1/3 of the base, or where the point of rotation was on the triangle (again a mistake in measurement).

*I did not take into account friction.

4-May 2015 Angular Acceleration

Purpose: Part 1: Draw some numerically specific conclusions about the effect on the acceleration of the system of increasing the hanging mass, the radius of the torque pulley, and the rotating mass. Part 2: Determine the experimenta values for the moments of inertia of the disks (or disk combinations) from the data of part 1.

Theory: Part one that a change in mass of the pully, hanging or spinning disk, as well as the radius of the pulley will effect to angular acceleration. Part 2 the moment of inertia for the apparatus can be determined using the data collected in Part 1

Procedure: Determine the mass and radius' of the various disks and pulleys of the apparatus, get the apparatus to spin with both disks and just one, attach a pulley with a string to the apparatus and tie the other end to a known mass, allow the mass to "bob" up and down from the apparatus, and measure for angular acceleration. Compare results for Part 1 and calculate the moments of inertia for Part 2.

Apparatus

Measurements

*The radius and mass of the top steel disk, 6.323 cm and 1361 gams respectively.

*The radius and mass of the bottom steel disk, 6.332 cm and 1348 gams respectively.

*The radius and mass of the top aluminum disk, 6.330 cm and 466 gams respectively.

*The radius and mass of the small pulley, 2.5 cm and 10 gams respectively.

*The radius and mass of the large pulley, 5 cm and 37 gams respectively.

*The hanging mass 25 grams.

Acceleration Graph (lower)

Data

Expt #		grams hanging	Pulley	Disk	rad/s^2 down	rad/s^2 up	average rad/s^2
1	hanging mass	25	small	top steel	0.6026	-0.6522	0.6274
2	hanging mass*2	50	small	top steel	1.207	-1.311	1.259
3	hanging mass*3	75	small	top steel	1.758	-1.949	1.8535
4	hanging mass	25	large	top steel	1.161	-1.272	1.2165
5	hanging mass	25	large	top Al	3.293	-3.595	3.444
6	hanging mass	25	large	Top+ Bottom steel	0.5888	-0.6408	0.6148

Graph of Velocity of spinning disk and hanging mass

"Calculations" Part 1

*Since the pivot is at the center of mass of the apparatus and the acceleration is constant, we know angular acceleration= acceleration/radius of pulley, so the smaller the radius of the pulley the larger angular acceleration.

*Greater tension on the sting should equal a larger torque, and since torque= moment of inertia*angular acceleration, we should conclude that a greater hanging mass (therefore greater tension in the string) should cause higher torque value which would mean a higher angular acceleration.

*From the Velocity graph's we can determine the acceleration and angular acceleration from the falling mass and apparatus respectively from looking at the slopes.

*The acceleration of the mass is 0.02874 m/s^2, and the angular acceleration 0.6124 rad/s^2, while using the large pulley (0.05 m radius). So Angular acceleration equals acceleration divided by the radius(0.02874/0.05), and we get 0.5748 rad/s^2

Conclusion Part 1.

 *The first "calculation" isn't confirmed from the data, when a trail was conducted using the same mass but different radius pulley, the greater radius produced a higher angular acceleration. This was not expected. I believe something must have gone wrong, but I don't know what.

*The second "calculation" was confirmed by the data, as a larger hanging mass did result in a greater angular acceleration.

*The final calculation was "close" to the actual measured angular acceleration, so that confirms the hanging mass' acceleration is the same to the tangential acceleration of the apparatus.

Calculation Part 2

 *Sample

*Results

calculated from Data

0.004663131

0.004642941

0.004870745

0.005034936

0.001448455

-0.000735616

 Conclusion Part 2

*My numbers were far off, I can only assume serious error occurred during measurements of the pulley's and disks, therefore none of my data is useful.

*Ideally the moment of inertia should have matched with the calculations, but they did not.

*Also my calculations didn't take into account friction in the system

6-May 2015 Moment of Inertia and Frictional Torque

Purpose: Predicting, with less than 4% error, how long it would take for a 503 gram dynamics cart to roll down a 1 meter of sloped track, while it is tied to an apparatus made of three cylinders that slows its descent.

Theory: Because the apparatus is made of three cylinders, and we know the total mass of the apparatus, the moment of inertia can be determined with careful measurements of the length and radius of the three cylinders to determine the volume(V=Length*Pi*Radius^2) of each individual cylinder. We can determine the mass of each cylinder with this information, and therefore calculate the total moment of inertia (the moment of inertia for a cylinder is =0.5*Mass*Radius^2) by adding the individual moments of inertia of each cylinder. The angular acceleration can be determined by finding the angular velocity (omega=W), which can be found by determining the velocity of the apparatus as it spins. With angular acceleration and the moment of inertia we can determine the torque on the apparatus (Torque=Inertia*Angular acceleration), due to the force of the cart pulling on the apparatus as it descends (there will be friction), with these now known forces (using force diagrams) determine the acceleration of the cart, and use the kinematics equation (Distance=0.5*acceleration*time^2) to calculate the time it would take for the cart to travel down the sloped track with less than 4% error.

Procedure: Determine the volume of the apparatus by taking measurements, realize the percentage of volume will correlate with percentage of mass of each cylinder of the apparatus. Calculate the moment of inertia for the entire apparatus as I_total= I_1+I_2 +I_3. Use logger pro to determine the angular acceleration of the apparatus as it slows. Slant a 1 meter track against the table, place a 503 gram cart at the top of the track, and tie a string from the cart around a cylinder on the apparatus. Time how long it takes for the cart to descend 1 meter and compare the result with the calculated time that your measurements suggest the time would be.


Apparatus

Measurements of Apparatus

*Total mass 4808 grams

*Cylinder #1: Length=5.0 +/- 0.01 cm, Radius= 1.55+/- 0.01 cm, Volume=  37.7 cm^3

*Cylinder #2: Length=5.15 +/- 0.01 cm, Radius= 1.55+/- 0.01 cm, Volume=  38.9 cm^3

*Cylinder #3: Length=1.55 +/- 0.01 cm, Radius= 10.025+/- 0.01 cm, Volume=  489.38 cm^3

* Total Volume 565.98 cm^3

* Percentage of Volume of each cylinder to the total should be the equivalent of percentage of mass.


                                      Volume percentage= Volume individual/Volume total*100%

                                           Mass individual= (Volume percentage/100) *Total mass


*Cylinder #1 percentage of volume 6.66%, so mass equals 320.3 grams

*Cylinder #2 percentage of volume 6.87%, so mass equals 330.4 grams

*Cylinder #3 percentage of volume 86.5%, so mass equals 4157.3 grams

Caption of Apparatus as it decelerates using logger pro

Graph of X,Y Velocity and Omega vs time

*Omega can be calculated as = sqr(Velocity_x^2 +Velocity_y^2)

*Once graphed as Omega vs time, the slope will give us angular acceleration (a_a)

*Angular acceleration (a_a) is -0.4063 m/s^2

The Track and Force Diagrams

*Angle track makes with the horizontal is 47.5 degrees

*Force diagram on track

*Actual time it takes for cart to reach end of 1 meter track= 7.16 seconds.

Calculations

*Total of moment of inertia

*Calculating time

*Calculated time was 6.98 seconds

*Difference from actual (Actual time-Calculated time/Actual time *100%) was 2.93%

Conclusion

*Since we fell within the necessary 4% error margin, I say the experiment confirmed my methods of determining the moment of inertia of the apparatus and its angular acceleration.

*I used difference percentage, instead of propagated error, perhaps my result would've have been more accurate had I used this method, but I was satisfied with the 2.93% of the difference.

*Although the calculations took frictional torque into account, it did not take friction from air resistance into account or friction from the track.

15-April 2015 Collisions in Two Dimensions

Purpose: To show that momentum and energy are conserved during collisions, using the known formulas.

Theory: That a known mass going at a known velocity, colliding with another known mass, setting it and the original mass at other different but known velocities, proves that momentum is conserved (M_a*V_ao + M_b*V_bo = M_a*V_af + M_b*V_bf). Energy will also be conserved as the original kinetic energy from the first mass collides with another mass. Kinetic energy initial will equal the kinetic energy final of the new velocities of the original mass and the one it collided with (0.5*M_a*V_ao^2 = 0.5*M_a*V_af^2 + 0.5*M_b*V_bf^2).

Procedure: Set up a flat level surface, with camera to capture the collision in video, get three marbles (two steel and one aluminum), weigh for mass for each marble, set one marble in the center and roll another marble to collide with it, do two collisions (one steel marble colliding with the other steel marble and one steel marble colliding with the aluminum marble), use the video capture program to determine the velocity of the marbles before and after collision, and graph the results. Preform calculations to confirm energy and momentum are conserved.

Apparatus

*Video screen shot with Cartesian grid and a reference "bar" to determine the velocity of the marbles

Graph of two steel marbles colliding as Position in the x, y coordinates vs time

Graph of steel marble and aluminum marble colliding as Position in the x, y coordinates vs time

*The computer can also determine the velocity (in the x,y direction) of the marbles.

*We can take the average velocity in the x,y direction of each marble, before and after, the collision as well.


Graph of two steel marbles colliding as Vax, Vay, and Vbx vs time

Results of steel marble and aluminum

*Mass of steel marble 28.1 grams

*Mass of aluminum marble 10.1 grams

*Original Velocity in the x direction of rolling steel marble = 0.5961 m/s

*Original Velocity in the y direction of rolling steel marble = 0.005765 m/s

*Original Velocity in the x direction of aluminum marble = 0 m/s

*Original Velocity in the y direction of aluminum marble = 0 m/s

*Final Velocity in the x direction of rolling steel marble = 0.5622 m/s

*Final Velocity in the y direction of rolling steel marble = -0.08335 m/s

*Final Velocity in the x direction of aluminum marble = 0.1062 m/s

*Final Velocity in the y direction of aluminum marble = 0.1825 m/s

Results of two steel marbles

*Mass of both steel marbles 28.1 grams

*Original Velocity in the x direction of rolling steel marble = 0.4817 m/s

*Original Velocity in the y direction of rolling steel marble = 0.02122 m/s

*Original Velocity in the x direction of stationary steel marble = 0 m/s

*Original Velocity in the y direction of stationary steel marble = 0 m/s

*Final Velocity in the x direction of rolling steel marble = 0.2655 m/s

*Final Velocity in the y direction of rolling steel marble = -0.1624 m/s

*Final Velocity in the x direction of stationary steel marble = 0.2020 m/s

*Final Velocity in the y direction of stationary steel marble = 0.1360 m/s

Finding actual velocity

*We must remember that actual velocity can be found using the Pythagorean theorem

*So actual velocity can be found as (Actual velocity= sqr(Vx^2 + Vy^2))

Velocities of steel marble and aluminum

*Original Velocity of rolling steel marble = 0.5961 m/s

*Original Velocity of aluminum marble = 0 m/s

*Final Velocity of rolling steel marble = 0.5683 m/s

*Final Velocity of aluminum marble = 0.2135 m/s


Velocities of two steel marbles

*Original Velocity of rolling steel marble = 0.48211 m/s

*Original Velocity of stationary steel marble= 0 m/s

*Final Velocity of rolling steel marble = 0.3112 m/s

*Final Velocity of stationary steel marble = 0.2435 m/s


 Calculations of Momentum sample

*We set them equal to each other, because the amount of momentum "going in" should be the same as the momentum "going out" from the collision.

Results Momentum

*First Row Two steel

*Second Row steel and aluminum

Initial	Final	Error
0.0135473	0.015587	15.05673
0.0167504	0.018126	8.209769

 Calculations of Energy sample

*We set them equal to each other, because the amount of energy "going in" should be the same as the energy "going out" from the collision.

Results Energy

*First Row Two steel

*Second Row steel and aluminum

Initial	Final	Error
0.009985	0.009357	-6.28943
0.0049925	0.004768	-4.49904

Conclusion

*It would seem the conservation of momentum of and energy has been confirmed (the numbers are close), though the amount of error is a little troubling

*Most likely because of how much rounding was required throughout the calculations, more and more rounding error was accumulated.

*Not only that but the trials don't account for energy lost due to friction(no matter how smooth the surface was) or the production of sound and heat from the collision.

*The positions of the marble before and after collision the camera used to determine velocity was manually placed, the points were placed frame by frame by where they were seen, this could have made the velocities less accurate.

27-April 2015 Balistic Pendulem

Purpose: To demonstrate the conservation of momentum and energy principle, using the known formulas, and using propagated uncertainty will account for any error in the results.

Theory: The formulas of conservation of momentum (Mass original*Velocity original= Mass final*Velocity final) and the conservation of energy (Kinetic energy initial= Potential energy due to gravity) can be used to find the velocity of a steel ball as it collides into a ballistic pendulum, and any error will be accounted for using propagated uncertainty.

Procedure: Take measurements of the ballistic pendulum apparatus, weigh steel ball, give the ball an initial velocity so it collides with the ballistic pendulum, preform calculations for this initial velocity using the measurements and account for any error.

Apparatus

Measurements

*Steel Ball mass 0.00767 kg +/- 0.0001 kg

*Ballistic Pendulum mass 0.0809 kg +/- 0.0001 kg

*Angle formed from pendulum after collision 0.314 rad +/- 0.00873 rad

*Length of Pendulum 0.02 meters +/-  0.0005 meters

Calculations

* 0.5*M_total*V_final^2= M_total*g*H

*H is calculated using trigonometry

*H=L-L cos (angle)

* Mass' cancel out and

                                                                  0.5*V_f^2=g*H

                                                                   V_f=sqr(2*g*H)

                                                                    V_f=sqr(2*9.8*(0.02-0.02 cos 0.314))

                                                                    V_f= 0.138 m/s

*Mass_ball*V_i=Mass_total*V_f

                                                             V_i=(Mass_total*V_f)/Mass_ball

                                                             V_i=(0.00767+0.0809)*0.138/0.00767

                                                             V_i= 1.59 m/s

*Propagated error

                                                dv= ((M_b+M_p)*sqr(2*g*L-L cos (angle))/M_b

                                               dv/dM_b=sqr(2*g*L-L cos (angle))/(-M_b^2) *dM_b

                                               dv/dM_p=(M_b+1/M_b)*sqr(2*g*L-L cos (angle))*dM_p

                                               dv/dL= ((M_b+M_p)/M_b)*sqr(2*g*1-cos (angle))/2sqr(L) *dL

                                 dv/dang= ((M_b+M_p)/M_b)*sqr(2*g*L))*(sin (angle)/2sqr(1-cos angle) *dang


*Add all the values and dv=+/-0.301 m/s

*Initial velocity is 1.59 m/s +/- 0.301 m/s

Conclusion

*The initial velocity of the steel ball as it collided with the pendulum was 1.59 m/s giver or take 0.301 m/s.