Lab 12 - Atwoods Machine PDF

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Activity 12 Atwood’s Machine Objective: To get more practice drawing free-body diagrams and applying Newton’s laws to mechanical systems. Activity 12.1. Analysis of Atwood’s Machine The Atwood machine (George Atwood, 1784) is a device that provides a simple illustration of Newton’s Laws, and allows a careful study of motion with constant acceleration, since the acceleration can be made arbitrarily small, making fairly precise measurements possible. Nevertheless, as is always the case, the “real life” system will be more complicated than the simplified model that you will be studying here. To be precise, the model in the simulation neglects: • The mass (inertia) of the pulley • The friction acting on the pulley • The mass of the string The inertia of the pulley can be taken into account using the techniques to be introduced in Chapter 9, and friction could be included as an “ad hoc” term in the equations of motion (to be determined experimentally). The mass of the string is much harder to include in the calculations, requiring the methods of advanced (Lagrangian) dynamics in order to do it properly. 12.1.1. Look at the model The simulation you will be using is at https://www.physicsclassroom.com/PhysicsInteractives/Newtons-Laws/Atwoods-Machine/Atwoods-Machine-Interactive. Go there now and enlarge the demo window so you can see better. The default system shown is not the traditional Atwood machine; rather, it is an illustration of the system studied in detail back in Chapter 6 of your textbook (beginning with Section 6.3.1). You may want to go back to that chapter and study that system carefully now; you will need to use it before this activity is over! For the moment, however, let us bring up the traditional Atwood machine. For that, you need to click on one of the red hanging masses on the lower left of the screen. Start by clicking on the smallest one. 12.1.2. Free-body diagram

Now you can see the Atwood machine, with equal masses on both sides. To make sure that the masses are equal, grab the red mass and move it upwards a little, then click start (click anywhere in the window to stop the animation). What happened? The pulley didn’t move because the masses are equal

This is a result that many people find surprising. Our intuition seems to be telling us that when you let go, the system should move in the direction of the lowest hanging mass, even if the two masses are equal. To convince yourself that this should not be the case (under the assumptions we have made), draw a free-body diagram for each block in the space below using the Free Body Diagram Powerpoint from the last activity. Please rewatch the instructional video for that activity if you do not remember how to use it. Make sure you follow all the conventions we have introduced in class regarding labeling of forces, etc.

+𝑥

𝑇 𝐹𝑃,𝑅

𝑁 𝐹𝑃,𝑅

+𝑦

𝐺 𝐹𝐸,𝑅 12.1.3. Analysis of system at rest Now answer the following questions: If the system is not accelerating, what does that tell you about the net force on each mass? (Newton’s first law):

According to newtons first law, this means that an object at rest with no acceleration resulting in no force will stay at rest with no acceleration resulting in no force.

If the string is massless, how do the tensions on both ends of the string compare (in magnitude)? They’re equal because the mass of the blocks are equal. T = ma + mg which is going to be equal to both because they have the same mass, are at rest, and gravity is 9.8.

Based on this, the information on the screen, and your free-body diagram, what are the masses of the two hanging weights? Make sure you show your work and explain your reasoning. You may do your work on paper, then take a picture of it.

𝑚𝑙𝑒𝑓𝑡 =

𝑚𝑟𝑖𝑔ℎ𝑡 =

(So, if you thought the system should move in the direction of the lower hanging weight, was your intuition completely wrong? Actually, no. If you include the mass of the string, the system really is unbalanced towards the side where more of the string is hanging. In real life, however, this is typically a very small effect, and the system is often prevented to move by the small static friction acting on the pulley at the axle. Somewhat ironically, our model manages to give the observed result, by neglecting two real-life effects—the mass of the string and the static friction—that, in this situation, tend to cancel each other out.)

12.1.4. Analysis of system in motion Now, click on the medium-size hanging weight, raise it, and click “start”. Record below the time the motion took, and the displacement: ∆𝑡 = 0.98 s ∆𝑦 = -1.58 m Based on these data, and assuming the acceleration is constant, calculate the acceleration of either one of the masses (one of them is accelerating upwards, the other is accelerating

downwards; make sure you report the correct sign for the one you chose). Show all your equations. For the red mass, v = -1.58/0.98 = -1.61 A= v/t = -1.61/0.98 = -1.64 m/s2 Now, based on your free-body diagram (modified as needed to account for the net acceleration and the larger mass), derive an equation for the acceleration, in symbolic form, in terms of the two masses and the constant 𝑔. Again, show all your work below. 𝑎 = m2g – m1g /m1 +m2 NEGLECTING FRICTION

Based on this and your data above, what is the mass of the medium weight? (Again, show your work: solve the equation you just found for the unknown mass, algebraically, then substitute the known value of 𝑔, the value of the known mass, and the value of 𝑎 from the experiment.)

𝑚𝑚𝑒𝑑𝑖𝑢𝑚 =

As a further check on this, derive an equation for the tension in the rope that applies when the system is in motion, in symbolic form, in terms of the two masses and the constant 𝑔. Again, show all your work below. At the end, substitute all the values you have and predict the numerical value of 𝐹𝑡 for this setup: 𝐹𝑡 = m2g + m2a = m1g – m1a

Now run the system again and watch the value of the tension that is shown while the masses are in motion. Record this value below. How does it compare with your prediction? 𝐹𝑡 (simulation) = 13.1 N

𝐹𝑡 (theory) = 13 N

Finally, in your own words, explain why the tension in the rope is different when the system is moving than when it is at rest (either with a mass resting on the ground, or “your hand” holding it). A free-body diagram, particularly for the mass that is on the ground or being held, would be very helpful here. When it’s a rest, theres no acceleration to be added or subtracted to the weight (mg)

12.2. Modified Atwood Bring back the “modified Atwood” setup by clicking on one of the red blocks on the left (there are red carts, which presumably roll with negligible friction, and red blocks, which do experience friction). Choose the middle block and click “start.” Note the time elapsed, the displacement, and the tension in the rope while the system is in motion: ∆𝑡 = 2.48 s ∆𝑥 = 2.01 m 𝐹𝑡 = 9.1 N You can use this to calculate the acceleration, as before. 𝑎 = 0.32 m/s2

Now repeat this with a different suspended mass on the right. You can change the suspended mass by clicking on the blue weights at the bottom right. It is safe to assume that the middle blue weight has the same mass as the middle red weight that you used earlier. Different suspended mass (state which one): ∆𝑡 = 1.17 s ∆𝑥 = 2 m 𝐹𝑡 = 13.7 N 𝑎 =1.46 m/s2

Based on this information, calculate the mass of the block, and the coefficient of kinetic friction between the block and the surface. Show all your work, and explain your reasoning. (You may use your textbook as a reference, but nothing else.)

𝑚𝑏𝑙𝑜𝑐𝑘 = 𝜇𝑘 =...