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What is the angular momentum of an object that is spinning at 30degrees per second with radius of Zm and weighs 50kg?What is the angular inertia for carousel that i...

Question

What is the angular momentum of an object that is spinning at 30degrees per second with radius of Zm and weighs 50kg?What is the angular inertia for carousel that is 3 meters in radius and weighs 100kg?What is the angular acceleration of an object if it starts rotating at 10 degrees per second and increases to 100 degrees per second in 5 seconds, if it has radius of Sm and mass Of 1Okg?If the effort arm is 20% closer than the point of insertion which is . 12m from the axis and the resistance is

What is the angular momentum of an object that is spinning at 30degrees per second with radius of Zm and weighs 50kg? What is the angular inertia for carousel that is 3 meters in radius and weighs 100kg? What is the angular acceleration of an object if it starts rotating at 10 degrees per second and increases to 100 degrees per second in 5 seconds, if it has radius of Sm and mass Of 1Okg? If the effort arm is 20% closer than the point of insertion which is . 12m from the axis and the resistance is 1Okg with resistance arm ofl.45, how much force needs to be produced? If an object starts traveling at 20 meters per second and deaccelerates at _ meter per second squared how far will it have gone in scconds?



Answers

$\cdot$ The rotor (flywheel) of a toy gyroscope has a mass of 0.140 kg. Its moment of inertia about its axis is $1.20 \times$ $10^{-4} \mathrm{kg} \cdot \mathrm{m}^{2} .$ The mass of the frame is 0.0250 $\mathrm{kg} .$ The gyro- scope is supported on a single pivot (see Figure 10.68 ) with its center of mass a horizontal distance of 4.00 $\mathrm{cm}$ from the pivot. The gyroscope is precessing in a horizontal plane at the rate of 1 revolution in 2.20 s. (a) Find the upward force exerted by the pivot. (b) Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min. (c) Copy the diagram and draw vectors to show the angular momentum of the rotor and the torque acting on it.

Okay, So in this question, we're dealing with a gyroscope processing around a pivot point. We're told that the mass of the rotating wheel of the gyroscope is 0.140 kg. We're told that the wheels moment of inertia is 1.20 times 10 to the negative fourth kilograms meters squared and we're told of the mass of the frame of the gyroscope, which I'm gonna call MF is 0.0 to 50 kg. We're also told that as its processing around the pivot point, the center of gravity of gyroscope is that a horizontal distance of four centimeters from the pivot point? So we can say that the radius, uh is 4.0 centimeters which we can rewrite as 0.0 for 00 meters. We're also told that the period of the procession is 2.2 seconds, so we can write t easy for 2.20 seconds and then in part, a were asked to find what the force supporting the gyroscope problem. The pivotal point is so we have to find the normal forests provide about a pivot point. So this is a little bit different one you might expect from this kind of problem. Um, but we're just analyzing the forces acting on the gyroscope. So weaken, you know, do a quick drawing of the axis of the drive scope here. Eso The normal force is all the way the left here. You know, if we're mimicking the diagram so we can say that this is the normal force and then the force of gravity is acting here at the center of mass of the access of the gyroscope. Or will the senator massive the folder a scope? So these are really the only two forces acting on it, the normal force and the force of gravity. And we know that the gyroscope isn't moving up or down. Um, it's just processing, and it also has the rotational flywheel, but it it is not accelerating linearly. So we can say that the normal force and the force of gravity balance So we can say that F n is equal to F G. And we have the mass of the gyroscope so we can go ahead and say that F G is going to be equal to the mass of the individual components. So the mask, the wheel plus the mass, the frame times acceleration due to gravity. So then the normal force will be equal to 0.140 plus 0.0 to 50 times 9.80 And so we get that are normal. Force is 1.62 Newton's So that's the answer for part A. And then we can move on to Part B, which is closer to what we been doing for the rest of the chapter. Um, here, we're supposed to find the angular speed of the flywheel off periscope. So for this, we want to use the equation that was given at the end of the last section of the chapter Ah, which relates the speed of the procession capital maker to a bunch of other values. Um, that are a little easier to find. Um, And so it says that capital mega is equal to the applied torque, uh, times the angular momentum of, or divided by the angular momentum of the gyroscope, which we can rewrite as the force of gravity times the lever arm. Ah, because the torque is being created by the force of gravity being perpendicular to the lever arm, which is the axis of the gyroscope. And then we can divide that by I Omega, the angular momentum off the gyroscope wheel. So, um, the other thing that we need to know here is what Capital Omega is, and it's actually pretty much given to us. We can say that capital omega will be equal to two pi divided by the period of the procession and were given a period of procession at the beginning because it does one revolution in, ah, one revolution every period, which is two pi radiance. And then the period is the time that it takes to do that. So we can say that omega is equal to pi over tea so that we can set Ah, these two equations against each other his equal. So we have to buy over tea left hand side. On the right hand side, we have f g or divided by I omega and we want to solve for little omega so we can multiply both sides by omega and the device both sides by two pi over tea and we'll get the equation. Omega is equal to F G or see over. I times too high. So we actually have. All of these values were given will be calculated f g. In the previous part of the problem, we actually calculated FN but happen and after your equal to each other. So that works were given our were given tea and were given I and then two pi, of course are constants. So we can go ahead and start plugging. Everything in omega will be equal to 1.62 Newtons times 0.40 m times the period of the procession which is to 0.20 seconds. All of that is divided by the moment of inertia of the flywheel which this 1.20 times 10 to the negative fourth kilograms meters squared and that's multiplied by two high. And then this will give us the angular speed of the flywheel in radiance per second, which is ah, 189 radiance for second. But the question asks us to give her answer in revolutions per minute. So we just take this answer and well, multiply it by, um 60 over two pi to convert it into revolutions per minute. And so we get the answer of 1.80 times 10 to the third revolutions per minutes for our little Omega. And that's our answer for Part B and then part C. We're basically just copying that were just asked to copy the diagram and mark, um, the torque and angular momentum vectors on the diagram. Um, so I'm going to do a top down view because it's easier than trying to draw things in perspective. Um, so I'll do a top down view. So we have the pivot point here, and then in the diagram, the, uh, access of the gyroscope points out to the right. And then I would say that I'll make this little rectangle Here are flywheel and in the diagram we have that the gyroscope is processing counterclockwise. So, um, for this to happen, we need to think it'll be easier to think first about our torque. So we have our vector pointing out. So it's the right because the center of the rotation is the, uh, pivot points on the left. So the actual points out towards the rights are our vector points out to the right. We're doing a top down view, so gravity is going to point into the screen into the page. So if we do our right hand rule, um, and we sweep from the right into the page, um then are right thumb will be pointing upward, so our torque vector is going to be here. And since our torque vector and R D L Victor are pointing in the same direction in order to create a counterclockwise rotation, that means that our angular momentum vector must be pointing out to the right here. Because if we are going to make the axle, uh, rotate counterclockwise, um, the actual tries to line up with the angular momentum vector and in order to make counterclockwise rotation, then we need the angular momentum to start pointing out to the right cause we're constantly adding, um, a small change in the angular momentum that is perpendicular to left of the angular momentum beggar. So are changing the angular momentum vector will be a little bit upward, and then as it rotate, then the axle line up with that, and then as it rotates, it will be a little bit this way, and then as it rotates, it will be a little bit further this way

Hello. So for the first part is gonna be the nature of the hope that the nature of the square, Okay? So that that of the hoop is given by three miles squared over to the square is going to be two times um R squared over 3% R squared. Okay? So since you can give it a mask you can just put it in and then uh we also have our, okay so this is gonna be a final answer then our uh angular momentum, it's gonna be i omega. Okay. You already found I so you know, omega is uh two pi on T. So there's gonna be a final answer very much.

Problem. 11.58. We have a discus rotating horizontal plane that has a massive 150 kilograms a radius, two meters and the given moment of inertia. Leo person walking from the rim thio half a meter away from the center. And given the initial frequency, your angular speed with which it's moving, we'd like to find what its final angular speed is. So this is ah, and conservation of momentum. Angular momentum problem. Yeah, make a final. And so, um, the total moment of inertia is just the sub of the that of the disc. So this is the initial so, um, r squared, invited by my our final squared. So the moment of inertia is just the sum of the moment of inertia the disc, plus the point, the point mass that the this person represent Erupt that represents his person. Uh huh. And so then we get 2.6 radiance per second is the final angular speed

Negotiation. It was told that I uniform. This is rotated about its symmetry exists. This goes from risk to an angular speed of 11, radian per second, Angular speed of 11. Or you're a radian per second. In a time interval does a. T equals two Of .20 2nd. With a constant angular acceleration. We assumed to be this alpha. The constant angular acceleration and the energy they rotational inertia and radius of the tests are the rotational inertia. I close to that is given 1.5 kg into meter square In two. Major Square and the radius is 11.5 centimeter. The radius is 11.5 centimeter. So the first question I was asked, where is the angular acceleration During the .20 and terrible. So define the acceleration. We will just use them. Love angular motion which is omega final. That is omega final. Close to omega initial fast house. Far into the time interval. So alpha equals two. Yes, this was um This was in rest. So Omega initial zero will also be omega final divided by the time and terrible. So mega finally 11. I'm still already in 1st, 2nd and Delta, she is your .20 seconds. So the answer we get in 55 55 radiant per second square. The angular acceleration in question being I was asked that what is gonna talk on the disk during this time. So during this time the network could be, this is tara cow B moment of inertia into the angular acceleration. The moment of inertia is given here. It is given here, which is 1.5 Kg Centim. We write it 1.5 into Alpha, which is the angular acceleration, which is 55 million per second. Newton and two m. So the tower could be 1.5 to 55 So it is 82 .5 million in two m in question steve that was as bad after the applied stalks tops. The frictional Charles remains. This term causes an angular acceleration of magnitude 9.8 million per second square. So of the two calls to -9 eight because it is fixed frictional. That that's why a minus sign is added at the first of its brilliant per second squared. Uh So through what total angle starting from time teaser or does it is appropriate before coming to rest. So again, just use the formula of angular dynamics. So data final would be 80 P fast half into alpha two into He's squared. This is an the first the first formula we use, but we have to find T. To find the we have to use another formula which is Omega final equals two omega initial Plus Alpha two into T. From here we get equal to Make any shell divided by Alpha two. That would be omega. Any shell would be found that here have found 11 points it already in per second? Live in funds here already in 1st 2nd divided by not 9.8 radiant per second squared still. So the answer is the time interval is 1.1-2 seconds. For here paid out. We assumed it and our TV zero. So it is half In two, Alpha 2, Half into Alpha two, which is -9.8 into t squared, which is 1.1 to 2 square. Okay, uh the the final would be let's just calculated. Did you find the best? six 6.168 radiant. We ignored the minus sign here. So it would be 6.168 radiant. So in question, d I have to find out what is the speed of a point halfway between the rim and the disk of exploitation exist. 0.20 seconds. After the applied talk is removed. After the applied talk is removed. After the applied chart is removed. Uh after .202 seconds, it could be the velo city would be uh yes, it is omega not. And it is the mega initial. My ass, al fatih alpha two T. So it would be, It is 11.0 million per second minus alpha two equals to 9.8 million per second at a time interval is 0.20 2nd. The radiant per second, 11 -9.8 and 2.20. The angular velocity would be 9.04 radiant for a second. So in the Midway the radius could be are by two. We found our here is 11.5 cm, so 11.5 centimeter divided by two, which is close to 5.75 Centimetre in major. It would be 0.0575 m. The well city would be B equals two Omega into Are by two at the midway. So mega calls to 9.04 into our by twinkles with 0.05 75 major and it'll be me here for a second. So and funds 04 into 0.0575. That equals to your point 5198 major per. Stick it at the midway.


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