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A solid cylinder with radius $0.140 \mathrm{~m}$ is mounted on a frictionless, stationary axle that lies along the cylinder axis. The cylinder is initially at rest. Then starting at $t=0$ a constant horizontal force of $3.00 \mathrm{~N}$ is applied tangential to the surface of the cylinder. You measure the angular displacement $\theta-\theta_{0}$ of the cylinder as a function of the time $t$ since the force was first applied. When you plot $\theta-\theta_{0}$ (in radians) as a function of $t^{2}\left(\right.$ in $\left.\mathrm{s}^{2}\right),$ your data lie close to a straight line. If the slope of this line is $16.0 \mathrm{rad} / \mathrm{s}^{2},$ what is the moment of inertia of the cylinder for rotation about the axle?

And this problem. We're looking at a cylinder, uh, mounted on an axel. It's with a three Newton four supply tangentially to its surface. Uh ah. It's radius is 0.14 m. And after it starts spinning from the force, uh, the angular displacement is measured and graft against t squared. And it makes a straight line, uh, with a slope of 16 radiance per second squared. So, given this information, we're supposed to find the moment of inertia of this cylinder, Uh ah. And using our to torque equations, we can we can figure that out. So seeing that these are both equal to torque, we can set them equal to each other. Moment of inertia. Times Alfa equals R f Sign Phi and then isolate the moment of inertia. R f sign five over Alfa. Okay, and now we have everything on top. All we need to find is Alfa uh, which we can do using this information about the graph that's presented to us. So this is the slope of a graph off five versus T squared. So if we plug in this slope here, we know and give it that it's a linear graph we know that this graph is going to look something like five equals B plus 16 t squared. And now just looking at this, we can see that it bears a striking simulate similarity to, um, our equation of motion. They it is equal to date or not, plus, v not t Plus, not a not V, actually, Omega, not t plus Alfa, not t squared. And given that it's starting at rest, this is gonna be zero, so we can see the similarity here. I've got the one half sorry about that. Uh, s so we can see the similarity here and see that this 16 is going to go where this is so 16 must be equal to one half. Alfa or Alfa must be equal to 32. But if just saying that, uh, these things look similar isn't good enough for you. Ah, I can just you do the math and show you that this is going to be the case by the way, we do it. So once again, we know that this is what the equation is gonna look like. Data equals B plus 16 t squared. And we know that Alfa is equal to the second derivative off data with respect to time. So all we do is take the second derivative of this. Take the first derivative first leave one here. So the first derivative of this is going to be zero plus bring down the too. So this will be 32 t on DSO now take the second derivative zero is gonna stay zero and then this derivative will just remove the tea and we will get Alfa is equal to 32 radiance per second squared and now all we need to do is plug in our numbers. Moment of inertia is equal to R 0.14 times f three I'm signify it's tangential to the surface, which means it's going to make a right angle with the radius. And the sign of a right angle is of course, one. And that will all be over 32 which will give us a final moment of inertia of 0.131 kilogram meters squared

In this problem. A rope is wrapped around a cylinder with the radius R and a mass Big M. The rope has a mass of little M, and the rope and the cylinder are rotating about the cylinder central axis at an, um, at an angular frequency of omega, Not we are asked to find the final angular frequency after one rotation of the cylinder where the rope is hanging vertically downward and traveling downward at some speed V. We need to figure out what the final angular frequency, which I will call Omega Final is. And we also need to figure out what V is in this case. So, um, this is a conservation of energy problem. Um, as we go from our initial set up to our final set up, there is a loss of gravitational energy associate ID with the changing position of the center of mass of the rope. And, um, because of that, there is a gain in the total kinetic energy off the system. So what we're gonna do is set up in equation that will equate the energy of our initial set up to the energy of our final set up, and then we can use that to solve for Omega Final. Alright, So first, our initial set up the energy is entirely given by, um, the rotational energy of our rope and of our cylinder. Um, there's some gravitational energy associated with the position here, but because we only care about difference of gravitational energy, I'm going to include a negative potential energy in this term here, and you'll see that later I'll show you. But first, the total rotational energy of our initial set up is given by one half times I omega squared for the rope and plus one half I omega square for the cylinder. All right, we can ignore the thickness of the rope. We're told that we can ignore the thickness of the rope and so we can approximate that the mass of the rope is entirely concentrated along the surface of the cylinder. And so the and I'm sorry. The moment of inertia of our rope is then given by a little m r squared. All right. And the initial angular frequency is Omega. Not then for our cylinder. The moment of inertia is given by one half. Begin R squared and then omega not squared All right, So this is the total rotational energy of our initial system. All right, so now let's figure out the energy of our final system. In this case, we have some rotational kinetic energy associated with the cylinder. So I omega squared plus one half m v squared the kinetic energy of the rope with mass m minus. And this one I do minus. Because in this situation are gravitational energy is less than in our previous than in our initial, um, in our initial set up. So the gravitational energy here is lost. We have a the mass of the rope. The center of mass of the rope descends by a distance of pie Are we know that because the whole distance of the rope is two times pi r. It's the circumference of the cylinder. And so the center of the mass, which is halfway along the rope, descends at a height, descends a height of pi r. And that means the associate ID change in potential energy is mg pyre. All right, so now we replace All right, so this I just replaced I with the known moment of inertia of the cylinder for V v has a given relationship between between itself and are in Omega Final. We know that this relationship exists for Omega Final and our because that's the relationship for tan, gentle, tan, gentle velocity of the of the surface of the cylinder, which is going to be the same as the velocity of the rope. In this case, because the rope moves around the cylinder without slipping. So we can say that one half MV squared is one half m r squared omega f squared and then we subtract mg pi r Yeah, yeah. Now we set these equations equal to each other and we can solve. So first, my first step is to rewrite both of these equations. Oh, and by the way, you'll see that now we have one equation and our Onley unknown value in this equation is Omega final. So we can use this equation to to solve for mega Finally just need to do a lot of complicated algebra two isolated on one side and express it in terms of, um, other quantities. All right, so this whole complicated equation we just need to solve this in terms of Omega. Not so in my first step here. I'm going to do two things. I'm going to multiply each term Times four. And I'm also going to divide each term by R squared. All right, so I'm multiplying each term by four divided by r squared. And this term becomes to em Omega squared. Plus this term and Omega squared equals and I'll make a final square toe Plus to em make a final squared minus four m g pie over are all right. So to repeat myself in this first step, I divided each term by r squared multiplied each term by four. In this next step, I'm going Thio I'm going to combine like terms and on the left side and on the right side. Alright. And our next step is to divide every term of this equation by to m plus m so that becomes omega. Not squared equals omega Final squared minus four mg pie over our times. Too little implicit. Big in perfect. All right, so in this last in this last bit, I'm going to do two steps. First I'm going to add this term to both sides of the equation, and then I'm going to take the square root of both sides toe get omega final by itself. And what we'll see is that Omega final equals the square root of omega not squared. Plus for mg pie over are to and plus yeah, all right. And this is our answer. Wait. I have toe put a box around that. So before we move forward, um, one good thing to check at this point is toe make sure that the units are going to be appropriate. We see, on the inside we have, uh, a frequency squared here and then on the top, it's mass times acceleration on the bottom. We have mass and a distance there. So this whole term here will also come out to give us, uh, units of frequency squared and we take the square root of that. We get units of frequency, and so our units do match up, and that's good. Um, the last step is to figure out what V is. And I already said that V is given by this relationship here, so we can simply take what we just found out. Multiply times are so the equals, our times, omega final, and we just show that omega final is square root of Omega, not squared. Plus or mg pie over are to m plus and all. Right. And there you have it. That's the whole problem. Good job, everyone.

So here we need to find the link angular speed of the cylinder and then linear speed of the rope. So we should say GP equals mg Y c m. Where why cm is a vertical distance of on Jax center of Mass. We know that the kinetic rotation Connecticut rotational kinetic energy initially plus the kinetic the translational connect energy initially plus the gravitational potential energy equals the kinetic energy, the occasional kinetic energy final plus the translational kinetic energy final and weaken substitute So we have half I omega squared Plus, we say hi Omega initial squared plus half and the initial squared plus mg. Why cm This is going to be equal to have, um, I omega final spared. Plus, uh, um, these final squared. And we know that the center of mass for this is going to be all over two or simply the length of the rope over, too. And this is going to be too pai r of the circumference of the of the cylinder. You know that guy too, and we know that theological CNN equals pi R. So we can use this in order to Saul for this, we can use this and just substitute for the why of the center of Mass. So at this point, we can say, Okay, one over two. And be careful when you're substituting because there's some outbreak manipulation. It's, ah, very easy to get lost or confused as to what exactly you're writing down. So just make sure that you are diligently plugging in your variables. And on the other side we have half half r squared. We'LL make a final squared plus half and R squared Omega Final Squared. So at this point, we can say, OK, this is going to be capital we can say Okay, let's try to solve for Omega Final So we can say that Um plus I'm over four plus small him over too. Times are square Omega initial square Plus I'm Jean pi r equals And then I'll be off for us and over too R squared Omega Final squared and we find that omega final It's going to be equal to Omega Initial Square plus for pie mg all divided by our times M plus two em to that all to the half power and then for because we know that V equals our Omega F that means that this is going to be our Omega initial squared plus for pie and she all over Ah, and plus tio to the half hour. So these would be your answers. And that's the end of the solution. Thank you for watching.

Number 35. We have the cylinder. I've listed all the information we were given. Everything was in the correct unit. So I didn't write them. The other thing. We're told that this one this break. This is a red thing. When this break things applied, it Reduced the speed by a factor of two. So if I consider this the starting velocity, the ending velocity, some final angular velocity on behalf of that. So 38. Okay. First we wonder what the angular acceleration was. Just the magnitude. You know, it's gonna be negative because it's slowing down. Um So I've rearranged the first cinematic equation. So the angular acceleration is going to be that final velocity minus initial velocity. But by the time, So my final velocity was 38. My initial velocity was 76. And the problem told me the time was 6.4 seconds. So for that I get negative 5.94 accelerations of the bureau radiance per second squared. And then port p. Ask what the force was. The friction force that this brake applied. I'm going to use the rotational version of Newton's second law. So ethical dilemma. But the rotational versions, it's work is the moment of inertia times the angular acceleration. And we know that torque is force times uh the radius or the distance to the pivot point. In this case it's going to be that distance which is the radius that were given. So I can plug in these values. I'm looking for this forest because it's gonna be the friction force, the radius. I was given .083 the moment of inertia of the cylinder. I was given .615 and I just calculated the angular acceleration. I'm just gonna put in the magnitude here. I get 44 point out. That's a force, so that's Newton's.