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(35 points) unifon solid disk of mass 30 kg and radius 20 m is free t0 rotate about frictionless exle Forces of 90 and 125 N are applied t0 the disk, 4s indicated i...

Question

(35 points) unifon solid disk of mass 30 kg and radius 20 m is free t0 rotate about frictionless exle Forces of 90 and 125 N are applied t0 the disk, 4s indicated in the drawing:A (5 pts) What moment of inertia of the uniform solid disk? I: 3 Mc? 5 (80ks)L2ra)" Ks475 "clcdes 02 900B (10 pts) What the net torque and angular acceleration produced by the two forces? 16,Ao)(ion- 1sw)l 9rut Trt N'm [0 6 Ilg 'm 1,67 Ks m"/s(C) (10 pts) Calculate the time required t0 complete

(35 points) unifon solid disk of mass 30 kg and radius 20 m is free t0 rotate about frictionless exle Forces of 90 and 125 N are applied t0 the disk, 4s indicated in the drawing: A (5 pts) What moment of inertia of the uniform solid disk? I: 3 Mc? 5 (80ks)L2ra)" Ks 475 " clcdes 02 900 B (10 pts) What the net torque and angular acceleration produced by the two forces? 16,Ao)(ion- 1sw)l 9rut Trt N'm [0 6 Ilg 'm 1,67 Ks m"/s (C) (10 pts) Calculate the time required t0 complete the 60 revolutions, if the wheel starts from rest ?tr 60 1307 60 = Lo + 24 Io 0(+) 6 - 120 = (0.67 15 "mt/s?) (4J: '0] - - 2.04s (D) (10 pts) Starting at rest, what is the time and number of revolutions required t0 reach the [0 rev/s angular speed?



Answers

A uniform disk is rotated about its symmetry axis. The disk goes from rest to an angular speed of $11 \mathrm{rad} / \mathrm{s}$ in a time of $0.20 \mathrm{s}$ with constant angular acceleration. The rotational inertia and radius of the disk are $1.5 \mathrm{kg} \cdot \mathrm{m}^{2}$ and $11.5 \mathrm{cm},$ respectively. (a) What is the angular acceleration during the $0.20 \mathrm{s}$ interval? $(\mathrm{b})$ What is the net torque on the disk during this time? (c) After the applied torque stops, a frictional torque remains. This torque causes an angular acceleration of magnitude $9.8 \mathrm{rad} / \mathrm{s}^{2} .$ Through what total angle $\theta($ starting from time $t=0$ ) does the disk rotate before coming to rest? (d) What is the speed of a point halfway between the rim of the disk and its rotation axis $0.20 \mathrm{s}$ after the applied torque is removed?

The magnitude of force, if too, can be found using equation the network, which is a cool toe I don't offer and the equation that is our times Forced to minus our times. Force One Physical toe High times off I will combine. These equations were all for physical tau Omega Times T ah, we made a nod is equal to zero holding for If I do, it's too we find after to be just due to be. M times are times omega divided by two tea, plus if one some student value. 0.2 times, 0.2 times 2 50 Divided by two times one point 25 plus 0.1 This views us the magnitude off to be 0.14 old Newton.

All right now, I can't believe this, I forgot two. Press the second button to start recording. So I went through this whole thing and it went really well but I forgot to record. So um let me go over this again. I drew the diagram here and then I wrote down all the information that was given this was given first up here, but then we were told that initially the disk is at rest And then after 1.5 seconds Its speed increases with the constant acceleration to 250 radiance per second. So um I wrote then that force is going to be f to minus F one, that's going to be in the positive direction which is counter clockwise times the radius and then that torque is also I times alpha. Since the disk I equals one half M R squared, putting these two together and solving for alpha, I get this, I simplified by cancelling out are in the numerator and the denominator. Okay, so that gets me to hear this top part. What I then did is I took this and I used this equation just one of the equations of motion, angular motion, solved it for alpha and then I substitute it in here. Oh this is just further simplifying. So that should have been with the top part there. Okay, substituting that in for alpha. Okay, now then I just kept solving this equation right here becomes this and then I put it into a calculator and I was so excited because I figured out how to do a subscript in dez mose. F sub one Is 0.1. So now I've got f sub one plus all this stuff over to t. So I got to show you because this is so exciting that I figured out how to put in a sub script. Oh wait you know what I'm gonna do? I'm gonna put it up here. Uh huh. And then I'm gonna click here, click here. Subscript. F some too equals. Now. I don't know if there's a shortcut for this. I'm going to check for a shortcut, but I didn't know how to do this. In the past. Right? There's the subscript. Um I wonder what else I have not known. Um All right, so this is the correct answer, So thank you for watching.

I have. Everyone here is given this a having the months six getting on initial Anguilla Validity 3 60 rpm Gloved by direction Radius off is given if you millimeters march This be having the mosque ticketea on initially. Yet it is that wrist that this are brought together by Apply force off until you can So the excel of noticed it So this is the force applied on a origin Please We'll bring it and be together If a picture the picture, it's white bun fight neglecting very friction Fight by a angular acceleration off each final and globality off each there to see it. Why Sim Ping occur A friction Fourth up blood It's applied. So that is and have a gun. But so this could be moment off. Energy off a This will be How am I? A Are a squid six in tow 60.8 the square. So it will be Why do you know 19 toe? You mean square forced a polite on it is be this is the force off beat. Continue Tunisia plight So here it will be a normal election on this will be the friction this is in so submission off course. If the big will get n Scotto, be that it's continued. Friction will be mewing too. Well, newest 0.1 pipe you will get. Did you get no moment of points on a conflict robbed By taking positive? You will get hair into our A B I a. In tow for a substitute. The value and fine course we have, major, you can radius is a 800. That is point, but it is all the time. So it why didn't you eat meat? Moment off. Inner Soviet calculated 0.0 192 in tow Bar A. So from here, angular acceleration off You will get well when fight Radian per second Not similarly on B. This is that re This is the direction off normally action and he edited in order to see you. It's a moment often say you may calculate, huh? Marcel T in tow, RV square mask off. He's given Katie releases while 06 but it's a moment. Don't taste. Why did you do it? You know 54 Caity Peters Square. No moment. Ophuls off about three point contract dropped by sticking positive in tow. R B movies. Curto Ivy in tow. For me, so is three neutral. Radius is 30.6 moment. Often it's already BM calculated. So angular acceleration off beaches, you will get 30.3 of radiant farcically square. This is the answer part. This problem. Thanks for watching.

So here they're going to find the resultant exploration after it has after the uniforms, solid disc has turned through point one revolutions. So we can first write down our givens. We have a radius of this uniform Solid disc point four zero zero meters. You have a mass of thirty kilograms. You know, we also have a fatality of one point one zero radiance per seconds. And this will be times time. This will be plot plus six point three zero ratings per second Squared times t squared. Let's first buy this first try to find the ah, Angular exploration is simply derivative of thie angular displacement. So this will be one point one Zahra radiance per second plus twelve point six zero radiance per second squared times T. And then we can say that the Alfa T will simply be the derivative of the angular velocity. And this will be twelve point six zero radiance per second squared. So at this point, we can say that, um, the change and Seita is point one one hundred revolutions. We also know that there's two pi radiance per one revolution. So this is going to be point six to eight radiance. And at this point, we confined thie the tangential acceleration. This is something to be our Alfa. So this would be point for times twelve point six or five point zero four meters per second squared. We also need to find a time so we can set this delta theta equal to the the equation for theta. And we can appoint six to eight equals one point one t plus six point three t squared. And we can use a T I eighty four to eighty five ninety nine, which everyone you'd like in order to solve for tea and t equals point two four seconds. And then we're going to find the angular velocity. A T equals point two four seconds. This is going to be equal to one point one plus twelve point six times point two four. And this is going to be equal to four point one two four radiance per second. At this point, we confined the radio acceleration and find that there's going to be able to our Omega Square. So this will be point for times four point one two four square. And that's going to be equal two six point eight zero three meters per second squared at this point, a resultant. It's going to be equal to a tangential squared, plus the square root of a tangential squared, plus a radio squared, and we find that this is going to be equal to five point zero four squared plus six point eight zero three squared, and this is going to be equal to eight point four six meters per second squared. So this would be your answer. This would be the result since acceleration after an angular displacement of point one revolutions, that is the end of the solution. Thank you for watching.


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