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Principles of PhysicsPHYS 2211K(CH 12 HWProblem 12.4616 of 20ReviewA22kg 20-cm-diameler lumtable ratales 140 rpm (riclianless bearings; Two 510 g blocks (all rom ab...

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Principles of PhysicsPHYS 2211K(CH 12 HWProblem 12.4616 of 20ReviewA22kg 20-cm-diameler lumtable ratales 140 rpm (riclianless bearings; Two 510 g blocks (all rom above, hit the turntable simultaneously at opposite ends diameter; and stick:Part AWhat is tne tumtable'5 angulni velocity: TDM Just alter tnis event? Express your answel significant figures and include the appropriate units_70rpmSubmitPrevlou: Answer Request AnswerIncorrect; Try Again; attempts remainingProvide FeedbackNext

Principles of Physics PHYS 2211K (CH 12 HW Problem 12.46 16 of 20 Review A22kg 20-cm-diameler lumtable ratales 140 rpm (riclianless bearings; Two 510 g blocks (all rom above, hit the turntable simultaneously at opposite ends diameter; and stick: Part A What is tne tumtable'5 angulni velocity: TDM Just alter tnis event? Express your answel significant figures and include the appropriate units_ 70 rpm Submit Prevlou: Answer Request Answer Incorrect; Try Again; attempts remaining Provide Feedback Next



Answers

Consider again the turntable described in the last problem. Determine the magnitudes of the rotational acceleration in each of the following situations. Indicate the assumptions you made for
each case. (a) When on and rotating at 33 rpm, it is turned off and slows and stops in 60 s. (b) When off and you push the play button, the turntable attains a speed of 33 rpm in 15 s. (c) You
switch the turntable from 33 rpm to 45 rpm, and it takes about 2.0 s for the speed to change. (d) In the situation in part (c), what is the magnitude of the average tangential acceleration of a point
on the turntable that is 15 cm from the axis of rotation?

First of all, let's to part a well, we need to find out rotational X relation. We know that rotational X relation is represented by Alfa and L far is equal to change in angular velocity with respect to time and changing. And the velocity, uh, is written is omega F minus omega I well, Omega if is the final rotational velocity and omega is the initial professional velocity divided by time. Uh, in this case, Omega F is zero and omega I, which is initial rotational speed, right? Uh, that is equal to 3.45 Radian per second Radian per second, divided by Delta T Well, Delta T is 60 seconds, therefore rotational X relation is equal to zero point 06 ready and per second squared ready in for second square. Now we do part B. Well, in part, being Omega, I is zero Omega f is given, which is 33 r p m. Well, 33 rpm is equal to 3.45 radian per second. Now let's find out rotational X relation. Rotational acceleration is omega half minus omega. I divided by time. Let's plug in numbers. Omega F is 3.45 Well, Omega is zero. And here, Delta T. That's time, which is equal to 15 seconds. So rotational X relation in this case is zero point 23 Radian 1st 2nd square. All right now let's do part C Well, in part c Omega I is 33 are p m. Well, 33 rpm is 3.45 ready in for a second since one rotation equals two pi. Therefore, we have 3.5 million per second and Omega F is equal to 45. 4500 PM 45 rpm equals 4.71 ready and per second and we are given time. Time is equal to 2.0 2nd. Now let's find out traditional X relation. So we have 4.71 minus 3.45 divided by two. So we have 0.63 ready in we're second square. Let's do final part that well, we want to find out 10 dental X relation here so tend intellects relation is equal to our times Alfa right, We have our we have Alfa is well so zero point 682 right, so 3.682 multiplied by the radius, which is 0.15 m. So tend intellects relation is equal to 0.102 um, Major for a second square.

For a turntable were given the initial angular velocity, which is Omega and is 1.8 revolutions per second, which I convert to radiance per second by multiplying by two pies. Let's 11.3 radiance per second. The moment of inertia of the turntable I sub t is 0.2 kilograms meters squared. The party that we're gonna drop on the term table has a mass m of 200 grams or 2000.2 kilograms. So we convert everything two kilograms here because we want to use S I units and the distance away from the center that the party has dropped his 20.15 liters. So for part A, we are asked to find the initial kinetic energy of the turntable. So since it's all rotational, the initial kinetic energy of the turntable which we call Katie I is equal to 1/2 Iomega initial squares. This is a make, I swear. So we just simply have to plug those values in when we find that the initial kinetic energy of the turntable is equal to 1.2 a jewels so we can box set in as their solution. For part a part B says, What is the final rotational speed of the system with the lump of the putty now on the turntable? So to do that, we're gonna consider the fact that the end of their momentum is gonna be conserved to the final angular momentum. Elsa, death is equal. The initial angular momentum will The final angular momentum is going to be the total moment of inertia of the putty plus turntable system. So this is the moment of inertia of the turntable I 70 plus m r squared to get the moment of inertia of the putty. Uh, where m is the mass and R is the distance away from the center that the putty lies and then omega f is the final rotational velocity. What we're trying to find this is equal to the initial moment in the momentum of the turntable, which is I so tee times, omega chi. So we know everything here except for omega F. So we simply saw for Omega death we find that this is equal to I sub t times Omega chi divided by Isom t plus him are square. So simply plug these values into this expression and we find that the, uh there. Um, rotational speed in the end is 9.23 radiance per second. So we can box that in is our solution for B Heart C has asked us to find the final linear speed of the lump of putty. Okay, So defined the linear speed of the lump of putty. You simply have to use the definition that v final. The linear speed is equal to go make a final times the distance the putty is away from the centre are so just play these values in and we find that this is equal to 1.38 meters per second. Okay, so now for the next three questions D, e and F, what we're going to be doing is we're going to find the change in kinetic energy of these different parts of the system. So first, we're gonna find the change in kinetic energy of the turntable. Okay, so this is, um, parte de we'll call this Delta Katie for the change in kinetic energy of the table. So this is just gonna be Katie final minus Katie initial. But we already found we go up. We already found Katie. Initial right so we just need Katie final. So this is gonna be Katie Final is gonna be 1/2 Iomega final squared where I is the moment of inertia of the table. We could also write out the expression for Katie initial, but we already have a numerical value for Katie initial cause we found that in part a party. So we can just write this again as Katie I because we already have that numerical value. Otherwise, you could write it as 1/2 isom t omega initial squared. So plugging these values in, we find that Delta Katie is equal to negative 0.43 jewels. So it actually loses some energy, which makes sense because it's been slowed down due to the putty being dropped on it. Which is why we have that negative sign. Okay, now, party were asked to find the change in kinetic energy of the putty. We'll call this Don't decay, P. Okay, but this is equal to K p f minus k p i. But the party had no initial energy, so that goes to zero. So this is just KPs So K p f is 1/2 times the moment of inertia of the Putty, which is just m r squared times that and your speed of the putty in the final state Omega F Square. That's the same Omega value is the turntable since their rotating at the same speed. So playing this in, we find that this is equal to 0.19 jewels. So it actually gains kinetic energy, Which makes sense because it started out with none. And now it's rotating. So it has some. And then lastly for F, we're asked to do it for the entire system. So, um, so for the entire system, we'll just call this don't decay. This is gonna be K final minus k initial. Okay, so que final minus minus K initial. Well, what is K final minus K initial. So this is K for Final is que t final plus KP final minus Katie initial because there is no KP initial. So this comes out to be equal to negative 0.24 jewels so we can make a little statement about this so we can say that the the total kinetic energy there is a total change in kinetic energy, which we're going abbreviate with Katie in this system is equal to the sum of the kinetic energy changes in the individual constituents that make up the system. It's a little bit bigger and weaken box all of that in as our solution to Part F, which is the final part of the question.

Problem. 11.38. We have a sanding this with this rotational inertia about its axis, and we have attached to a drill that defies the work of 16 Newton meters about that same axis. And so if we apply this pork for 33 milliseconds, we want to find the magnitude of angular momentum and the angular speed of the disc. So from the angular version of Newton's second Law, we know that office how over I and we know if you have constant, angular acceleration, that Omega is the angular acceleration times the time and so that this will be how times T over I was putting this here. And then I Omega is angular momentum. So I omega angular. Momentum is equal to pork time. It's time this works out. B zero point by three kilogram meters squared person. Yeah, So Megha has already said Over here it is out T o r I. So this was going to be 440 radiance per second and if we like, we can convert that rpm and it's 4200 rpm

Okay, so here you have a wedge shaped walk jumping that which it blocked mats. One it, That's I am one. And this wedge is M two. And this whole thing is heights h here. So in this scenario, the first party of a mass that sliding down this wage then after this mass is now moving to the right with the velocity for meters the second and then the wedge m two just call it in to point out to simplify things is now also moving to the left. With this, certain velocity be to So in this case, we want to find first the two and then use V two to find a church. You can so first things first. Since this is, um, you can see that this is some kind of interaction between two masses. Every time you have interaction between two masses and it's an isolated system, you can always think of eventual prosecution. So if we take our system to be this whole in one into combination here, we can see that p I should be. You go to P F because there's no external force in this system. So in that case, then we see also that before this mass flies off, this block is not moving this month. It's a sliding down. So that means that the government is actually zero. And so that means that that zero is going to be the same. I don't you know, the final Mendham again. Get writing one one Yeah, plus him to to be. Yeah, and in this case, off the air after zero. So in one is block them to his wedge. Oh, so now we see that we have given everything So you know that this in one is zero Oh, this wedge is three kilograms. So we already and we know that the finals for me this was so This is is your time for plus three times need to final That should give you zero so nervous. Just move this to the other side. And once you do that, you find that the two final He's just negative zero 0.6 six. Here's the second It's negative again because the block moves in the opposite direction or positive here. Now, once you have this velocity now you want to find this height here We know a lot the energy that this has after this collision happens, history have come from somewhere within the system for what happened. So the only thing that changes the engine energy of the second this year so 19 seconds we can write equations with that contribution of energy that says that which intro an injury that everyone has, which is this put into any here should be equal to the kinetic energy that m one as after this long flights off. So you want squared, plus the kind of energy of wedge going in the opposite direction. So I said before this ever to have won the squared so noticed that we can calculate this. We calculate that, which means you can actually find that this height year so symbolically you will see that this high is actually good off him. One the one square plus oh, him to feet too square. So that's a two year That's a to over Oh, him one g here. So this says that all of the kinetic energy is equal to the potential energy that has had before I started sliding down. So we have or these numbers. This is your five. That's three. This is the four meters per second. This is 67 they just calculated. And 1.5 like that in you find at the height its expertise you appoint no I fried shoot meters.


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