Atwoodmachinelab20

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Houston Community College *

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1401

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Physics

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Dec 6, 2023

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docx

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5

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Lap report 20: Atwood’s Machine Name: Trung Bao Truong Student ID: W215647597 Instructor: Todd Zapata ABSTRACT: In this experiment, we aim to understand an Atwood’s Machine apparatus, which built with a PASCO super Pulley. The Photogate (Figure 2) is used to measure the velocity of both masses as one moves up and the other moves down. Based on photogate’s data, we will have the graph of slope of the velocity vs time, that is the acceleration of the system. INTRODUCTION: We will experiment and theoretically determine the acceleration of the weights of an Atwood's Machine (figure 1). We will try to prove Newton's Second Law, which is a mathematical statement that relates force, mass, and acceleration. According to Newton's Second Law, acceleration, a, is directly related to net force, F, and inversely related to mass, m. Naturally, this results in F=ma. Using the Atwood's Machine, experimental acceleration data will be collected and compared to the theoretical acceleration predicted by Newton's Second Law using a modified version of F=ma. The two masses are the only two variables in this system system that we will control. We also know the gravity value, which was determined 9.8m/s 2 . Figure 1. Atwood’s Machine 1
Figure 2. Pulley and Photogate DISCUSSION: As shown in Figure 4, Atwood's Machine consists of two unequal masses connected by a single string that passes over an ideally massless and frictionless pulley. When the heavier object is released, it accelerates downward, while the lighter object accelerates upward. The forces acting on each of the masses are depicted in the free-body diagram (figure 3). T denotes the string's tension. With m1> m2, m1 represents the descending mass and m2 represents the ascending mass. Each mass experiences the same magnitude of acceleration, a, but they accelerate in opposite directions. When Newton's 2nd Law (Fnet = ma) is applied to the descending mass m1 and ascending mass m2, we get: m1g – T = m1a m2g – T = m2(-a) a = (m1 -m2)g/(m1 + m2) (1) Figure 3. Free body diagram 2
In the first experiment, We added a single 100g mass and a single 20g mass to one mass hanger (5g) for a total mass of 130g (m1), and second mass hanger (m2) add single 100 g mass for a total of 110 g. and hang the masses on the string over the pulley (figure 4). Figure 4. Atwood’s Machine Moved m2 to its lowest point and release, m1 descended and m2 ascended. We have obtained data from photogates and graph from computer (figure 5). 3
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Figure 5. Mass Speed The slope of the graph (figure 5) of the speed versus time graph is the acceleration. The units of the slope are m/s 2 . acceleration measured = 0.688 m/s 2 . acceleration theoretical = (m1 -m2)g/(m1 + m2) = (130 – 110)x9.8/(130+110)= 0.81 m/s 2 This difference is because there is friction between the rope and the pulley in the actual experiment. In the next experiment, we added an additional 20g mass to m2 (to make it balance with m1). If there were no friction, and we gave m1 a push downward, it would continue at constant speed. Moved m1 to its highest point, give m1 a gentle push downward and release it. The speed is decreased, so we add more mass to make it keep the constant speed ( m2 = 130g, m1 = 131g), we obtain the data from photogates and have graph from computer (figure 6). From here we have found the mass of friction to be 1g (131 g – 130g = 1g), the weight of this mass is the extra force needed to overcome the frictional force, and thus does not contribute to the acceleration. 4
Figure 6. Friction compensation runs Re-calculate the equation (1) but subtract off the frictional mass from m1 in the numerator we have acceleration theoretical = (m1 -m2- 1)/(m1 + m2) = (130 – 110-1)x9.8/(130+110+1) = 0.77 m/s 2 %Error = (Measured – Accepted)x100 / Accepted = (0.81 – 0.77)x100/0.77= 5% The correction for friction improved the final result. The experimental acceleration is also slightly low due to neglecting the mass of the string and the rotational inertia of the pulley. CONCLUSION: Through this experiment, we can understand more about Atwood’s Machine and the equation: a = (m1 -m2)g/(m1 + m2). Acceleration is directly proportional to the net force acting on the system. This was also shown by the inversely proportional relationship between the acceleration of the masses and the sum of those masses. This is shown clearly in our graph. The acceleration of both objects is equal but in opposite directions. During the experiment, there will be errors compared to theory because there is friction between the rope and the pulley. To handle that problem, we added mass to m1 so that this mass creates a force enough to eliminate friction, and from there we got better results. 5

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