wkst_RotationalMotion

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Western Michigan University *

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113

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Mechanical Engineering

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Dec 6, 2023

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Lab Worksheet PS113L – Rotational Dynamics Name: Part 1 — Rotational Inertia Middle Radius of Axle Pulley (cm): 1.65 Axle Pulley Mass (g): 7 Outer Radius of Axle Pulley (cm): 2.7 Plate Mass (g): 65 Plate side a (cm): 13 Plate side b (cm): 12.5 Lg Ring Mass (g): 466 Lg Ring R outer (cm): 6 Lg Ring R inner (cm): 5.3 Table 1: Angular acceleration of objects around a central axis due to string tension from a falling mass. † Friction mass is the amount of hanging mass needed to spin the object at constant angular velocity (net torque = zero, acceleration = zero). Under this condition the torque from friction is equal and opposite the torque from the hanging ‘friction mass’. Trial Hanging Mass (g) Angular Acceleration α (rad/s 2 ) Friction Mass † (g) Final Angular Velocity (rad/s) Lg Ring + Plate 28 ± 2.27 ±0.00 51 2 1.962 Plate + Sensor 28 ± 19.9 ±0.13 2 8.3 Sensor only 20 ± 245 ±3.4 222.6 For all calculations assume g=9.79264m/s 2 . i 1. Using the Torque, Force, and angular acceleration equations from the lab manual Theory section, derive an equation for Moment of Inertia I of any object that is based only on directly measured variables: hanging mass, pulley radius, and angular acceleration. Show the steps of your derivation. I= gr(m-mf)/ α-(m-mf)r^2 2. Using your equation from above, calculate the experimental value of the rotational inertia of the Ring + Plate + Sensor together. Remember to subtract off the mass required to overcome friction (friction mass) from the hanging mass. a. Experimental Rotational Inertia of (Lg Ring + Plate + Sensor): M=2g r= 5.3g g=9.79264 α= 2.27 rad/s^2 I large ring+plate = 2/1000*5.3/100(9.79264/2.27-5.3/100)=0.000451659kgm^2 i Surface Gravity Prediction made available by the National Geodetic Survey (NGS), an office of the National Oceanic and Atmospheric Administration (NOAA). More NGS tools and calculators can be found at the NGS Geodetic Tool Kit Program .

Lab Worksheet PS113L – Rotational Dynamics M=20g α=245 I sensor = 20/1000*5.3/100(9.79264/245-5.3/100)=-1.38118*10^-5 I lg ring+ plate + sensor = 0.000437847kgm^2 3. Calculate the experimental value of the rotational inertia of the Plate + Sensor together. Remember to subtract off the mass required to overcome friction from the hanging mass. a. Experimental Rotational Inertia of (Plate + Sensor): α=19.9 9.79264*5.3cm(26g)/19.9-26g*19.9^2 =0.007519kgm^2 4. Calculate the experimental value of the rotational inertia of the Rotary Motion Sensor alone. a. Experimental Rotational Inertia of (Sensor): M=20g α=245 I sensor = 20/1000*5.3/100(9.79264/245-5.3/100)=-1.38118*10^-5 -0.0000138118 kgm^2 5. Subtract the rotational inertia of the Rotary Motion Sensor from the total rotational inertia of combination of the (Plate + Sensor). This will be the rotational inertia of the Plate alone. a. Experimental Rotational Inertia of (Plate): 0.007519+0.0000138118=0.00753081 kgm^2 6. Subtract the rotational inertia of the combination of the (Plate + Sensor) from the total rotational inertia of the combination of the (Ring + Plate + Sensor). This will be the rotational inertia of the Ring alone. a. Experimental Rotational Inertia of (Lg Ring): 0.000437847-0.007519=-0.00707916 kgm^2 7. Calculate the individual theoretical values of the rotational inertias of the ring and plate using Table 1 in the lab manual Theory Section and your measurements of their masses and geometries. a. Theoretical Rotational Inertia Lg Ring: ½*.466(.06^2+.053^2)=0.0014933 kgm^2 -2-

Lab Worksheet PS113L – Rotational Dynamics b. Theoretical Rotational Inertia Plate: ½*.065(0.13^2+0.125^2)=0.00105706 kgm^2 -3-

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Lab Worksheet PS113L – Rotational Dynamics 8. Complete Table 2 below. Use percent differences to compare the moment of inertia experimental values to the theoretical values. %difference = Experimental − Theoretical Theoretical × 100% Step 5 Acc. 1.09 0.308 19.4 Table 2: Summary Moments of Inertia data from angular acceleration experiments Object Experimental Rotational Inertia (gcm 2 ) Theoretical Rotational Inertia (gcm 2 ) Rotational Inertia % Difference Lg Ring + Plate 0.01460997 0.0014933 8.75% Plate + Sensor 0.00754462 0.00105706 6.14% Sensor only -0.0000138118 9. Which single object had the greatest rotational inertia? Without changing the type of object (plate, ring, etc) or its total mass, what could you change to make its moment of inertia smaller? Ring and plate. Changing the shape of object or axis of rotation could reduce the moment of inertia. 10.Which object was hardest to accelerate? Discuss how an object’s moment of inertia affects how difficult it is to accelerate. Sensor only. Since it has the least mass. 11.Why did we need to measure the friction in the system separately for each set of objects? Why could we not just measure friction once and use the same value for any object? For accurate predictions of the motion. -4-

Lab Worksheet PS113L – Rotational Dynamics -5-

Lab Worksheet PS113L – Rotational Dynamics Part 2 — Rotational Collisions & Conservation of Angular Momentum Pre-experiment Questions 12.After the ring collides with the plate, will the final angular speed be more than, less than, or the same as the initial angular speed of the plate? Explain your reasoning. Angular speed of the plate after collision would be less than initial speed since energy would be lost after collision. 13.After the ring collides with the plate, will the final angular momentum be more than, less than, or the same as the initial angular momentum of the plate? Explain your reasoning. Final angular momentum of the plate would be less than initial angular momentum. The initial angular momentum would be transferred to the ring after collision, resulting in decrease in momentum. 14.Do you think this will be an elastic or inelastic collision? What may happen to the rotational kinetic energy of the system? Inelastic collision. The final rotational kinetic energy of the system would be less than its initial. Data Axle Pulley Mass (g): 7 Outer Radius of Axle Pulley (cm): 2.7 Plate Mass (g): 65 Plate side a (cm): 13 Plate side b (cm): 12.5 Ring Mass (g): 466.5 Ring R outer (cm): 3.85 Ring R inner (cm): 2.75 Axle Pulley Inertia (gcm 2 ): Plate Inertia (gcm 2 ): 1736 Ring Inertia (gcm 2 ): 17475 Table 3: Angular collision data between a dropped ring and an initially rotating plate. Trial Initial Rotation al Inertia (gcm 2 ) Final Rotation al Inertia (gcm 2 ) Ring offset x (cm) Initial Angula r Velocit y (rad/s) Final Angula r Velocit y (rad/s) Initial Angular Momentu m (gcm 2 /s) Final Angular Momentu m (gcm 2 /s) % Difference Momentu m 1 1736 17767 1.4 25.853 6.709 4865.13 131465.3 -0.962% 2 1736 17601 0.5 26.697 7.069 5046.31 135462.8 -0.962% 3 1736 17716 1.4 29.184 7.343 5031.42 136541.1 -0.963% 4 1736 17506 0.4 25.499 6.8 5014.38 134487.4 -0.963% -6-

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Lab Worksheet PS113L – Rotational Dynamics Table 4: Kinetic Energy before and after an angular collision between a dropped ring and rotating plate. Trial Initial Kinetic Energy (gcm 2 /s 2 ) Final Kinetic Energy (gcm 2 /s 2 ) % Difference Kinetic Energy 1 726153.8 486213.7 0.493% 2 713546.3 513214.8 0.390% 3 734762.6 523476.7 0.404% 4 739084.0 531486.9 0.390% Post-experiment Questions 15.In class, the Pasco Capstone file calculated the Inertia’s for you for each situation. Show that you can correctly calculate the Initial and Final rotational Inertias for Trial 1 of the Rotational Collisions based on the initial measurements of each object. Ii=1/2*0.065g(0.13^2+0.125^2)+1/2*0.007*0.027^2=0.00105961 If=0.00105961+1/2*0.4665(0.0385^2+0.0275)+0.4665*0.014^2=0.00106023 16.What affect should each of the following have on the value you calculate for the final angular momentum? State if each would cause the real final angular momentum to be lower, higher, or unchanged from the value calculated in Table 3 and explain why . a. If the Rotary Motion Sensor itself (like the center axle) has a small rotational inertia (in addition to the axle pulley)? Angular momentum would remain constant then. After angular momentum is conserved, the moment of inertia increases, which causes the angular speed to decrease. b. If the frictional drag on the bearings during the collision cannot be ignored? Frictional drag is action reaction force. 17.Explain: Does the experimental result support the Law of Conservation of Angular Momentum? Initial and final angular momentum are found different, so the experiment did not support the law of conservation of angular momentum -7- Use the space below to show calculations for Trial 1’s angular momenta and kinetic energies. %Difference should be (final – initial). 0.058495% difference

Lab Worksheet PS113L – Rotational Dynamics 18.Was Kinetic Energy conserved in the collisions? Explain how you know. Kinetic energy was lost on final kinetic energy so kinetic energy is not conserved. 19.Discuss why you might expect to see a negative %Difference for momentum, but be very surprised to see a positive percent difference. According to the principle of momentum conservation, a system's overall momentum will remain constant in the absence of outside stimuli. Any variation from the calculated value of momentum is therefore most likely the result of outside forces or other elements that were not taken into consideration. In the absence of any external causes that may create such a force, a positive % difference would indicate the presence of a net external force, which would be startling and unexpected. 20.How can angular momentum be conserved, but energy not be conserved? When there are no external torques operating on a closed system, angular momentum is conserved, which means the system's overall angular momentum stays constant. However, energy can change forms and isn't always conserved because of things like friction, air resistance, and other energy losses. As a result, while energy cannot be preserved, angular momentum may. -8-