two-masses-are-attached-to-a-1-m-long-massless-bar-mass-1-is-3-kg-and-is-attached-to-the-far-left-si

Question

Two masses are attached to a 1 m long massless bar. Mass 1 is 3 kg and is attached to the far left side of the bar. Mass 2 is 5 kg and is attached to the far right side of the bar. If a third mass that is 2 kg is added to the middle of the bar, how does the center of mass of the system change?
(A) The center of mass shifts to the left by 0.025 m.
(B) The center of mass shifts to the right by 0.025 m.
(C) The center of mass shifts to the left by 0.075 m.
(D) The center of mass shifts to the right by 0.075 m.

Pritesh Ranjan · Accepted Answer

So we have I thought. And this rod, let&#x27;s say it is of a meter. And the two monsters are attached At the two ends. So this mass is three. Katie had. The other moss attached is a white kitchen. Okay, now. Okay, A 3rd masses addict. So where the third month is at it And in the middle, that means if I think this is the reference point, so exactly at .5 m from this end, the third mass is attached to the centre and it is a do pizza. Now, before this thermos was added and after this third mass is added, we need to figure out the difference the center of moss difference in central mask that needs to be computed. So let us first compute the center of mass in this situation. So when this mass is added, when the third master is two cages at it, the new center of mass. That will be next dash. So we use here effectively X. S. Submission. Am I excited over submission? Mhm. So the new center of mass will be three times its position. So this is the reference point ticket and then two times .5 Plus five times 1. And this will be divided by three plus four plus five. And this gives the new center of process one plus 55 Then and this is .6 m. So now the center of mass is at .6 m and earlier When only the two masses were there. So we don&#x27;t need to include the two KK bus. So in this case the center of boss that is X will be three times 0 was five times 1 over the two masses at least to treat this fight. So that will be five x 8 and this will be .625 m. So now we have the center of mass when the thermoses at it and the center of mass when only two months is were present. So we take the difference in central mass that is x minus extension, and that comes out to be 0.65 m minus 0.6 m, and this is point zero june five m. So this is the difference in central Mosque. After This, two KK mass is added to this year system.

Kristela Garcia · Answer

in this problem. You have this long massless bar and on one end you have a mass at his two kilograms. On the other end, you have a mess, that is four kilograms and then right in the middle, you have another mass of his four kilograms and we wanna hang a string on this side so that this whole situation of this thing is balanced. So where this thing needs to be placed is on the position of the center of mess. So if it&#x27;s on the center of Mass than this thing will have the same torque on each side and then our situation will rotate. It will just be balanced so we can apply the position of the center of mass equation, which tells us that the some of the mass times its position away from our center of mass, all divided by my total mass, is going to give me the location. So we want to define our position as a position from a reference point. Now we know that the mass lists rod is 1.5 meters, so that tells us that this distance is 0.75 and this distance is 0.75 So we can say that our location and I&#x27;m gonna call the far left side. We&#x27;ll call this position zero. So the two kilogram mass is at position zero and then we&#x27;re going to add our four kilogram mass, which is now a position 40.75 meters away, and then our other four kilogram mouse is 1.5 meters away, and then our total mess will just be the sum of Oliver masses. So here I get a location of the center of mass with respect to what I have deemed as position zero as 0.9 meters. So that means somewhere here at their appointment meters from the left is where my string needs to be placed so that this thing is bounced.

Ben Nicholson · Answer

problem. 11.1. We have a bar that&#x27;s 50 centimetres long and has a mass with a couple of other masses. Intuit, whose size is such that we don&#x27;t need to worry about how big they are, we can treat them as his point masses. And so we want to know if we want to balance this rod with some sort of full crew where along it we would want to put the fulcrum, which is the same was asking, Where is its center of gravity? And we want. It is a length from the left. So when the variation in the force of gravity with you know elevation could be ignored, like in most cases the center of gravity in the center of Mass were the same and the make that look like a signal, the, um, center of mass is just the weighted average of all of the positions of things in the in the system. No. So you some of the massive, each object times it&#x27;s positioned. That&#x27;s all divided by the total mass. So in this case since we&#x27;re measuring, we wanted from the left end. So let&#x27;s go ahead and just, you know, measure things from from the left end. So here it would be zero, the center of mass of the bar, which we can use as its position is in the middle of it exits uniform. And then this is 50 centimeters away. So we have zero for this one. We don&#x27;t need to really worry too much about that right now. Uh and then we have are what, 0.120 kilograms of the bar times the 25 centimeters it is away from the left end. And then our 0.11 kilograms of the red mass here times half a meter away. It iss. And then all of this will be divided by the sum of the masses. 12 kilograms. Scare of playing girl, five by kilograms, plus 0.11 kilograms. And so this works out to be 29.8 centimeters. Since this is originally given in centimeters, might might as well make sure answers and seven meters to it doesn&#x27;t really matter for this. It&#x27;s as long as you don&#x27;t get confused. With what? You know how you&#x27;re using the units? Yeah, you could. Didn&#x27;t you report this in meters? Are do this calculation and centimeters. You can convince yourself that, you know, you&#x27;ll get the same answer if all of these air in centimeters instead of meters, because you&#x27;re just multiplying by 100 or dividing by 100 good. Anyway, So, yeah, So you would want to put it, you know, somewhere about here ish? Yeah. To the right of the center of the bar.

Ze-Han Lee · Answer

All right, So in this problem, uh, we have two blocks, so there&#x27;s a fuck room, and the lengths off this sari is too Well, the lens of this side is also too well, and their two blocks is Ah ah, at the center and the mess of the first ones and one and does them too. And we also know that I&#x27;m to equal to everyone. So I want to know that which choice? We&#x27;re not balanced. Uh, we&#x27;re not. Balance the sister. So for parties, is that Adam s equal to m two on the right on the far left side and the mess equal and one the far right side. So that means me at two, we had an to over here and away at I&#x27;m won over here. Okay, so this actually, we&#x27;re not balanced the system, because ah, uh, try to calculate the torque of a system of the new system. So torque induced by end to buy this one is ah to g times to our A and torque induced by this one is a plus and one g times out, and talk induced by this one is Ah, I&#x27;m too g times l the pork induced by this one&#x27;s. I&#x27;m one g times two out Ray and I&#x27;m to equal to M one. So this we&#x27;re not be equal. Okay, so it&#x27;s not Bannister.

Nathan Silvano · Answer

Okay, so in this problem, we have to rotating systems. So let&#x27;s first draw these two rotating systems. The 1st 1 consists with a mall off must m attach to Araj. So as we can see, we have the rods in here in two different situations. And we have the ball of mass M that goes from this points in half, off the lens off the raj to this point. And here also to half of the lens off. They&#x27;re Raj. Sorry. Spiriting The second system we have is this system like this one. So we have here the same rods in the same mess. But this time, if everything of curse a different way So this is the same rod. But this time this is the ball that more is at the end off their Raj and we have three different situations here. Three different points off the movement. This is the point A of the movement, this is the point B and this is the point c of the movement. And the only thing we need to do in this problems to predict that when the mess off the first system let&#x27;s put here system one system to when the mess off the first system and finish his movement. That means when he reaches the vertical point, when we that we represent here, we want to know. Ah, what position the second system will be knowing that the e voting movement goes from this point to this point. So this is the beginning point. Here&#x27;s the initial ball. Okay, Well, the only thing we need to know to answer this problem, let&#x27;s put here the only thing we need to know to answer this problem is that we must know that&#x27;s the angular acceleration is going to be quoted talk divided by the moment over in your shirt, off the system. So as we can see, we know that the talk in this problem that&#x27;s put in here the talk is going to be Hef off to the first system. Actually, let&#x27;s put his way to get more organized. The first ceased in the first season. The talk is going to be half off l because it is the position off the mess, the multiplies, MGI and the moment of our nurture of the first system. This is a moment of inertia is going to be quote uh am half off L Square, which gives us am divided by four Elway&#x27;s square. Now let&#x27;s do this same for the second system. So the second system system to the let&#x27;s see did talk is going to be I am G l. And the moment over a near share is going to be M l Square. Therefore, from here in the system one, we can conclude that the angular speed actually the angular acceleration is going to be just talk divided by the moment over in Ayrshire, which gives us to G divided by L. And on this side we have that the angular acceleration, which also is the talk divided by the the moment of inertia gives us on Lee G divided my l. Therefore, we can see that the left system So the first system has the larger angular acceleration. Okay, because of the two g l and not only G l. And we can conclude that when the rods to the left reaches, the first system reaches the vertical point there. Rods to do right is not yet in the vertical position. Therefore, we can see that the second rod will be in position A and that&#x27;s the answer. The rod will be in this position in here because she was not reached the vertical position yet. And that&#x27;s the final answer. Thanks for watching.

Khoobchandra Agrawal · Answer

here. Two white masses are held in a place do distance apart. So here this is the mask. And this is another much on they are Toby held after distance deep at the midpoint between them. Oh, this is a This is You may say maybe it&#x27;s called toe be Oh, is the mid point. So it&#x27;s called Tau V. It&#x27;s called Divi to m is then displaced. Here it is capital. Um and it is displaced. Yeah, a small distance x perpendicular to the line, So o p you may call it This is P point. So opiates called toe X, which is much less than the please. Uh huh, Yeah, in the first part showed that live additional force on capital and it&#x27;s 16 g m in tow upon d Square. What is the direction? Off course We show that And actually big, angular frequency off four y d Hold up gm by D and beat it up Friday by toe. Do you want in c part? Okay, what would be the period? Our population? What? Um, it&#x27;s ableto 100 cages on these goto 25 cent emitted dick well, and our village, if it is displaced from center to watch either off the fixed months, let us to start solving very lending question. First part. We&#x27;re starting solving the first part. This particular having the mass m. This is also him Who is the midpoint already? I have written ever. This is capitalism. This value rejects. This will be Have fun. This will be have toe both having the equal magnitude Egypt G M m. Upon this distance on this distance will be access square flys The square by foot are Thank you. Here two year already. This is miss points of the way too. I have written already. Ever. I would no networks on it have full consider. Have fun. Consider are concerned. Yeah. Have one scientist Your neck force will be tow. Watch the midpoint. Oh, on its valuable be toe one science theater. Or you may write to F to science theater. That is your choice, but original will cancel out so net ports will be towards along si o no substitute Develop in tow. Sign up, Peter From the figure sign of data will be X upon rooftop access square plus the square by food. So total force will be and to G m M x D upon access square, plus the square by foot to the power three way too. But actually much less than the access square can be neglected. Some networks you will get 16 the m M x upon. Did you? Yeah. So this is the answer of a part Direction will lead towards oh towards midpoint. Oh, fixed judge. So he fixed particles Now be point mhm Net force on am not let force on em is directly because not to x And it directed the watch equilibrium position. So it performs as a chip with constant cape 60 g m into him upon the tube. So it&#x27;s angular velocity off as a symbol. We root off k upon him on solving it. It&#x27;s valuable we for by the root off G I forget to write here Just a moment. Let me correct it. Mhm. You will get g a small m upon the no Omega is going toe choir. Ponting The time period office estimates Bye bye, amigo. So to party by foot de upon gm, it can be written. It&#x27;s bye bye toe. So this is the answer up. Be part now we have to find the answer. Topsy, I&#x27;m afraid we have to calculate 53.14 d is given while to 5 m upon to cool while 25 6.67 10 to the power minus 11 and masses given 100 KT. So time period off as a champion, We two white food and tow 10 to the power three seconds. That is 40 minutes deeper if em is to be displaced. But what? One off fixed judge. So he fixed particle. There is a net force on em towards some US led force. Watch up fixed particles so it would not perform at the chip but continues to move. So what&#x27;s up? Fixed particles. That&#x27;s all for it. Thanks for watching it.

Problem

A 5 kg box is connected to a pulley with rope in …

Problem 31 Easy Difficulty

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