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A machine part has the shape of solid cylinder of mass 500 g, diameter of 4.00 cm and length of 10.0 cm It is spinning about a frictionless axle through its coaxial...

Question

A machine part has the shape of solid cylinder of mass 500 g, diameter of 4.00 cm and length of 10.0 cm It is spinning about a frictionless axle through its coaxial center; but Its rim is scraping against metal, resulting in a friction force of 0.0375 Nat that point. The moment of inertia of a solid cylinder rotating about its coaxial center is [ MR?Question 1 of 2 What is the tangential acceleration of the machine at a point on the rim?

A machine part has the shape of solid cylinder of mass 500 g, diameter of 4.00 cm and length of 10.0 cm It is spinning about a frictionless axle through its coaxial center; but Its rim is scraping against metal, resulting in a friction force of 0.0375 Nat that point. The moment of inertia of a solid cylinder rotating about its coaxial center is [ MR? Question 1 of 2 What is the tangential acceleration of the machine at a point on the rim?



Answers

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 250. N applied to its edge causes the wheel to have an angular acceleration of 0.940 $\operatorname{rad} / \mathrm{s}^{2}.$ (a) What is the moment of inertia of the wheel? (b) What is the mass of the wheel? (c) If the wheel starts from rest, what is its angular velocity after 5.00 s have elapsed, assuming the force is acting during that time?

Here, we're going to be using version of Newton's second law for rotational motion, along with some kinda Matics to figure out the frictional force working on a grindstone and nobody is touching this with a tool or anything. So it is grinding to a halt on its own. Um So we know the initial angular speed is going clockwise and we know that it's going to grind to a halt. So we know the final angular speed. And the question is um if it takes Let's see 45 seconds to stop, what is the angular acceleration of the wheel? And what is the force of friction acting on the wheel? So kim Maddox. We have the usual situation with ordinary acceleration except the rotational analog. That angular acceleration is the change in angular speed in time and that is the final angular speed minus the initial over that time period. So we'll go ahead and write down the angular acceleration as negative. But really what that means is the direction of that acceleration is counterclockwise. Okay. And if we use our um Omega initial in the correct units, so notice that we want to convert 2000 rpm into rads per second. Um So that's an S. I. Unit that we can use, we cannot use revolutions, permit it. We have to use rads per second and that will give us an angular acceleration in rats per second squared. Now what's true about that is it is negative or we could just say it is counter clockwise. What's true about that angular acceleration though is it is the same throughout the entire wheel. So we may be interested in the tangential acceleration at the edge of the wheel and that is coming about through our rotational to linear conversion, which just uses the radius in question. Um So the tangential point on the rim um we just normalize our angular acceleration By the radius of the Rim, which is given to us as .0375 m. And we can then convert to a linear acceleration, which is also in the counterclockwise direction. Probably easier to say that than to keep leaving a negative sign. The next step is to figure out the force of friction acting on the wheel, and there's a couple things that we need to do that. Um First of all, we have to know what is causing the friction and it's not somebody touching the outer rim, so it's actually friction on the axle, the central axle, which is the small are, and what will use is the rotational analog of Newton's second law, which says that um the sum of torques acting on a rotating object is equal to its moment of inertia times alpha. And we're going to assume there is only one torque coming from the force of friction and it is acting perpendicular to the axle, and again the force of friction will be going counterclockwise, so all leave off the negative signs, which maybe a little bit confusing. So they tell us everything the size of the axle is given to us as a diameter of one centim, so radius of half a centimeter. Um And we know the moment of inertia. We have figured out the angular acceleration in radiance per second squared, and we know the radius in S. I. Units. So we should come up with a force of friction in newtons. And again, that is a counterclockwise force touching the wheel tangentially. And finally, um we're asked to find out how much work that this friction does, and we do know that the work done by friction is going to be negative, so yeah, we'll have another negative sign. But we're going to use the work energy theorem for rotations that that work done is equal to the change in kinetic energy or K final minus K. Initial. So yes, we have mm hmm. Negative work done makes a lot of sense because it is friction and that force always acts opposite the displacement in this case an angular displacement. Okay, so we have all we need. They gave us the moment of inertia of this wheel in S. I. Units. We have the angular velocity in S. I. Units. So our work should come out in jewels and it's quite a large amount of jewels. So yes, that axle is going to have to dissipate lot of energy um over that short 45 seconds. So it's no surprise grinding wheels can get hot running them.

For this problem. We have ah, component. Disc was different mess densities, um, in different radio. So the moment of inertia consists of two parts, um, one coming from the smaller disc where the d m um, can be reading this road. Devi a name? Polar coordinates, DVD simply defy Arkham CR immigration. Elif, I gives you two pi, and, um And then there's this original off from zero to the smaller radius R R one. The second term takes care of the the second part of the disc, which has the same form. Except that now we have a different density. Row two and also the original law goes from our one toe are too. Which is this growing shuttled region. Okay, apply integration will obtain it for my expression. I equals piled for two, um, times no one all under the force. Go to our two to the floors. Ministro to all want to force on Dhe. This, uh, we can plugging the numerical values for no one row too, and are well, going to be careful to change this, um, going are seem to the center units. So 70 centimeters, 0.7 meters. Similarly, 50 centimeters is Sierra 500.5 meters, Um, three gram per centimeter scored is sturdy kilogram per meter squared. Similarly, you go to he's 20 kilogram per meter squared blood gayness, numerical values. We can find that the total moment of inertia is 8.52 kilogram meters squared.

In this problem, we want to figure out with moment of inertia of the combined system is. And so to do that we can sum the individual moments of inertia. So the total moon inertia is the more inertia that this, plus the moment of inertia of the ring and then the more inertia of the disc is just 1/2 times the mass of this times, the rays of the square and the mourners of the ring. It's just 1/2 that was the mass of the ring. Times are ones where plus R. C squared or the radio. We're given in the problem, thes to forms I got from Table 92 and they're one of the common forms of the moan of the nursery. And so the next office fear of what? The mass of the disc in the mass the ring are well, the mass of the disc is equal to the density times the area well, the density is three grams percent in your square. It's given in the problem, and in the area of the disc is Pi r square, so pine radius of square. Now, at this point, you have to be careful about units. We want the mass to come out and kilograms so that we can use it in this and get our standard assault units back, which means we have to be careful with this Grams. We might want to convert two kilograms, and we also have to make sure that the rays of the disc your is given in centimetres. So when we square, we get sending your squared and then it cancels with this unit here. In the end, if you make sure to plug in the correct values and do the unit conversions correctly, it will get a massive the disc equal to 23.56 kilograms, employing this into there, converting the radius of the disc into meters, and it's wearing it. I get that I disc is equal to 2.945 kilograms. Time's meter squared, and so now we find the mass of the ring so the mass of the ring is the same thing. It's the density, which is too right. As for cinema, you're sporting time's thie area of the ring, which is pie. And then I have the larger radius squared minus the small radius. Word this theory of a rain. Same thing here. You have to be careful with Convert this grams two kilograms and they have to be careful to use a radio radio values here of centimeters so they cancel here. If you do this correctly, you should get a mass of a ring 15 0.8 kilograms and then playing this mass into this formula along with the radio gives that the moment. Our show the rain is equal to 5.58 kilograms. Major sporting. At this point, we have a value here and we have a value here. So we simply some them to get the total. When you do that, you get the total inertia. Lone inertia is 8952 kilograms. You're square and that's the final answer.

Hi, everyone. Here it is. Given cylindrical discs having the mass. Don't take a G and release 22.4 centimeters. That is the point to go for a meeting in the first part we have to fight. It's the moment of inertia. Moment of inertia is to find this am I. The square by two contents into point to cook food. So it is to weep. Why? Why did you go to Cajun battery script? No second part If it is rotating with angular speed of 1200 rpm dirties 1200 upon 60 RPs she to to party so it becomes a radiant per second. So it is to be equal to 45 or radiant per second. It it is stopped. So find the veracity. Zero in the time it develop 60 seconds that we have to perforate 13th. So don't will be calculated by finding that angular acceleration, angular acceleration will be. Do you find it? We're not less terrified. Duty be Finally zero b notice 45 and far into 16. So I'm gonna reiteration. You will get minus to wait three pipe A radiant person can hence, regarding talk will be high into Airfone moment of inertia is like 502 I had to. Mhm. So you will get my reasonable 0.0 four. Number six. Nuclear do be different, that's all. Thanks for watching it.


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