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Adisk with an angle of zero at t = 0 8 rotates clockwise at 25 rad with a counterclockwise angular rad acceleration of 0.50 for 2.08.What is the final angular posit...

Question

Adisk with an angle of zero at t = 0 8 rotates clockwise at 25 rad with a counterclockwise angular rad acceleration of 0.50 for 2.08.What is the final angular position of the disk?Answer using coordinate system where counterclockwise is positive.Round answer to two significant digits:rad

Adisk with an angle of zero at t = 0 8 rotates clockwise at 25 rad with a counterclockwise angular rad acceleration of 0.50 for 2.08. What is the final angular position of the disk? Answer using coordinate system where counterclockwise is positive. Round answer to two significant digits: rad



Answers

In the previous problem, suppose the disk has an angular acceleration of $232 \mathrm{rad} / \mathrm{s}^{2}$ when its angular speed is $640 \mathrm{rad} / \mathrm{s}$ Find both the tangential and centripetal accelerations of a point on the rim of the disk.

So the question tells us that we have a circular disk with a radius of 0.1 meters, and it's accelerating with an angular acceleration of one radiant per second squared and has a, uh, angular velocity of two radiance per second. The first part of problem wants us to figure out what the angular velocity is after a time of five seconds. So to do this, we can use our kingdom attic equations to. So for this unknown variable that we have, the one that we're going to use for part A is the one that states that the, uh the final angular velocity is equal to the initial angular velocity, plus the angular acceleration times the time we know that the final, uh, the initial angular lost the is to Iranians per second, and we know that the acceleration is one radiant per second squared and the time is five seconds. So this means that the final angular velocity is seven radiance per second for part being were asked to find what the total displacement is, uh, are angular. Displacement is for the wheel. So to do this, we need to use the cinematic equation which states that The change in angular displacement is equal to the initial angular velocity times a time plus 1/2 times the angular acceleration times the times squared. And so we know that the initial angular velocity is equal to two radiance per second multiplied by the time, which is five seconds plus 1/2 times the acceleration, which is one radian per second squared times the time squared. And so if we put this into a calculator, I do it in your head. That's 10 plus 25 divided by two, which gives us 22 0.5 radiations as the total displacement of the wheel. And lastly were asked to figure out what the tangential acceleration is anywhere on this wheel. And so to do this, we take our linear acceleration. Are we take our angular acceleration and multiply it by the radius of the wheel and that I give us the tangential acceleration. So the radius of the wheel is 0.1 meters, and the angular acceleration is one radiant per second squared. So the tangential acceleration is 10.1 meter per second squared, and this is the final answer

Hi. So we're given question that at this point rotates Point A rotates on circular disk and didn't show angle is set to quote deserve At the time of zero and deep in the jungle, A velocity is equal to 0.1 Red the certain and the acceleration angular acceleration equal to. And we want to find the velocity and acceleration. The and E. I want to find evolutionary acceleration when the time is because 1/2. So first of all, we need to find the I didn't find the angle of a lot of Angela velocity when the times on 1/2. And we're going to use this formula. So question we're not. It was powerful. That's the culture europe on one plus the angle acceleration is too times. He Right now is the time. Which is one because we're going from the time of 0 to 1 second. Time interval one second and A result is going to be 2.1. We returned the second. Next we're going to find the angle At the time. At the time it cost to one when the time is one. So to try and the any shovel in our shangla, velocity 10 time plus half my glasses very sometimes times square. So this dessert .1 terms one with one of our three Times two times 1 square. And this is a quote 1.1 Regions. So down to find you philosophy V when he is from Latvia close to our omega and draw a triangle 40 side. So let's this will be the angle But when there's 1/2, which is 1.1 regions. And so this side Vertical said sign 1.1 santos and you're fine to coarse 1.1 Careful Tennis .2 m Which is 200 millim like we're given in the question converted to me to All right. So next level is to be we turn yourself to this line here and perpendicular to this line here. And this will be the direction. So it's going to be in the up and left direction. So her dental first. Let's take the horizontal component. And uh we're just is going to be sheriff went to of course I want yeah. Angela velocity and just 2.1 and this is this is going into negative X direction. So it meant -1 plus is going in a positive direction to us Plus G. And you were just here when it's going up, it's going to be surely this should we sign This should be signed because when it's going to be left, this will be the radius. Yeah. When when it's going to when is going up This will be the radius. This will be zero. Times 2.1. Okay. And our results is going to be -0.37 for all right. Plus zero point 19- five G. We transfer seconds. And now for B Destininations in two parts the condition exploration. And that formula for that is our alpha. And since the solution is going the same direction and velocity, we're just going to use the same idea used here. And suggestions the values for Yeah, But this angle velocity two young girl up the direction. So this would be my not 0.2 Sign 1.1 times two. All right. Yes. Several points Because one playing for one change to ju next who do the normal acceleration. And that's the coincidence. Angelopoulos regis stating And low velocity squared. Yeah. And yeah, I'm just come over here and show you. So, the normal acceleration was directed towards center or the circle, wow. And the direction bring that into components. We get down and leapt. So when when it's going left, the horizontal first one's going left, that's minus i minus. And we're just here is going to be SharePoint two costs 1.1. So I am love lost the Square 2.1. It's quite Yeah. All right. And this one day for the protocol is going down so that's negative. Why negative retro? I want to sign 1.1 times 2.1. Great. Okay. And so the final acceleration. The acceleration. Total acceleration is going to be a C plus A N. We had the normal and tangential accelerations. It was going to be called my nurse. 0.75 seven. All right -0.625 two. The unit is made up except on the score. So this acceleration just offer lost too.

So here we're going to use Equation 10 15 forecourt, eh? And we can say that we have 60.0 radiance which would equal 1/2 times omega sub one plus omega sub too multiplied by the time of 6.0 seconds. So, essentially, the angular displacement is gonna be equal to the average angular velocity multiplied by the time, uh, knowing that Omega sub two equals 15.0 radiance per second we confined Omega sub one is equaling 5.0 radiance per second. This would be our answer for part a four part B. We're going to use Equation 10 12. So Equation 10 12 gives that the angular acceleration would then be equal to 15.0 radiance per second, minus 5.0 radiance per second, divided by the time of 6.0 seconds. This is giving us one 0.67 radiance per second squared. This would be our answer for part B four parts. See, we're going to interpret omega equaling Omega Sub one and Beta simply equal state of someone which equals 10.0 radiance. And we can say that the initial angular velocity would be zero radiance for second. Therefore, we can use Equation 10 14 and say that stada initial would be equal to negative Omega sub one square to hide about two times Alfa plus fade us of one and we can then solve and we find that the initial angular displacement is 2.50 radiance. This would be our answer for part C. That is the end of the solution. Thank you for watching.

Okay. And this problem, we have a wheel. It's spinning and angular. Um, velocity changes after undergoes an acceleration. And our goal is to get the final angular velocities. Let's get down the givens, um, and then I'll talk about the approach for the problem. So it's initially has an angular velocity. Ah, 140 radiance per second and it's pointing east so we can use North is up. South is down. Huestis left and Alfa Angular Acceleration 35 radiance per second squared 68 degrees west of north. So, um so here's our Alfa, and that's applied for five seconds. So and we want to make a final, so make a yeah, So I guess it's first pointing up. That's kind of getting pulled over to the left, and we want to figure out where it ends up after this. A lot of time. So this one seems a little complicated, but fortunately, we have a vector math to help us out. Um and so we just want to go with a formula that relates all these and salt for Omega final. So the formula is Alfa is equal to Omega final minus omega initial divided by time. And, um, if we want to solve for Omega Final, we can just built by both sides by time and at Omega Initial. So that means to make a final is gonna be Alfa T plus omega initial. So it's just gonna be some vector addition. So what we're gonna do is add the vector groups may put that victor symbol so we want to add Omega initial, which is going like this and then Alfa. But we want to multiply it by time, so we're gonna make it five times longer. So something like this, that's not really five times, But it's bigger at least. And we just really want to add these two vectors. Soto, add two vectors. You need to, um I like to do the tip to tail method. So I just moved omega over here. So now it's tip to tail with this and in the some. All right, let me label this to be as clear as possible. So here is Al Fatih. And then the sum is going to be represented in blue, which was I go to the tale of the 1st 1 to the tip of the 2nd 1 And that's gonna be our Omega final. The some of these two. So, um but what? To add vectors, we need to break them up into their components. And so, um, let's go ahead and break Alfa. So if this is from a drop off over here, So here's our Alfa. Oh, Elfatih, actually, um, let's we can put a number to it too. So Al, Fatih or I could do that. One of my head? I think so. 35 times five. So that's 71 40 1 75 So 1 75 radiance per 2nd 0 and then it actually ends up being about the same size of the only initial, but whatever. So this is 68 degrees. Um, let's go ahead and write down what this is. So that's gonna be, um, So I have to add up to 1 80 No, sorry. 90. So this one's gonna be 22 degrees, and therefore then we're gonna get the part that's along the North end as, um, Alfa T. Boone running out of room so it's gonna be off a T signed 22 and then this one's gonna be Alfa T Co signed 22. Um And so if we weaken kind of call, we can use the standard north. Um, I had J hat, so this is like, the J hat direction up and down. And then this is the I had. So therefore, we can rewrite this factor as, um offorty times. Um, so it's up. So I heard the J part's gonna be positive. And then it's to the left of the eyes. Gonna be negative. It's gonna be Alfa Tee times negative co signed 22 I hot plus, um, signed 22 and that's in the J hot direction. Then we want to add that to make initial What direction is Omega initial? It's in the positive. I had direction. So that's gonna be plus omega initial some kind of mixing numbers and symbols. But oh, and that's fully in the I had direction. And so, um, then we can say that the full vector just kind of collect the terms we have the one that I hot the terms that are in that I had directions that's gonna be Omega initial minus elfatih. And then the terms in the J hot direction, which is just Al Fatiha signed 22 and then the magnitude of will make a final is therefore So if you have a vector, we just need to find the like We just basically, we just found this in this and so to find they'll make a fine always use the Pythagorean theorem with this triangle we've created and say that it's the horizontal is term squared plus the bottom and then plus the vertical term squared, all square rooted. So then what we want to do is do omega initial minus elfatih squared. Plus, um, Al Fatih. Oh, I forgot my co signed 22 here Coast 22. And we just want to go ahead and evaluate this. So I'm gonna go ahead and put that into a calculator. So 1 40 year ham's gonna and pause the video idea that Okay, with that, I get, um, term role 66 Iranians for a second, and we probably want a direction. Yeah, magnitude and direction. OK, so again, we found, um, this horizontal peace, and we found this vertical piece. And so, um, so the directions was gonna kind of be I'm just kind of redrawing the Omega Final Vector is going like this. And we know that this horizontal side is this omega I minus elfatih Coast 22 in the vertical is, um Al. Fatih signed 22. And in this angle, Um, so we were our goals to find this angle. And so what we want to dio is say, um and then we can Yeah. So what we want to do is then the tan of this angle is opposite over adjacent. So then, um, the angle is then the arc tanne of Alfa T signed 22 over Omega Initial plus up a T coast 22. So go ahead and put that into a calculator. Um, So I got set, though, that 74 degrees I had to take the, um sorry. Opposite. Over adjacent. Okay, I've confused myself. Me pop. Okay, um, so I needed you to take the magnitude of this because I kind of in making this trying, but I sort of assume all the sides are positive. So anyway, I got that. This angle when I plugged into a calculator was 74.5 degrees. So 75 degrees. So then, um, me write that down. And so if we're talking about the angle with respect to the East access like the positive X axis, the conventional orientation we need to do, um 1 80 minus 75. So then I would say that, um that the angle is 105 degrees from the east access, so I'll go ahead. Write that down 105 degrees from yeast access.


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