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10dm diameter pullev - of moment strings to it on either of inertia side and OOX1O the one exerts force kg m" has weights attached by other exerts force of of-...

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10dm diameter pullev - of moment strings to it on either of inertia side and OOX1O the one exerts force kg m" has weights attached by other exerts force of of-10, 20.0 N Nkon rest and position these and torques position and no others act on The pulley was velocity = for 200 initially at ofthe pulley at the end of the 200 milliseconds What milliseconds? the argularWhat is the rotational kinetic energv ofthe pulley in (3) atthe end of the 200 milliseconds?

10dm diameter pullev - of moment strings to it on either of inertia side and OOX1O the one exerts force kg m" has weights attached by other exerts force of of-10, 20.0 N Nkon rest and position these and torques position and no others act on The pulley was velocity = for 200 initially at ofthe pulley at the end of the 200 milliseconds What milliseconds? the argular What is the rotational kinetic energv ofthe pulley in (3) atthe end of the 200 milliseconds?



Answers

(III) Two blocks are connected by a light string passing over a pulley of radius 0.15 m and moment of inertia $I$. The blocks move (towards the right) with an acceleration of $1.00 m/s^2$ along their frictionless inclines (see Fig. 8-51). $(a)$ Draw free-body diagrams for each of the two blocks and the pulley. $(b)$ Determine $F_{TA}$ and $F_{TB}$, the tensions in the two parts of the string. $(c)$ Find the net torque acting on the pulley, and determine its moment of inertia, $I$.

So here for party, this would be the 33 body of three free body diagrams that we need to show for part B. Now we need to apply suit a Newton's second law for the two blocks. And we're going to say that, of course, the positive X direction, uh, is going to meet the right so we can say that for the mass. Hey, the top lock. We're going to say that the sum of forces in the ex direction would be equal to force tensions of a minus amps of a G sign of fate us of a. And this would be equal to the massive eight times the acceleration of block A. Uh, here here, we're going to just say that acceleration would be the total acceleration of the system. Eso therefore forced tensions of a would be equal to m sub a multiplied by G sign of fate us of a plus A. At this point, we can solve, So the force tension so okay ah would be equal to 8.0 kilograms multiplied by 9.80 meters per second squared times sign of 32 degrees plus 1.0 meters per second squared and we find that the force tension in a four sentence of a rather would be 49 0.6 newtons now four Massive b. We're doing the exact same analysis. Some of forces in the ex direction would be equal to mass sub the G sign of fate. Assad Bey. This would be minus forced tension, said B, and this would be equal to the master B times the acceleration of the system. Ah, we find that the force tension said B, would then be equal to the massive be times the acceleration of brother massively times g sign of status of B minus the acceleration of the system. And now we can solve so 10 kilograms times 9.80 sign of 61 degrees minus one 0.0 meters per second squared and we find that the force tension sub B would be equal to approximately 75.7 Newton's, So this would be our force tension sub A and are forced tension. So be for part B. Now, for part C, we net torque on the pulley is caused by these two. Tensions were going to say that clockwise is positive. So when me trying to catch someone, we're trying to calculate the torque. These some of the torques would be equal to the force Tension sub B minus the force tension sub a times radius are so this would be 75.71 minus 49.55 Let's not round into the very end. You newtons, Times 1.5 meet up. Sorry, My apologies. 0.15 meters on. We find that the sum of the torque it's going to be equal to 3.92 Newton meters again because this is positive this would be clockwise. Now we're gonna use Newton's second Law to find the rotational inertia of the pulley. We know that the chan gentle acceleration of the Police Grimm is the same as the linear acceleration of the blocks. Um, we're going to assume that here the string doesn't slip, so we can say that the sum of torques would be equal to ay times the angular acceleration which b which would be equal to ay times the tangential acceleration, eh? Divided by our and so this would be equal to the force. Attentions of be minister forced tensions of a multiplied by our and we confined I so I would then be equal to the force. Tension so B minus the force. Tensions of a times R squared divided by Hey, Let's solve 75.71 minus 49.55 units here would be Newton's multiplied by 0.15 meters quantity squared, divided by the tangential acceleration of 1.0 meters per second squared. And we find that the moment of inertia, I would then be equal 2.59 kilogram meter squared. So this would be our final answer for part C. That is the end of the solution. Thank you for watching.

For cinnamon 92. In this question, there is Ah, really? Okay, we're different Radio that must and do is hanging like this. This is and don't g And here the detention Here it is hanging a month and one that and one DS with this is tension P one. This is also attention, people. Distention is acting on the police creating a talk on the police The police coming in this direction with angular excellent channel The linear X elation off Muslim who is a on the linear exploration off must and one is a one. So now, like it's like that occasions for months. And when I am for must m one weekend, right and one g my nest d one should be quite and one a one similarly for must I m we can i d to minus m g should be quite it. Now the net door on the holy is equal to the one into our one minus they do in the I do and this note tortured. We quit movement off in a share of the police into its and clerics election. So no such shooting the village 31 92 the one is M one g minus m one a one into our one minus the two, Will you? Is it? And that's M do G in the I do, and they should be equal to a moment of inertia in tow. Angular X elation now expending this weekend and one G are the one minus and one and the linear desolation. A one can also be returned. Us are one Alfa, so it will become our one Esquire in dual par minus and a to I this a. So I can be the Dennis Alfa. Are you inspired? Because linear X elation is equal to I to Alfa, So instead off it to weaken Die Alfa into I do inspired minus and G I do. This is equal to I alpa, the relation between angular solution, a lunatic solution. It is now subscribing the values and when is one in 29.8 into 0.5 minus one into alpha in 20.5 you sweat minus M two, which is two into a far in tow. Zero point. Do we split my less do in tow? Nine point it into our 20 which is 0.2 m, They should be equal to movement off in a Chia into Alfa. Solving this, we will get Bangla Rex election of the holy s zero point for two, you know, six Radian. But second, it's where in the part B, we have to find the genetic isolation of mass M one. A one is equal to are one. This is a photo are one is 0.5 in tow. Alpha 0.4 two seed or states. This is a quote futile. Point 21 Meet that. But secondly, sweat similarly for mass to the ex Elation is a two equal to I 0.2 into Alfa, which is zero point fall to CDO's six. Solving this we will get that. Secondly, near X elation at zero point Zito it Ford 12 Meet that for a second. It's what this is down

Looking at the figure, we can see that some of the police some of the talks are being applied in the counter clockwise direction and some are being applied in the clockwise direction. And since Alfa, which is what we're trying to find here, the angular acceleration is equal to the net torque divided by the inertia of the pulley I, which were which were given in the question. We need to find the total network. So this is the counterclockwise torque minus the clockwise torque, so the counterclockwise torque is torque one. We're told that this is equal to 80 Newtons and it's three a away from the centre, right, So 38 times t one torque in the clockwise direction is tor T two and T three, where t two is a distance away and t three is to a distance away. So plugging in t two of 100 Newtons and T three of 50 Newtons, we find that this comes out to equal in terms of a 40 times a. Now that we have that value, we can calculate Alfa since we know I is equal to 10 kilograms meters squared. So we have t seven divided by I and this comes out to equal for a In the units are radiance per second squared. Now that we know that we can calculate the angular velocity after four seconds since the angular velocity Omega is just equal to Alfa Times, the time that's transpired this is Omega is equal to Alfa Times. The time that transpired Tea party is equal to four seconds. So plugging this in this comes out to equal 16 a radiance per second, Linkenbach said. It is our solution for part B.

So this question we're talking about a polio and me got some information about the radius for us and, um, some of the cinematics And our goal is to get the moment of inertia. So let's get the givens down, and then we can discuss how to do the problem. So the force is equal to 13.0, no ends. And the radius is 0.15 meters. And the, um, Omega initial is zero. Yeah, I made a final. A sequel, Teoh. Um 14 rev per man. Who? Those units, my heart like titans. Every time something's in a rev per minute radiance for second. Okay, And then 4.5 seconds. Um and so we want to get I so basically a lot of information. But if you have seen these equations a 1,000,000 times like I have, you can kind of see how to compose it. So whether two equations that help us, um, think about this we have because it's like, three, actually. So we have Turyk is our f and then that's equal toe I Alfa. So we need something about Alfa. Um Alfa, of course, is defined as the changed omega over the change in time. If if Alfa is ah, constant. And so then we have, um we have Omega final minus immig initial make initial zero. So leave it off, and then Delta Time is just the time before I get too far. Lets convert those two radiance per second because that C s I unit, and you want to use s i units in order to make your units workout. So there's two pi rod, per rev. And then oh, I might run out of room. Sorry. And then there is 60 seconds per men, and I didn't have to put that there. So, um, so you can cancel out there the revs and the men's and then we really just need to multiply 14 times to pie divided by 60 million times. You know, sometimes pie divided by 60. You got one point to get our omega. And, um, the units will be like, which is 1.4, um, seven radiance for second. So, you know, I guess I could have done this at a different time. Um, but after I have finished organizing this formula, but I felt compelled to do it now. So hopefully that was not too confusing. Um, okay. And I just wanted to double check my calculations for attempts to pie 60. Wonderful. Okay, so now I have Alphas Omega Fine, over T and I can just sort of plug it. And here, right? And so I can say this is equal to I Omega final divided by time. And now if I want to solve for I let's look at these two parts of the equations. I can just bring this t This will make a final t over to the side so I can say I is equal to our times t divided by Omega final. And look at that. I have all of these givens and I need to find I. So let's go ahead and get multiplying. So I want to do are just 0.15 Let me try to and wasn't actually diameter they gave me. I made that mistake recently. Yes, they did give you the radius. Wonderful. So I want to do 0.15 times 13 kinds. What peas. 4.5 and then you want to divide it by 1.47 because we really didn't know it to two Sig figs. I'm probably irrationally superstitious, and I like to just use this money. Six weeks is possible and then cut down the sig figs at the end by I think that's irrational. Um, so anyway, 0.19 it's what I would get for the moment of inertia. And then so 0.19 And I know it's kilogram meter squared, and I know that's true just because I used all s I So a meter forest would have kilogram meter per second squared. I'm gonna prove it to you, though, because this is I'm a presumably a high school class where you should really still be practicing this kind of stuff. So, um, let's see. So you're so you have a divided by second squared time seconds that gives you one over a second and then you're dividing by this one over second. So then that kind of kills the seconds, and then you just help with kilogram you prescribed so we know it him with the right formula. So I was looking this over. I realized I made a typo in my populations. Um, so I read a type of a slash instead of a dot, but I have I know Moment of initials aren't traditionally, Hugh. Not traditionally, but And the problems I've seen, they're not huge numbers. So the number did kind of make sense, but as I re did this, I actually got a slightly different number. Um, which was how many? 66 do we have? Oh, yeah. We actually, actually, you do have 366 I'm really glad I kept my 36 things around. Um, but now I can see that it's 5.9 seven kilogram meters squared. Tell me, Go ahead and fix that.


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