Angular Momentum Lab - Lab report, received letter grade A PDF

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Lab report, received letter grade A...

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STONY BROOK UNIVERSITY DEPARTMENT OF PHYSICS AND ASTRONOMY PHYSICS 133 SECTION 12

Angular Momentum

Experiment Date: 4/10/2020 Report Date: 4/19/2020

Introduction: In this angular momentum lab, we are tasked with studying the principle of conservation of angular momentum. To perform this lab, we obtained a rotating platform to determine the angular momentum. We begin the lab by using the photogate and find the angular acceleration to measure the frictional torque, α fr . Secondly, we measured the radius of the cylinder in the center of the rotating platform. Next, we would attach a hanging mass and measure the net angular acceleration. Once we determined all this data using the photogate, we were able to calculate the moment of inertia for the platform, I

platform

for the first part. To complete the

second part, we determined the radius and mass of the disk and while using the hanging mass, we dropped the metal disk on the platform to determine its effect on angular momentum. Next, we dropped the plate and measured the change in angular velocity, ω . We then calculate the moment of inertia of the plate that was dropped and find the total moment of inertia. We use the angular velocities and the total moment of inertia to determine the angular momentum.

Theory: (Derivation of Moment of Inertia) To derive the moment of inertia equation, I =

mr(g−rα net ) α f r+α net

from the sum of torques and the sum of

forces equations, we must start with torques by using the equation, τ We can determine that τ

fr =

equation then becomes τ

ext

I α fr and that τ

ext =

net

=τ

ext

−τ

fr

= I α net .

Fr=mr(g-a). I n this situation, a=ra , and the

=mr(g- rα net ).  After determining this, we can use the previous

equations to find that mr(g-r α net )- I α fr = I α net . Once, we rearrange the equation to get a simpler equation we get mr(g − rα net ) = I (α net + α fr ) , we can end up with I =

mr(g−rα net ) . α f r+α net

Calculations: See Attached Excel Sheet

Results: Once we did all the calculations, as seen on the excel sheet, we calculated a value of 0.919 +/0.011 k g · m 2 /s for the initial angular momentum. As for our final angular momentum, we received a value of 0.632 +/-0.009 k g · m 2 /s . The angular momentum was not conserved within the uncertainty.

Error Analysis: (Handle of the Disk) The handle of the disk was not taken into account in this lab and if it was, we would have ended up with a greater moment of inertia since the moment of inertia for the whole system is the sum of moment of inertia of the handle and the disk. Overall, this would make a larger total moment of inertia for the system, thus decreasing the angular velocity of the system and also a decreased angular momentum. (Off-Center Drop) A systematic error we could have is an off center drop due to human error. If the disk is dropped off center, the axes that the disks rotate around would be parallel and the parallel axis theorem can be applied. This parallel axis theorem states that the distance between the axes equal to h  and the equation to find the moment of inertia would be I = 21M R

2

+ Mh 2 . This would increase the

moment of inertia and as a cause will decrease both the angular velocity and the angular momentum.

Conclusion: Our final results were 0.919 +/- 0.011 k g · m 2 /s for the initial angular momentum and 0.632 +/-0.009 k g · m 2 /s for our final angular momentum. The angular momentum was not conserved within the uncertainty. The final angular momentum was decreased greatly from our initial angular momentum and this could be due to both of the systemic errors stated earlier. The final results could have been better, but due to the possible errors, I am satisfied with my results....