PHY 221 Lab Report 7 PDF

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PHY 221 Lab Report 7...

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Sarah Jenkins, Vipsa Patel, and Summer Smith Section-3, March 1, 2016 Studio Session 7 March 1, 2016 Objective: To be able to interpret and describe rotational motion and translational motion of extended objects. To evaluate angular acceleration relation Scientific principles behind the experiment: To evaluate relationships between angular acceleration, moment of inertia, angular momentum, and torque. Data/Evidence: Experiment 1: Make the following measurements and record the data in your log: time it takes to complete 5 rotations: 11.25 seconds distance from the shoulder to the elbow:

0.030 meters

distance from the shoulder to the middle of the hand:

0.067 meters

Motion of the hand: How far in degrees did the hand travel during the five rotations? 360 * 5 = 1800 total degrees How far in radians did the hand travel during the five rotations? 1800 * pi / 180 = 31.42 radians How far in meters did the hand travel during the five rotations? 0.067 * 2= 0.134 m (diameter) 0.134 * pi=0.421 m (circumference) 0.421 * 5 = 2.10 m What was the average angular speed (deg/s and rad/s) of the hand? 31.42 rad / 11.25 s = 2.79 rad/s 1800 deg / 11.25 s = 160 deg/s What was the average linear speed (m/s) of the hand? 2.10 m /11.25 s = 0.1867 m/s What was the average angular acceleration (deg/s2 and rad/s2) of the hand? How do you know? 2.79 rad/s / 11.25 s = 0.2482 rad/s2

160 deg/s / 11.25 s = 14.22 deg/s2 What was the average centripetal acceleration (m/s2) of the hand? (0.1867 m/s)2 / 0.067 m = 0.52 m/s2 Motion of the elbow: How far in degrees did the elbow travel during the five rotations? 360 * 5 =1800 degrees How far in radians did the elbow travel during the five rotations? 1800 * pi / 180 = 31.42 radians How far in meters did the elbow travel during the five rotations? 0.030 * 2 = 0.06 m (diameter) 0.06 * pi = .1885 m (circumference) .1885 * 5 = 0.9425 m What was the average angular speed (deg/s and rad/s) of the elbow? 31.42 rad / 11.25 s = 2.79 rad/s 1800 deg / 11.25 s = 160 deg/s What was the average linear speed (m/s) of the elbow? 0.9425 m / 11.25 s = 0.08378 m/s What was the average angular acceleration (deg/s2 and rad/s2) of the elbow? How do you know? 2.79 rad/s / 11.25 s = 0.2482 rad/s2 160 deg/s / 11.25 s = 14.22 deg/s2 What was the average centripetal acceleration (m/s2) of the elbow? (0.08378 m/s)2 / 0.030 m = 0.2339 m/s2 Which quantities are different and which quantities are the same for the hand and the elbow? Discuss your answers with your partners and write down questions if you have any. Degrees and radians are the same, as well as the average angular speed and average angular acceleration. The distances in meters is different, which inpacts the average linear speed and the average centripetal acceleration. These were are all three different from the elbow versus the hand. Exploration:

Describe the direction of those arrows, while the angular speed of the wheel is increasing, constant, or decreasing. While increasing, the arrows are perpendicular and the acceleration points to the center of the circle. While constant, the arrows do not change. While decreasing, the arrows get smaller and move away from the center of the circle. Click Go let the simulation run for approximately 10 seconds. What is the magnitude and direction of the torque on the wheel? The magnitude of the torque is 6 Newton meters in a direction tangential to the circle. The direction of the torque is perpendicular to the center of the circle. What happens to the lady bug? The lady bug will eventually fly off of the wheel. What provides the centripetal force to keep the bug moving in a circle? Frictional forces are acting upon the merry-go-round, which keep it spinning due to centripetal forces. Why does this eventually fail? Frictional forces have their limits, so eventually the forces will plateau and at that point the merry-go-round will fail. Observe the acceleration vector after you click Go. How does it change? The acceleration vector starts small and not pointing towards the center, but after 2 rotations the vector points towards the center and increases in magnitude. Will the acceleration vector ever point directly to the center? Why or why not? The acceleration vector does initially point away from the center of the circle, but after only a few seconds, it turns and points directly to the center of the circle and remains pointing to the center. What happens to the acceleration vector? After a break force is applied to the rotating circle, the acceleration vector decreases in magnitude and moves in a direction away from the center of the circle. Use the ruler to measure the radius r of the boundary between the green and pink circles. Record r in your log. r = 2.9 meters Calculate what the tangential component of the applied force must have been. Record Ftang in your log.

Ftang = 2.8 Newtons Compare the torque τ and the angular acceleration α and calculate the moment of inertia I of the disk from τ = Iα. Record I in your log. 10=I(10)  I = 1 kg*m/s2 Compare with the moment of inertia displayed in the graph. Record the comparison in your log. 1 kg*m/s2 is the moment of inertia from the graph. Predict what will happen to the moment of inertia if you keep the mass of the platform the same, but you create a hole in the middle (increase inner radius). Record your prediction. The moment of inertia will be larger because all the mass will be concentration around the rim of the circle, because the radius of where the mass is located will be larger. Set the inner radius equal to 2. Find the moment of inertia for this shape. Record it. 1.25 kg*m/s2 Even when the force on the platform changes, the moment of inertia graph remains constant. Why? The torque and angular acceleration will increase proportionally to one another. The torque would have to change for the acceleration to change, and since they do not change, the inertia remains constant. While the disk is moving, change the inner radius to 2 m. What happens to the moment of inertia and the angular velocity? While the disk was moving, the angular velocity decreases by 0.2 radians/second and the moment of inertia increased by 0.25 kg*m/s2. Make some more changes to the inner radius, outer radius and mass of the disk. Describe what happens. Record your predictions, answers or explanations in your log. As we made changes, we began to see that as the mass of the disk increases, the angular velocity decreases and the moment of inertia increases. As the outer radius decreases, the angular velocity increases and the moment of inertia decreases. As the inner radius increases or the mass is more distributed to the outer rim, the moment of inertia increase and the angular velocity decreases. Experiment 2: Does it become harder to hold the stick horizontal? Why? You are holding up the same weight. Yes because it is away from the center of the mass. Use the force sensor and Capstone to measure the weight of the stick and the clips. -1.0

When you push down on the stick, what happens to the reading of the force sensor? The Force moves in the downward direction quite quickly.

force sensor reading -2.5 weight suspended from left clip -1.5 (optional) weight suspended from right clip weight of meter stick and clips -1 distance d from left clip to force sensor hook (pivot point) 20 cm distance from force sensor hook to CM of meter stick 30 cm distance from force sensor hook to right clip 80 cm Modeling the meter stick as a forearm and the force sensor as the biceps, compare the force that the biceps has to exert to keep the forearm horizontal to the force it has to exert to just support the weight of the forearm. Comment on the relative magnitude of these forces. The relative magnitude of the forces with the extra weights added is roughly one and a half times the weight of the meter stick alone.

List all the forces (magnitude and direction) acting on the forearm (meter stick) and calculate all the torques (magnitude and direction) exerted by those forces about the pivot point, when the stick is horizontal and in equilibrium. Do all the forces and torques cancel out? The masses apply a force of 1.5 in the downward direction. The center of gravity applies a force of 1.0 in the downward direction. The torque on the masses is -30 to the left. The torque on the center of gravity is -30 to the left. Yes, all the forces and torques cancel out. Discuss this simple model and your answers with your partners and write down questions if you have any. None Conclusion: In Experiment 1, we determined the time it took and distance traveled of the hand and elbow for one of us to complete 5 complete rotations of the arm. Using this, we were able to calculate the average speed, linear speed, angular acceleration, and centripetal acceleration. Because of the distance traveled in meters being different between the hand and elbow, we found that this resulted in different average linear speed and centripetal acceleration while the degrees, radians, average angular speed and average acceleration were the same. During the Exploration, we looked at how different forces can affect torque and how changing the moment of inertia can either increase or decrease the angular velocity. We also learned that for there to be no rotational or linear movement on an object all of the forces and torques must cancel out. In Experiment 2, using a meter stick and a force sensor we were able to see the relations between center of mass, added force/weight, and torque. Hanging the meter stick from the force sensor at the center of mass, the meter stick was balanced, with only the weight of the stick being applied. However, hanging the meter stick away from the center of mass resulted in it being unbalanced. Therefore we had to add weight to one end, creating more force to pull the opposite end up to balance out the stick again. There could have been possibility for error in experiment one from having an accurate time, or having slightly off measurements in the length of the arm....