"(II) An Atwood machine (Fig. 11–16) consists of two masses, m...

Question

"(II) An Atwood machine (Fig. 11–16) consists of two masses, m_A = 7.0kg and m_B = 8.2kg , connected by a cord that passes over a pulley free to rotate about a fixed axis. The pulley is a solid cylinder of radius R₀ = 0.45m and mass 0.80 kg.

(a) Determine the acceleration a of each mass.

(b) What percentage of error in a would be made if the moment of inertia of the pulley were ignored? Ignore friction in the pulley bearings. Use angular momentum.

"

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice from together. So first stop, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A building employs a dual counterweight system with two elevator cars. Elevator car, a has a mass of 1200 kg and starts at the 10th floor. Elevator car beat with a mass of 1250 kg. Begins at the second floor. They are connected by a steel cable that runs over a pulley that is free to rotate about a fixed access at the top of the shaft. The POY is a solid cylinder with a radius of 0.80 m and a mass of 100 kg. Find the acceleration of each elevator car as they move in the shaft. That's our end goal is ultimately we're trying to solve for the acceleration of each elevator car as they move in the shaft. OK. So with that in mind, let's quickly look at our diagram here that's provided to us by the prom itself. So let's assume that this gray rectangle at the far, at the very top of our diagram represents like the ceiling of this particular building. So the ceiling of this building has a orange ST so this orange little bar which is connects to our pole which is represented by a circle and the POY has some radius R which is from the center of the POY to the outer edge. And then we have a red bold red line on either side connecting our elevator car A which is capital, M subscript capital A denotes the massive car, A, an elevator car, A is represented by a blue rectangular prism and then like like our blue rectangular pris prism for elevator car A, we have elevator car B which is represented by a yellow rectangular prism. And we have a bold red line connecting both elevator car A and elevator car B to our pulley. And that is our steel cable. It's the bold red line. Awesome. And as we could see, elevator car A is gonna go up and it's gonna accelerate up and then elevator car B is gonna accelerate down. And then we have the weight force denoted as a downwards force pointing downwards from the rectangular prism for both elevator car A and B respectfully and recall that the weight force is the mass of the object multiplied by the acceleration due to gravity. Awesome. And looking at our multiple choice answers now, as we could see all of our multiple choice answers are in units of meters per second square. So let's read them off to see what our final answer might be. A is 0.10 B is 0.15 C is 0.20 and D is 0.25. OK. So moving right along here. So first off, let us draw a diagram to help us better visualize the forces at work on the POY system itself. So looking at the poly here, we can see since we are using a circle, did you know our poli, we can see that we have some angular speed omega going to the right. So going clockwise and then we have at 0.0 which is the center of our POY going out to the utmost edges some distance R which is the radius of the pole. So now we need to determine what the angular momentum of the system is first. And then after that, we will need to determine what the torque on the system is. So let us recall that the equation for torque is, we're gonna use the Greek symbol Tau to denote torque and Tau is equal to DL ID T. And let us note that angular momentum is denoted by L and T denotes time. So now at this point, we need to recall and use the angular momentum equation which states that capital L is equal to M A which remember that subscript capital A denotes our elevator car A and subscript capital B denotes our elevator car B. So ma multiplied by V multiplied by R plus MB multiplied by V multiplied by R plus I multiplied by Omega. So Omega once again is our angular speed. Awesome. And I, in this case, is our moment of inertia. OK. So now that we quickly reiterate, so ma is our massive elevator car. AM B is the massive elevator car BV is our velocity or speed. R is the radius of our poli I once again is our moment of inertia. And the, and Omega is our angular velocity more specifically than speed its velocity. And recall that Omega and this is very important. We need to recall that Omega is equal to V divided by R. So it's the velocity divided by the radius. Thus, we can, we can plug in our value for Omega into our equation for the angular momentum. And we can write that L is equal to B multiplied by R multiplied by M A plus MB plus I multiplied by V divided by R. Awesome. So with that in mind, we can move right along. And we need to, at this point, we need to recall and use the equation to help us solve for the net torque on the system about the axis O as seen in our diagram. And we need, by taking to do this, we need to take the clockwise rotation to be positive. So considering that the clockwise rotation is positive. We can write that Tau O is equal to MB multiplied by G where G is the acceleration due to gravity multiplied by R minus ma multiply by G multiplied by R. So now this is where it's really important to recall and use that Tau is equal to DL by DT. Thus, we can go ahead and write that MB multiplied by G multiplied by R minus ma multiplied by G multiplied by R is all equal to D by DT of V multiplied by R multiplied by MA plus MB plus I multiplied by V divided by R Awesome. So now we need to simplify and when we simplify, we will get G multiplied by R multiplied by MB minus MA in this case, which is equal to MA plus MB multiplied by R. And then we need to multiply this by DV, by DT plus I divided by R multiplied by DV by DT. So now at this point, we need to solve four and let's write this in blue, we need to recall a note that A, the acceleration is equal to DV ID T. OK. So in order to solve for A, which is our final answer, the acceleration value, we need to recall and use the following equation, we need to state that A is equal to G multiplied by MB minus MA. So MB minus MA all divided by MA plus MB. And this is all plus I divided by R squared. Awesome. And this is equal to, when we, this is, we can also say that this is equal to G multiplied by MB minus ma all divided by MA plus. And this is plus MB. And don't forget your parentheses plus. And now we need to write out the value for the moment of inertia in this case, which since we're dealing with a circle, we can write for a cylinder more specifically because you could think of the poli like a cylinder. So it's one half multiplied by M multiplied by R squared. So you may be able to recall this off the top of your head. But usually they have like a table with all your moment of inertia equations written out. And it's usually in your textbook or your professor will provide it to you. But each moment of inertia equation is different based off of whatever three dimensional shape you're dealing with. OK. So this is all equal to, since we're simplifying now. So G multiplied by MB minus ma all divided by MA plus MB. And this is all added to one half multiplied by M where lowercase M is our mass. OK. So now at this point, we need to plug in our known values into our equation for A and finally sol for the numerical value of A, which is our final answer. So as we should recall, we should recall that the acceleration due to gravity is 9.81 m per second squared. And then we need to multiply this by 1250 kilograms minus minus our value for our elevator car a which is 1200 kilograms. Awesome. And this is all divided by. So we need to take 1200 telegrams and we need to add that to 1250 kg. And then we need to add this to one half multiplied by our mass, which is given to us as 100 kg. Fantastic. So let me plug that into our calculator and we round to two decimal places. We will get 0.20 m per second squared and that's it we've solved for this and sol for our problem officially hooray, we did it because in our calculator, you'll get something like 0.1962 but we'll round to two decimal places as all of our multiple choice answers are rounded to two decimal places. Therefore, looking at our multiple choice answers, the correct answer has to be the letter C which is 0.20 m per second squared and that's it. That's how you solve for this problem. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Key Concepts

Atwood Machine

Moment of Inertia

Angular Momentum

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