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Your answer is partially correct: Try againperson of mass 88 kg rides on Ferris whee whose radius is 7.2 m. The person'$ speed is constant at 1.9 m/s: The pers...

Question

Your answer is partially correct: Try againperson of mass 88 kg rides on Ferris whee whose radius is 7.2 m. The person'$ speed is constant at 1.9 m/s: The person's location is shown by dot in the figure below_(a) What is the magnitude of the rate of change of the momentum of the person at the instant shown?(b) What is the direction of the rate of change of momentum of the person at the instant shown?inward to the center of the Ferris wheelThe magnitude is zero,outward from the center o

Your answer is partially correct: Try again person of mass 88 kg rides on Ferris whee whose radius is 7.2 m. The person'$ speed is constant at 1.9 m/s: The person's location is shown by dot in the figure below_ (a) What is the magnitude of the rate of change of the momentum of the person at the instant shown? (b) What is the direction of the rate of change of momentum of the person at the instant shown? inward to the center of the Ferris wheel The magnitude is zero, outward from the center of the Ferris wheee tangential to the person's path What Is the magnitude 0f the net force acting on the person at the instant shown?



Answers

As seen from above, a playground carousel is rotating counter-clockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. The carousel has a radius of $1.50 \mathrm{m},$ an initial angular speed of $3.14 \mathrm{rad} / \mathrm{s},$ and a moment of inertia of 125 $\mathrm{kg} \cdot \mathrm{m}^{2} .$ The mass of the person is 40.0 $\mathrm{kg}$ . Find the final angular speed of the carousel after the person climbs aboard.

All right, I wrote down the important information and I want to use conservation of angular momentum. So I omega at the beginning is going to equal to equal I omega at the end plus the person who gets on that person's inertia is M. R. Squared. And then the person is also going to have omega one the same at the end. So solving this, omega one equals I over I plus um R squared times omega zero. Putting in this, this in a calculator. You get omega one his 1.83 radiance per second. Thank you for watching.

Okay, so in the first part here, we need to figure out the rotational inertia of the disc. The merry go round and we are told what the uh The radius of gyration is. It's 91 cm or .91 m. And we also know are we figured out earlier in the text that the radius of generation is equal to the square root of the rotational inertia divided by the mass. So we can use this equation here. So if we solve this four, I we get an eye of the Merrigan round is equal to k squared times M or M K squared. So we can just plug in our values here. M is 180 kg are as one K is 0.91. So we get a rotational inertia Of the merry go round is equal to 149 kilogram meters square. Now part B. We want to figure out the angular momentum of the boy. Now the angular momentum is given by Iomega. So this is I boyo make a boy. And the rotational inertia of the boys. Just going to be, we can treat him as a particle. So it's going to be M. R. Squared. And his angular speed is going to be his tangential speed divided by the radius, the over arm. And so we've got an R squared over R. One of those parts is going to cancel when we left with them. The r All of which we know. And if we plug our values in for his masses, 44 kg is Tangential speed is three m/s, and the radius. The full race of the merry go round is 1.2. We'll get his angular momentum as 158 killing ram meters squared over seconds. And then finally part C. We want to know the new angular speed of the merry go round after he jumps on. So to do this, we are going to use a conservation of angular momentum. Where The merry go round starts with zero angular momentum and we want to know the final angular momentum. Well, the angular momentum at the beginning is just that the boy. So I think the momentum of the boy is going to be equal to the combined moment of inertia is of the merry go round and the boy times the new rotational speed. So new rotational speed is his England momentum 1 58 divided by the rotational inertia of the married around, plus his rotational inertia M. R. Squared. If you plug in those values, you'll get a result of 0.74 radiance per second.

In this question given that is that is in the playground, there is a small medical around is given having a radius that is Our is given, there is 1.20 m and uh the math of this merry go round is given. That is and one is equal to 1 80 kg. And further data is given that his days of corrosion is given for merry go round. That is K is equal to 91 Centim or I can see it in the entire unit, it will return is that is 0.91 m. And now in the question is saying that there is a child whose mass is given, that is M2 is equal to 44 kg. So this child is run at a speed given that is we is equal to three m/s along a path that is tangent to dream of initially stationary merry go around and then it is jumped on. So in the kitchen it isn't that the neglect friction between the bearing and the soft of merry go round and follow the parts given. So here you can see that three parts given. So let's start asking every part one by one. So in the party it is asking that we have to find the rotational inertia of the medical around, about it axis of rotation. So I considered that will witness that is I is equal to and one that means myself merry go around, multiplied by areas of cooperation is square. That means K square. So now for the data that is given. So I will get the answer. Power party, january internet series I is equal to 1 80 kg. Multiplied by 0.91 Weather is Squire, so not simplified. The answer is That is 149 Kg meter square. So this is the answer for part. Now come to the next part. That is part B in which it is asking that we have to find the magnitude of angular momentum of the running child about the axis of rotation of merry go round. So I can see that that will recognize that is angular momentum of the child about the axis of rotation. That will be equal to I am too. That means mass of child multiplied by is plenty of spill. That is we multiplied by. Yeah, eight years old. A small merry go round. So now next step is years after the data. That is human. So I'm too is human in the equation that is math of child, that is 44 kg and the animal spirits given that is we, that is three m per second And our is human that is 1.20 m. That means status of medical around so subdued here, then I will get the angular momentum of the child about the axis of rotation of the merry. Go round. So it will be a nice that is held. Child is equal to 44 times of 44 kg terms of three m per second, times of 1.20 m. So after falling where you will be, that is 158 kg meter square per second. So this is the answer for part B. Now come to the last word that is far to see in which it is has hindered that we have to find the angular speed of merry go round and child after the child has jumped onto the merry go round. So I will consider that the anglo spirit of medical around and child after the child has jumped daddy's omega. So that means omega of maybe go down. So in the party it is asking that we have to find the angular speed of America around and child after the child has jumped onto the merry go around. So I will consider that angular speed will be Daddy's omega. So for finding this omega that is angular speed here. I will use the angular momentum conservation forum because there is no external torque is acting. So the according to angular momentum conservation. The next step will be euthanized. That is initial angular momentum will be equal to Angela. Finally, final angular momentum. So initial angular momentum will return is that is M. V. R Plus zero. So 0 is showing that the angular momentum of maybe go around. So because this is a stationary so it will return to zero. And and we are seeing the angular momentum of the child about the axis of rotation. So that will be equal to Now I'm writing the on the right side that is final angular momentum that will return as it is. I system times of oh my God! Here I systems will be witnessed. There is total rotational inertia about the axis of rotation, so that will return us. There is equal to I considered rotation initiator of medical down plus rotational inertia of the child. Mhm. Multiplied by now. Right, the remaining term that is remaining factor that is omega. And this omega we have to find out that is angular speed of the maybe go around and child after the child has jumped onto the merry go round. And now left side will witness that is after simplifying. It will return issues and we are so No next step is I considered on the right side. I might go That means the rotational inertia of merry go round will be organized by using the answer from party. That is 149 daddy metal square plus. Now the additional inertia of child about the axis of rotation will create a nice that is. I can see their families. There is M2. I was squared. Are the square multiplied by now. Right? The remaining factor that is omega that will be called to. I consider it on the left side. I would write NvR mm. Yeah we have to write the and that is showing the muscles child. So that means and we are now next to be subdued. The data that is human. Em to his mass of child that is 44 kg. Multiplied by these human that is three m per second. Multiplied what it is is given that is equal to 1.20 m. That will be equal to 149 plus two is one that is much softer, that is 44 kg, multiplied by art is square. That means 1.2 is square times of omega. Now simplify it, then I will get the answer for parts seat that is equal to 0.744 radian per second, so this is the answer of party. Now you can see that I had solved every part that is part, having answered that is 149 kg meter square. Then part of me having answered that is 158 kg meters squared per second and parsi, having answered that is 0.744 radian per second, Thank you.

So here, of course, we have a perfectly an elastic salute collision, and we know that the conservation of angular momentum applies. So we can say I sub one omega someone equals ice up to Omega sub, too. We're trying to find the final angular velocity. So Mega Sub two unequal ice of one. The moment of inertia Initially times the initial angular velocity divided by the final moment of inertia. So this would be equal to omega. Initial times thie Moment of inertia, of the merry go round divided by the moment of inertia of the merry go round plus the moment of inertia of the people on top of the merry go round. And so this could be equal to Omega Initial those I sub married around, divided by I sub married around Plus here will have four people times the mass of one person times the radius of the merry go round squared And so we find that omega sub two equals point eight zero radiance per second some time times thirteen sixty and then for the denominator thirteen, sixty plus four times sixty five times two point one squared And so this will be equal to point for three four radiance per second. So this will be your final answer for party. And then for part B. They're asking us if the people were on initially and jumped off Radio Lee. So people were on and America around initially jumped off Radio Lee. This would mean that the change and angular momentum equals zero because they're jumping off radio Lee. They're not changing the angular momentum in any way, which means that the Omega Final would simply equal a mega initial. And again, this would equal point eight zero radiance per second. So this would be your final answer for apart being. That is the end of the solution. Thank you for watching.


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